First mechanics final

Laws of Motion | 101

16. The leaves fall when a tree is shaken. Give reason. [HSEB 2056]
Ans:
This is due to inertia of rest. Before the tree is shaken, both the tree and the leaves are at rest. When the
17.
Ans: tree is shaken, the lower part of the tree comes in motion suddenly but its upper part and leaves tend to

18. remain at rest due to inertia of rest. This breaks the connection between tree and leaves, and the leaves
Ans:
fall from the tree.
19.
Ans: If action and reaction are always equal and opposite, why don't they always cancel each and leave

20. no force for acceleration of the body? [HSEB 2052]
Ans:
When two equal and opposite forces are applied on a body at a time then they cancel to each other and
21.
Ans: the resultant force becomes zero i.e. the body comes in equilibrium. If the forces act on two different

22. bodies, then they do not cancel each other. The action and reaction forces defined by Newton's third law
Ans:
of motion are always equal and opposite but they never cancel to each other because they act on two
23.
Ans: different bodies. This makes the motion possible.

State Newton's second law of motion. [HSEB 2052, 2050]

Newton's second law of motion states that, "The time rate of change of linear momentum of a body is

directly proportional to the external force applied to it and the change in linear momentum takes place

in the direction of force applied." This law gives the measurement of force. Mathematically, F = ma.
The S.I. unit of force is Newton (N) and its dimensional formula is [MLT–2]

When a balloon filled with air and its mouth downwards is released, it moves upward, why?

[HSEB 2051]

This is according to Newton's third law of motion. Third law states, that; "to an every action there is

always equal and opposite reaction". When a balloon filled with air and its mouth downwards is

released, then the air escapes with a large velocity in downward direction which is action. Due to this,

the balloon experiences an equal and opposite reactional force in upward direction. Hence, the balloon

moves in the upward direction.

The acceleration of a falling body is measured in elevator travelling at constant speed of 9.8 m/s.

What result is obtained? [HSEB 2050]

We have, for downward motion, R – mg = ma, where m is the mass of falling body and R is the reaction

of the elevator. Since, the velocity is uniform, a = 0

∴ R – mg = 0

or, R = mg

or, mg' = mg

∴ g' = g

Therefore the effective value of acceleration due to gravity (g') is equal to value of g.

Is momentum of a moving body a vector or a scalar? [HSEB 2050]

Total quantity of motion contained in a body is called its linear momentum and it is measured as the product

of mass and velocity. That is, linear momentum P is, P = mv

Since, mass is a scalar and velocity is a vector quantity. The product of vector with a scalar always gives a

vector quantity. Hence, linear momentum (P) is a vector quantity.

A car is driven up a steep hill at a constant speed. Discuss all the forces acting on the car. What pushes

it up in the hill? [HSEB sample question] R F
When a car having mass m is on a steep hill (plane) inclined at an angle α with

the horizontal, then its weight mg acts vertically downwards, which has two

rectangular components: mgcosα is balanced by normal reaction R and mgsinα Ff
acts downward parallel to the inclined plane. As the car is moving up at αα
constant speed, so, the frictional force Ff acts in the direction of mgsinα as
shown in figure. The car needs to apply a force equal to sum of the two forces mg

(Ff + mg sin α) which pushes the car up the steep hill at constant speed. [HSEB 2060]
Why do we slip on a rainy day?

The force of interlock between projections and depressions of two surfaces in contact is the cause of friction.

On a rainy day, the water on a muddy road provides a thin layer in between our feet and road. This water

layer breaks interlocking between projections and depressions, and hence the friction is reduced. The

reduction in friction decreases the reaction of the road. Hence, we slip on a muddy road on a rainy day.

102 | Essential Physics

24. A bullet of mass 20 gm is fired with a velocity of 200 ms–1 from a gun. Find the momentum of
Ans:
recoil. Also, state the principle on which the estimation is based. [HSEB 2062]
25.
Ans: Mass of bullet (m1) = 20 × 10–3 kg, velocity of bullet (v1) = 200 ms–1

26. Let, mass of gun = m2, velocity of recoil of gun = v2
Ans: Before firing, both gun and bullet were at rest. So, total linear momentum before collision was zero.
27.
Ans: From linear momentum conservation principle,

28. m1v1 + m2v2 = 0
Ans: or, m2v2 = – m1v1 = – 20 × 10–3 × 200

29. ∴ m2v2 = – 4 Ns.
Ans:
So, the momentum of recoil of gun (m2v2) is 4 Ns. The negative sign suggests that the motion of gun is

opposite to the motion of bullet during firing.

When rain falls from the sky, what happens to its momentum as its hits the ground? Is your

answer also valid for Newton's famous apple? [HSEB 2068 Supp.]

When the rain falling from the sky hits the ground, the system includes rain and earth. The total

momentum of the system remains constant. The momentum of rain is transferred to earth. The mass of

earth is very large in comparison to that of rain and the momentum transferred to earth doesn't impact

the observable velocity of recoil of earth.

Yes, the answer is also valid for Newton's famous apple because the total linear momentum of an

isolated system is always conserved and in case of apple too, there is no violation of the principle.

A bird is in a wire cage hanging from a spring balance. Is the reading indicated on the balance
when the bird is flying about greater than, less than or the same as that when the bird is sitting in
the cage?

Here, whole the system (cage + bird + air) is being considered, is isolated from outside and the external
forces on the system remain unchanged. The action and reaction of internal parts neutralize each other.
Hence, the balance indicates the same reading in both the cases, either the bird is sitting in the cage or it
is flying in the cage.

What is free–body diagram?

A diagram showing all the forces exerted on each body of a system by the remaining part of the system
is called a free body diagram. The free body diagram method is very helpful in solving problems in
mechanics. When a system consists of a number of bodies connected together by strings, ropes, etc,
then we can consider each body at a time by taking account of all the forces acting on it. Then, we can
write the equations of motion for each body of the system by using Newton's laws of motion. Finally,
we can solve the equations of motion obtained for different bodies of the system to determine the values
of unknown quantities.

Why is it necessary to bend knees while jumping from greater height?

We have, Impulse = Force × time of impact

i.e. Impulse = F × t

For constant impulse (change in momentum), force is inversely proportional to time of impact of the

force. While bending the knees, the time of impact to stop on the ground increases and hence the force
(reaction of ground) on the person decreases. Hence it is necessary to bend knees while jumping from
greater height to reduce hurt (reactional force) on his feet.

Why is sand thrown on tracks covered with snow? [HSEB model question]

On the tracks covered with snow, the friction of the roads is considerably reduced. The vehicles may
slip on such tracks due to lack of sufficient frictional force. The frictional force increases by throwing
sand on tracks covered with snow. Hence, driving becomes safe.

Laws of Motion | 103

Numerical Examples

1. A little wagon with mass 7 kg moves in a straight line on a frictionless horizontal surface. It has an

initial speed of 4 m/s and is then pushed 3m in the direction of the initial velocity by a force of 10N.

Calculate the wagon's final speed and acceleration produced by the force. [HSEB 2073 C]

Solution:

Mass of wagon (m) = 7 kg Distance travelled (s) = 3m

Initial velocity (u) = 4 m/s Applied force (F) = 10N

Final velocity (v) = ? Acceleration (a) = ?

We have, F = ma

or, 10 = 7a

∴ 10 m/s2
a= 7

Also, using v2 = u2 + 2as

or, v2 = 42 + 2 × 10 60
7 × 3 = 16 + 7 = 16 + 8.57 = 24.57

or, v = 24.57

∴ v = 4.96 m/s.

2. A 550N Physics student stands on a bathroom scale in an elevator. As the elevator starts moving, the

scale reads 450N. Draw free body diagram of the problem and find the magnitude and direction of

the acceleration of the elevator. [HSEB 2072, C]

Solution: R = 450 N
Weight of student (W) = 550N

When elevator is in motion, the scale reading gives reaction (R). So, a
Reaction (R) = 450N

Acceleration (a) = ?

Here, R < W. So, the elevator is moving downward.

W 550 W = 550 N
Mass of the student (m) = g = 10 = 55 kg
Considering the free body diagram for the downward motion of the elevator,

W – R = ma

or, 550 – 450 = 55 × a

∴ 100 = 1.82 m/s2 (downward)
a = 55

3. Suppose you try to move a create by tying a rope around it and pulling on the rope at angle of 30°

above the horizontal. What is the tension required to keep the create moving with constant velocity?

Assume weight of the create 'W' = 500N and coefficient of dynamic friction µk = 0.40
[HSEB 2071, S]

Solution: Tsinθ F=T
Given, θ = 30°; W = 500N; µk = 0.40; T = ? R
Since, the velocity is constant. So

Ff = Tcosθ . . . (1) m θ Tcosθ
Where, Ff = Force of friction Ff
Also, in vertical direction,

R + Tsinθ = W

or, R = W – Tsinθ . . . (2) W

We know, µk = Ff
R

or, µk = Tcosθ . Tcos30°
W – Tsinθ or,0.40 = 500 – Tsin30°

Therefore, T = 187.61N

104 | Essential Physics

4. A light rope is attached to a block with mass 4 kg that rests on a frictionless horizontal surface. The

horizontal rope passes over a frictionless pulley and a block with mass m is suspended from the other

end. When the blocks are released, the tension in the rope is 10N. Draw free body diagrams and

calculate the acceleration of either block and the mass m of the hanging block. [HSEB 2070 C]

Solution:

Let, a = common acceleration of the system = ? T
m = mass of suspended block = ?
T = Tension in the rope = 10N 4 kg a

Ff = 0

Considering free body diagram of 4kg mass on T
frictionless horizontal surface,

T – Ff = 4 × a a
m
or, 10 – 0 = 4×a

∴ 10 = 2.5 ms–2.
a= 4

Also, considering free body diagram of suspended mg

block m,

mg – T = ma

or, m ×10 – 10 = m × 2.5

or, m[10 – 2.5] = 10

∴ m = 10 = 1.33 kg
7.5

5. In a physics lab experiment, a 6kg box is pushed across a flat table by a horizontal force F. If the box

is moving at a constant speed of 0.35 m/s and the coefficient of kinetic friction is 0.12, find the

magnitude of force F. What is the magnitude of force F if the box is moving with a constant

acceleration of 0.18 ms–2. [HSEB 2070 D]

Solution: R

Mass of block (m) = 6kg, Horizontal force F = ?
Coefficient of kinetic friction (µk) = 0.12, g = 10ms–2
First case:

When the block is moving with uniform m F=?

velocity of 0.35 m/s, then, acceleration (a) = 0 Ff

From Newton's second law, the equation of motion becomes

F – Ff = ma . . .(1)
or, F – Ff = 0
or, F = Ff [ since, a = 0] mg

or, F = µkR [ since, µk = Ff ]
R

or, F = µk mg = 0.12 × 6 × 10 = 7.2N ∴ F = 7.2N
Second case: When the block is moving with uniform acceleration of 0.18 ms–2, i.e. a = 0.18 ms–2.

Then, from (1), F = Ff + ma = 7.2 + 6 × 0.18 = 7.2 + 1.08 ∴ F = 8.28N

6. What would be the acceleration of a block sliding down an inclined plane that makes an angle of

45° with the horizontal if the coefficient of sliding friction between two surfaces is 0.3?

Solution: R
Acceleration of block along the inclined plane (a) = ? Ff
Angle of inclination (θ) = 45°

Coefficient of sliding friction (µ) = 0.3 θ
g = 10ms–2

The force down the inclined plane is,

F = mgsinθ – Ff [since, F = ma]. θ A
or, ma = mgsinθ – Ff [since,Ff = µR] O mg
or, ma = mgsinθ – µR

Laws of Motion | 105

or, ma = mgsinθ – µ(mgcosθ) [since, R = mg cosθ]

or, ma = m (gsinθ – µgcosθ)

or, a = gsinθ – µgcosθ

= 10 × sin45° – 0.3 × 10 × cos45° = 7.07 – 2.12 = 4.95
∴ a = 4.95 ms–2

7. A cricket ball of mass 145 gm is moving with a velocity of 14 m/s and is being hit by a bat, so that the

ball is turned back with a velocity of 22 m/s. The force of blow acts on the ball for 0.015 sec. Find the

average force exerted by the bat on the ball. [HSEB 2069 A]

Solution:

Mass of cricket ball (m) = 145 gm = 0.145 kg.

Initial velocity of cricket ball (u) = 14 m/s.

After hitting by a bat, the ball is turned black. So, final velocity of cricket ball (v) = – 22m/s

The negative sign is for opposite direction

Time of impact (t) = 0.015 sec

Average force exerted by bat on ball (F) = ?

From Newton's second law of motion,

m(u – v) 0.145 [14 – (–22)] 0.145 × 36
F = ma = t = = 0.015 = 348
0.015

∴ F = 348 N.

8. A 15 kg load of bricks hangs from one end of a rope that passes over a small,

frictionless pulley. A 28 kg counterweight is suspended from the other end of the

rope as shown in figure. The system is released from rest. Using free body diagram

method, find the magnitude of upward acceleration of the load and the tension in 28 kg

the rope while the load is moving. [HSEB 2069]

Solution:

Mass of load of bricks (m1) = 15 kg 15 kg
Mass of counterweight (m2) = 28 kg
Upward acceleration of load (a) =?

Tension in the rope (T) =? a
Value of g = 10ms–2

Considering free body diagram for upward motion of 15 kg load of

bricks,

T – m1g = m1a T
or, T – 15 × 10 = 15 × a

∴ T – 150 = 15a . . . (1)

Again, considering free body diagram for downward motion of 28 kg m2 = 28 kg
T
counterweight,

m2g – T = m2a a m2g a
or, 28 × 10 – T = 28a

∴ 280 – T = 28a . . . (2) m1 =15 kg

Adding equations (1) and (2), we get m1g
280 – 150 = 15a + 28a

or, 130 = 43a

∴ 130 = 3.02 ms–2
a = 43

From (1), T – 150 = 15 × 3.02

or, T = 45.30 + 150

∴ T = 195.30

106 | Essential Physics

9. Two masses 7 kg and 12kg are connected at the two ends of a light inextensible

string that passes over a frictionless pulley. Using free body diagram method, a

find the acceleration of masses and the tension in the string, when the masses

are released. [HSEB 2067 S]

Solution: T
m1 = 7kg, m2 = 12kg, g = 10ms–2 T 12 kg
Common acceleration of the masses (a) = ?

Tension in the string (T) = ?

Considering free body diagram for the upward motion of mass m1, 7 kg
T – m1g = m1a

or, T – 7 × 10 = 7a

∴ T – 70 = 7a . . . (1)

Again, considering free body diagram of the downward motion of mass m2, a
m2g – T = m2a

or, 12 × 10 – T = 12a

∴ 120 – T = 12a . . . (2)

Adding equations (1) and (2), we get T

120 – 70 = 7a + 12a m2 = 12 kg

or, 50 = 19a T
a
∴ a = 50 = 2.63 ms–2 a
19
m2g

From (1), T – 70 = 7×2.63 m1 = 7 kg

or, T = 18.42 + 70

∴ T = 88.42N m1g

10. A lift moves (i) up and (ii) down with an acceleration of 2 ms–2. In each case, calculate the reaction

of the floor on a man of mass 50kg standing in the lift. [HSEB 2059]

Solution: RR
Acceleration of lift (a) = 2ms–2

Mass of the man (m) = 50kg aa
g = 10ms–2

Reaction (R) =?

(i) Upward motion (R > mg), equation of motion is,

R – mg = ma

or, R = m(g + a) = 50(10 + 2) = 50 × 12

∴ R = 600 N. mg mg
(ii) Downward motion, (R < mg), equation of motion is, (i) (ii)

mg – R = ma

or, R = m(g – a) = 50(10–2) = 50 × 8

∴ R = 400N
11. The mass of gas emitted from the rear of toy rocket is initially 0.2 kgs–1. If the speed of the gas

relative to the rocket is 40 ms––1, and the mass of rocket is 4 kg, what is the initial acceleration of the

rocket? a [HSEB 2057]
Solution:

Mass of gas emitted per second mt  = 0.2kgs–1 rocket

Final speed of gas relative to rocket (v) = 40ms–1

Initial speed of gas relative to rocket (u) = 0

Mass of the rocket (M) = 4kg

Initial acceleration of rocket (a) = ?

Here F = Force on rocket

or, F = Force applied by gas; from Newton's third law. gas

Laws of Motion | 107

or, F = change in momentum of gas
time

mv – mu m(v – u) m
or, F = t = t = t (v – u) = 0.2 (40 – 0) = 8

∴ F = 8N

Also, Force on rocket, F = Ma

or, 8 = 4 × a

∴ 8 = 2 ms–2
a=4

12. A vehicle having a mass of 500kg is moving with a speed of 10ms–1. Sand is dropped into it at the rate

of 10 kg/min. What force is needed to keep the vehicle moving with a uniform speed?[HSEB 2066 old]

Solution:

Mass of sand dropped per second mt  10 kgs–1 = 1 kgs–1
= 60 6

Initial speed of sand relative to vehicle (u) = 0
Final speed of sand (v) = 10 ms–1

Mass of vehicle (M) = 500 kg

Force needed to move vehicle with constant speed (F) = ?

Here, Force needed = rate of change of momentum of sand.

or, F = change of momentum of sand

time

mv – mu m 1 10
or, F = t = t (v – u) = 6 (10 – 0) = 6

∴ F = 1.67N

13. An iron block of mass 10kg rests on a wooden plane inclined at 30° to the horizontal. It is found that

the least force parallel to the plane, which causes the block to slide up the plane, is 100N. Calculate

the coefficient of sliding friction between the wood and iron. [HSEB Sample Question]

Solution: RF
Mass of iron block (m) = 10kg

Angle of inclination (θ) = 30°

Least force parallel to the plane required to slide up the plane (F) =

100N. Ff θ
Coefficient of sliding friction (µ) = ?

g = 10 ms–2

Here, F = Ff + mgsinθ . . . (1) θ A
and, R = mgcosθ . . . (2) O

Also,µ = Ff F – mgsinθ [Using equation (1) and (2)] mg
R = mgcosθ

100 – 10 × 10 × sin30° 50
= 10 ×10 ×cos30° = 86.60

∴ µ = 0.577

14. A box rests on a frozen pond, which serves as a frictionless horizontal surface. If a fisherman applies
a horizontal force with magnitude 48N to the box and produces an acceleration of magnitude 3 m/s2,

what is the mass of the box? [HSEB sample question]

Solution:

Horizontal Force (F) = 48N Fm a
Acceleration (a) = 3 m/s2

Mass of box (m) = ? Ff

For a frictionless horizontal surface, Ff = 0
From Newton's second law, the equation of motion is,

F – Ff = ma

108 | Essential Physics

or, 48 – 0 = m×3

∴ 48
a = 3 = 16 kg

15. An 8 kg block of ice, released from rest at the top of a 1.5 m long frictionless ramp, sliding downhill,

reaching a speed of 2.5m/s at the bottom. What is the angle between the ramp and the horizontal?

[HSEB Sample Question]

Solution:

Mass of block (m) = 8kg

Initial velocity (u) = 0

Length of inclined plane i.e. ramp (s) = 1.5m R
Final velocity (v) = 2.5 m/s.

Angle between ramp and horizontal (θ) = ? Ff
g = 10 ms–2

For frictionless ramp, Ff = 0 θθ
Using, v2 = u2 + 2as
or, (2.5)2 = 02 + 2 × a ×1.5

6.25 mg
or, a = 3.0

∴ a = 2.08 m/s2

Now, for the downhill motion of the ramp, the equation of motion is,

mgsinθ – Ff = ma
or, 8×10 × sinθ – 0 = 8 × 2.08

or, sinθ = 16.67 = 0.208
80

∴ θ = sin–1 (0.208) = 12.02°.

16. Calculate the acceleration of a block sliding down a slope of 30°. Given coefficient of friction is 0.25

and g = 9.8 ms–2. [HSEB 2062 S]

Solution:
Acceleration down the slope (a) = ?, θ = 30°, µ = 0.25, g = 9.8 ms–2

As question no. (15), the force F down the slope is,

F = mgsinθ – Ff [since, Ff = µR]
or, ma = mgsinθ – µR
or, ma = mgsinθ – µmgcosθ [since, R = mgcosθ]

or, a = gsinθ – µgcosθ

or, a = 9.8 × sin30° – 0.25 × 9.8 × cos30°
∴ a = 2.78 ms–2

Important Numerical Problems

1. A block slides down a plane inclined at 30° to the horizontal. Find the acceleration of the block,

(a) if the plane is frictionless and (b) if the coefficient of kinetic friction is 0.2. R
Solution:

θ = 30° Ff
(a) a = ?

For frictionless surface Ff = 0
Force down the plane is, F = mgsinθ – Ff

or, ma = mgsinθ [since, Ff = 0] mg
∴ a = gsinθ = 10 × sin30° = 5 ms–2

(b) a = ?, µk = 0.2

Force down the plane is, F = mgsinθ – Ff [since, Ff = µkR]
or, ma = mgsinθ – µk R

Laws of Motion | 109

or, ma = mgsinθ – µk mgcosθ [ since, R = mgcosθ]

∴ a = gsinθ – µkg cosθ = 10 × sin30° – 0.2 × 10 × cos30° = 3.27 ms–2

2. An elevator and its load have a total mass of 800kg. Find the tension T in the supporting cable when

the elevator, originally moving downward at 10ms–1, is brought to rest with constant acceleration in a

distance of 25m.

Solution: T
Mass of elevator (m) = 800kg

Tension in the string (T) = ? a
Initial velocity (u) = 10 ms–1

Final velocity (v) = 0 F = ma
Distance traveled (s) = 25m, g = 9.8 ms–2

The resultant downward force is, F = W – T

or, ma = mg – T [since, F = ma] W = mg

or, T = mg – ma [since, v2 = u2 + 2as]

or, T = mg – m v22–s u2,

or, T = 800 × 9.8 – 800 022×–21502 = 7840 + 1600.

∴ T = 9440N

3. A body of mass 8 kg is acted upon by two perpendicular forces of 16N and 12N. Find the magnitude

and direction of the acceleration of the body.

Solution: F2 F2
m
Given, F1 = 16N, F2 = 12N, m = 8kg α
From figure, magnitude of resultant force is, F1

F = F12 + F22 = (16)2 + (12)2 = 20N

Also, let α be the angle made by resultant force F with F1, then

tanα = F2 12 = 0.75
F1 = 16

∴ α = tan––1 (0.75) = 36.9° with the 16N force.

F 20 = 2.5 ms–2
The magnitude of acceleration produced is, a = m =8

Hence, the magnitude of acceleration is 2.5 ms_2, which makes angle 36.9° with 16N force.

4. A ball A of mass 0.1 kg moving with a velocity of 6ms–1 collides directly with a ball B of mass 0.2 kg

at rest. Calculate their common velocity if both balls move off together. If A had rebounded with a
velocity of 2ms–1 in the opposite direction after collision, what would be the new velocity of B?

Solution:

Given, mass of ball A, m1 = 0.1 kg, mass of ball B, m2 = 0.2kg
Initial velocity of ball A, u1 = 6ms–1, Initial velocity of ball B, u2 = 0
Common velocity after collision, v = ?

From linear momentum conservation principle,

m1u1 + m2u2 = (m1 + m2)v.

or, v = m1u1 + m2u2 0.1 ×6 + 0.2 × 0 0.6
m1 + m2 = 0.1 + 0.2 = 0.3

∴ v = 2 ms–1

Second case:

After collision, final velocity of A, v1 = –2ms–1

Final velocity of B, v2 = ?

Since, m1v1 + m2v2 = m1u1 + m2u2

or, 0.1 × (–2) + 0.2 × v2 = 0.1 × 6 + 0.2 ×0

0.6 + 0.2 0.8 ∴ v2 = 4ms–1
or, v2 = 0.2 = 0.2

110 | Essential Physics

5. The masses m1, m2 and m3 of three bodies as shown in figure are 4kg, 2kg and 5 kg respectively.
Calculate the values of the tension T1, T2 and T3 when (i) the whole system is going upward with an
acceleration of 3 ms–2 and (ii) the whole system is stationary. Given g = 9.8 ms–2.

Solution:

Given, m1 = 4kg, m2 = 2kg, m3 = 5kg
Common acceleration (a) = 3 ms–2 (upward),

T1 = ?, T2 = ?, T3 = ? T1 a
(i) For the whoel system (m1 + m2 + m3) moving upward with acceleration 3 ms–2,

Resultant upward force = T1 – (m1 + m2 + m3) g

or, (m1 + m2 + m3) a = T1 – (m1 + m2 + m3) g m1
∴ T1 = (m1 + m2 + m3) (a + g) = (4 + 2 + 5) (3 + 9.8) = 11× 12.8 = 140.8 N

Similarly, for the upward motion of the system (m2 + m3), we have

F = T2 – (m2 + m3) g T2

or, (m2 + m3)a = T2 – (m2 + m3) g

∴ T2 = (m2 + m3) (a + g) = (2 + 5) (3 + 9.8) = 7 × 12.8= 89.6N

And, for the upward motion of mass m3, we have m2
F = T3 – m3g

or, m3a = T3 – m3g T3
∴ T3 = m3(a + g) = 5(3 + 9.8) = 5 × 12.8 = 64N

(ii) When whole system is stationary, then a = 0

Thus, T1 = (m1 + m2 + m3) g = (4 + 2 + 5) 9.8 = 11 × 9.8 = 107.8N m3

Also, T2 = (m2 + m3) g = (2 + 5) 9.8 = 7×9.8 = 68.6N

And, T3 = m3g = 5×9.8 = 49N

6. In the Atwood's machine in the given figure below, the system starts from rest. What is the speed and

distance moved by each mass at t = 2sec?

Solution:

Given, m1 = 10 kg, m2 = 15 kg, a
u = 0, g = 10ms–2,

t = 2sec,

v =?, s=?

Considering the free–body diagram for the upward motion of 10kg

mass, aT T a
T – m1g = m1a

or, T – 10 × 10 = 10a AB
m2=15 kg
or, T – 100 = 10a . . . (1)

Again, considering free–body diagram of the downward motion of

15 kg mass, m1= 10kg m2g
m1g
m2g – T = m2a
or, 15×10 – T = 15a

or, T – 150 = –15a . . . (2)
Solving (1) and (2), we get, T = 120N and a = 2ms–2

Now, using, v = u + at
or, v = 0 + 2 × 2 = 4ms–1

1 at2 = 0 × 2 + 1 × 2 × 22 = 4m
Also, distance moved s is, s = ut + 2 2

7. (a) What is the magnitude of the momentum of a 10,000 kg truck whose speed is 12.0 m/s?

(b) What speed would a 2000 kg sport utility vehicle have to attain in order to have (i) the same

momentum? (ii) the same kinetic energy?

Solution:

Given, (a) mass of truck (m1) = 10,000 kg
Speed of truck (v1) = 12 ms–1
Magnitude of momentum (p1) =?

Laws of Motion | 111

∴ P1 = m1v1 = 10,000 × 12 = 1,20,000 kgms–1

(b) Mass of vehicle (m2) = 2,000 kg

(i) Speed to achieve the same momentum (v2) = ?

Here, to have same momentum, m2v2 = m1v1

or, v2 = m1v1 1‚20‚000 = 60 ms–1
m2 = 2‚000

(ii) Speed to achieve the same k.E., v2 =?

1 m2v22 = 1 m1v12
Here, to have same kinetic energy, 2 2

∴  m1   10‚000  × 12 = 26.83 ms–1
v2 =  m2  v1 =  2‚000 

8. A chair of mass 12.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the

chair with a force F = 40.0N that is directed at an angle of 37.0° below the horizontal and the chair

slides along the floor. (a) Draw a clearly labeled free–body diagram for the chair. (b) Use your

diagram and Newton's laws to calculate the normal force that the floor exerts on the chair.

Solution: R=?

Given, mass of chair (m) = 12 kg

Applied force (F) = 40N

Normal Force (R) =?

Here, in the vertical direction, motion
Net upward force = Net downward force
or, R = mg + Fsin37° Ff m Fcos37°

37°

or, R = 12 × 9.8 + 40 × 0.61 = 142N

∴ Normal force (R) = 142N. F= 40 N

Practice Short Questions mg

1. Explain why a fielder lowers his hand while catching a cricket ball? [HSEB 2072 E]

2. Explain the difference between a "push" and a "pull" in the case of a heavy roller on a level rooad where the

force acts on the the roller making an angle with the horizontal road. [HSEB 2071 D]

3. It is easier to pull a heavy load than to push it, why? [HSEB 2069 S]

4. A cricketer lowers his hands while catching balls, why? [HSEB 2069, old]

5. It is easier to pull a lawn roller than to push it, why? [HSEB 2068, old]

6. Why do we slip on a rainy day? [HSEB 2060]

7. Why is it difficult to run fast on sand? [HSEB 2056]

8. Explain why a cricketer moves his hands downwards while catching a ball? [HSEB 2054]

9. Give reasons why a man getting out of moving bus must run in the same direction for a certain distance?

[HSEB 2052]

10. A women in an elevator lets go of her briefcase but it does not fall to the floor. How is the elevator moving?

[HSEB 2069]

11. If action and reaction are equal and opposite, why do they not cancel each other?

12. Why are we hurt less when we jump on a muddy floor in comparison to a hard floor?

13. How does friction helps in walking.

14. Why is Newton's first law of motion also called law of inertia ?

15. In a tug of war, the team that pushes harder against the ground wins, why?

16. A thief jumps from the upper stair of a house with a load on his back. What is the force of the load on his

back when the thief is in air?

17. Automobiles and bogies of the trains are provided with spring system. Explain why?

18. Can a single isolated force exist in nature?

19. A person falling from a certain height receives more severe injuries if he falls on a cemented floor than on a

heap of sand, why?

20. How can proper inflation of tyres save fuel?

Practice Long Questions

1. Define angle of friction and angle of repose. Establish a relationship between them.
2. State Newton's laws of motion. How does it lead to the principle of conservation of linear momentum?

112 | Essential Physics

[2072 D]

3. What do you understand by friction and mention its cause. Obtain expression for the relation between angle

of friction and angle of repose. [2072 E]

4. What do you mean by cold weldings? Show that angle of repose and angle of friction are equal. [2070 B]

5. State the principle of conservation of linear momentum. How does the Newton's third law of motion lead to

the principle of conservation of linear momentum? [2069 A]

6. State and prove the principle of conservation of linear momentum. [NEB 2074, HSEB 2069 A]

7. State Newton's laws of motion. Show that Newton's first law of motion defines force and second law of

motion defines the unit of force. [HSEB 2064]

8. What is the angle of repose? Show that when a body just begins to slide down on an inclined plane, the

coefficient of friction is equal to the tangent of inclination of the plane. [HSEB 2063]

9. State the laws of limiting friction. How would you measure the coefficient of friction between a body and an

inclined plane? [2055]

10. Show that the principle of conservation of linear momentum can be verified by using Newton's laws. [2052]

11. What are the laws of friction? How are they experimentally verified? [2051]

12. State the principle of the conservation of linear momentum and show how it follows from Newton's second

law of motion. [2050]

Practice Numerical Questions [g = 10 ms–2]

1. A 550 N student stands on a bathroom scale in an elevator. As the elevator starts moving, the scale reads 450

N. (a) Find the acceleration of the elevator (magnitude and direction). (b) What is the acceleration if the scale
[Ans: (a) 1.82 ms–2 downward, (b) 2.18 ms–2 upward]
reads 670 N?

2. The mass of gas emitted from the rear of a toy rocket is initially 0.1 kg/s. If the speed of the gas relative to the

rocket is 50 m/s, and the mass of the rocket is 2 kg, what is the initial acceleration of the rocket?
[HSEB sample question] Ans:2.5 ms–2

3. A box weighing 5 kg is pulled along a level floor at constant speed by a rope that makes an angle of 30° with

the floor. If the force applied on the rope is 10 N, what is the coefficient of sliding friction between the box

and the floor? [Ans: 1.9]
4. A bullet of mass 10 g travelling horizontally with a velocity of 300 ms–1 strikes a block of wood of mass 290

g, which rests on a rough horizontal floor. After impact, the block and the bullet move together and come to

rest when the block has travelled a distance of 15 m. Calculate the coefficient of sliding friction between the

block and the floor. [T.U. 2045] [Ans:0.33]

5. A block of wood of mass 1.5 kg rests on an inclined plane. If the coefficient of static friction between the

surfaces in contact is 0.3, find (i) greatest angle to which the plane may be tilted without the block slipping,

and (ii) the force parallel to plane necessary to prevent slipping when the angle of plane with horizontal is

30°. [TU 2056][Ans: 16.7°, 3.60 N]

6. A small object slides down an inclined plane inclined at 30° to the horizontal. What is (i) acceleration down

the plane, (ii) time taken to reach the bottom if the plane is 15 m long? Given µk = 0.25 and g = 10 ms–2.
[TU 2060] [Ans:2.8 ms–2, 3.2 sec]

7. Two forces have the same magnitude F. What is the angle between the two forces if their sum has a

magnitude of (a) 2F ? (b) 2 F? zero? Sketch the three vectors in each case. [Ans: 0°, 90°, 180°]

8. The dogs pull horizontally on ropes attached to a post; the angle between the ropes is 60.0°. If dog A exerts a

force of 270n N and dog B exerts a forces of 300 N, find the magnitude of the resultant force and the angle it

makes with dog A's rope. [Ans: 494 N, 31.7°]

9. A box rests on a frozen pond, which serves as a frictionless horizontal surface. If a fisherman applies a
horizontal force with magnitude 48.0N toe the box and produces an acceleration of magnitude 3.00 m/s2,

what is the mass of the box? [Ans: 16.0kg]
10. A student with mass 45 kg jumps off a high dividing board. Using 6.0×1024 kg for the mass of the earth, what

is the acceleration of the earth towards her as she accelerates towards the earth with an acceleration of 9.8
m/s2? Assume that the net force on the earth is the force of gravity she exerts on it. [Ans: 7.4 × 10–23ms–2]

11. A 8.00 kg of ice, released from rest at the top of a 1.50 m long frictionless ramp, reaching a speed of 2.50 m/s

at the bottom. What is the angel between the ramp and the horizontal? [Ans:12.3°]

12. A light rope is attached to a block with mass 4.00 kg that rests on a frictionless, horizontal surface. the

horizontal rope passes over a frictionless, massless pulley and a block with mass m is suspended from the

other end. When the blocks are released, the tension in the rope is 10.0N . (a) Draw two free–body diagrams,

one of the 4.00 kg block and one for the block with mass m. (b) What is the acceleration of either block? (c)

The mass m of the hanging block? [Ans: 2.5 ms–2 , 1.37 kg. moves downwards]

Laws of Motion | 113

13. You are to lower a safe with mass 260 kg at a constant speed down skids 20.0 m long, from a truck 2.00 m

high (a) If the coefficient of kinetic friction between safe and skids is 0.25, do you need to pull down or hold

back the safe? (b) How great a force parallel to the skids is needed? [Ans: (a) need to pull downward; (b) 279 N]

14. A ball of mass 0.2 kg falls from a height of 45m. On striking the ground, it rebounds in 0.l s with two–thirds

of the velocity with which it struck the ground. Calculate (i) the momentum change on hitting the ground and

(ii) the force on the ball due to the impact. [Ans: 10 Ns, 100 N]

15. A ball of mass 0.05 kg strikes a smooth wall normally four times in 2 seconds with a velocity of 10 ms–1.

Each time the ball rebounds with the same speed of 10 ms–1, calculate the average force on the wall.[Ans: 2 N]

16. Rain falls vertically onto a plane roof 1.5m square, which is inclined to the horizontal at an angle of 30°. The
raindrops strike the roof with a vertical velocity 3 ms–1 and a volume of 2.5 × 10–2 m3 of water is collected

from the roof in one minute. Assuming the conditions are steady and the velocity of the raindrops after impact

is zero, calculate (i) the vertical force exerted on the roof by the impact of falling rain and (ii) the pressure
normal to the roof due to the impact of the rain. (Density of water = 103 kgm–3). [Ans: (i) 1.25 N (ii) 0.48 N m–2]

MBBS, B.E., B.Sc. Nursing Entrance Preparation Questions

1. A balloon containing 2.5g of air is collapsed in two seconds. After the release of air through the hole with
constant velocity 2ms–1. Then average force is

a. 125 dyne b. 250 dyne c. 125N d. 250N

2. Whatever may be direction of two forces of 6N and 2N acting on a body of mass 2kg simultaneously. Then

minimum acceleration produced in the body is
a. 1m/s2 b. 2 m/s2 c. 3 m/s2 d. 0.5 m/s2

3. What surface force exerted by the passenger of mass 'm ' on the floor of lift when lift is moving upward

with acceleration a

a. mg b. m (g+a) c. m(g–a) d. none of this

4. A block of mass 1kg is suspended from rigid support by means of light rope. Then ratio of tension in the
rope when it is lifted up with acceleration 5m/s2 and lower down with acceleration of 5 m/s2 is

a. 1:3 b. 3:1 c. 1:5 d. 5:1

5. A books of masses 2kg and 3kg are placed on smooth horizontal surface in contact. Horizontal pushing

n

force of 10N is applied on block of mass 3kg. Then find contact force bet these two blocks.

a. 6N b. 4N c. 2N d. 3N

6. Two blocks of masses 6 kg and 4 kg held to ends of a light inextensible string passing over a frictionless

pulley , the acceleration of the system will be
a. 2m/s2 b. 2.5m/s2 c. 4m/s2 d. 5m/s2

7. A rocket with mass 600kg is set for a vertical firing . If the exhaust speed is 100m/s, the mass of the gas

ejected per second to supply the thrust needed to overcome the weight of the rocket is

a. 117.6kg/s b. 58.6 kg/s c. 60kg/s d. 76.4kg/s

8. A bullet of mass 10g is fired from a gun of mass 1kg with velocity of gun = 5m/s. The muzzle velocity will

be

a. 30km/min b. 60km/min c. 30km/min d. 500km/min

9. A ball of mass 0.5kg with a velocity of 2m/s strikes a wall normally and bounces with the same speed. If the

time of contact between the ball and the wall is 1 millisecond, then average force exerted by the wall on the

ball is

a. 125N b. 1000N c. 2000N d. 5000N

10. A boy having a mass 60kg holds in hands a school bag of weight 40N. With what force the floor will push
on his feet ? (g = 10m/s2)

a. 1000N b. 600N c. 640N d. 64N

11. A body of mass 5kg at rest explodes into three fragment is having masses in the ratio 2:2:1. The fragments
with equal masses fly in mutually perpendicular direction with speed 15m/s. What will be the velocity of
lighter segment?

a. 15m/s b. 15 2 m/s c. 30m/s d. 30 2 m/s

12. Which one the following group of three forces will not produce acceleration in a body acted by forces?

a. 4N,7N,15N b. 4N,7N,10N c. 4N,7N,12N d. 4N,7N,14N

13. A man is standing in a lift accelerating upward. Then apparent weight of the man is

a. Equal to actual weight b. More than the actual weight

c. Less than actual weight d. zero

14. A stretching force is applied at one end of a spring balance and an equal stretching force is applied at the

other end at same time the reading of the balance will be .

a. 5N b. 10N c. 2–5mg d. zero

114 | Essential Physics

g
15. A lift is going up with acceleration 2 . If a man of mass M kg is standing on a spring balance placed in the

lift. The reading of balance will be

a. mg b. 0.5mg c. 1.5mg d. zero

16. A block of mass 2kg on a horizontal surface begins to move when it is pulled at 30° with horizontal by 10N

force. Then coefficient of limiting friction for the block and surface is

2 b. 3 c. 2 1
a. d.

3 3

17. A 2kg block moves at constant acceleration of 2ms–2 when it is pulled horizontally by 10N. If it is pulled by

20N force on the same surface then acceleration will be 7ms–2
a. 4ms–2 b. 5ms–2 d. 10ms–2
c.

18. A block of 2kg slides at constant velocity at 20m/s on a horizontal surface if it is pulled horizontal by 8N,

the coefficient at sliding friction will be

a. 0.2 b. 0.4 c. 0.5 d. 1

19. A block is sliding down at 30° smooth inclined plane. Then coeff. of sliding friction will be

a. 0.2 b. 3 c. 1 d. zero
2 2

20. A body of mass 2kg rest on a rough inclined plane making an angle of 30° with the horizontal .The coeff.
of static friction between the block and the plane is 0.7. The frictional force on the block is

a. 9.8N b. 0.7×9.8× 3 N c. 9.8 3 N d. 0.7×9.8N

21. A body is moving with uniform velocity of 2ms–1 on a rough level surface. The frictional force on it is 10N.
If the body moves with velocity of 4ms–1, the force of friction will be

a. 2.5N b. 5N c. 10N d. 20 2 N

22. Frictional force is

a. conservative b. non –conservative

c. always constant in magnitude d. always zero when body is at rest

23. A body of mass 0.5kg is falling through air with an acceleration of 9m/s2. Then frictional force on the body

is

a. 4N b. 0.8N c. 0.4N d. 0.3N

24. A block of 10kg is pulled at constant velocity on rough horizontal surface by a constant force 19.6N. The

coefficient of friction for the block at the surface is

a. 0.1 b. 0.2 c. 0.3 d. 0.4

25. A block of mass 4kg is placed on a horizontal surface. The coeff of static friction is 0.4. If a force of 7N is

applied on the block, then frictional force is

a. 5N b. 20N c. 7N d. zero

26. The surface are frictionless, The ratio of T1:T2 is

a. 3 : 2 b. 1 : 3 c. 1:5 d. 5 : 1

50N

3kg T2 12kg T1 15kg 30º

Answer Key

1. b 2. b 3. b 4. b 5. b 6. a 7. c 8. b 9. c 10. c
11. d 12. b 13. b 14. a 15. c 16. d 17. c 18. b 19. d 20. a
21. c 22. b 23. c 24. b 25. c 26. d

Ch apter

5 Work and Energy

Teaching Manual Physics Grade – XI, Higher Secondary Education Board
Curriculum and Training Division Sanothimi, Bhaktapur

Syllabus:

Work, and Energy – Work; work done by a constant force and a variable force; Power;
Energy: Kinetic energy; work –energy theorem, Potential energy; conservation of
energy; Conservative and non–conservative forces; elastic and inelastic collision.

Objectives:

The objective of this sub unit is to make the students be able to understand work,
energy and power and utilize the knowledge in practice.

Activities (micro syllabus):

1. Discuss the meaning of work done in physics and in daily life.

2. Define work, energy and power; state their units.

3. Calculate work done by a constant force and variable force.

4. Explain work–energy theorem.

5. Define and derive kinetic and energy in terms of 1 mv2.
2

6. Define and derive gravitational and elastic potential energy.

7. State and explain principle of conservation of energy.

8. Define and differentiate conservative and non–conservative forces, giving
examples.

9 . Discuss elastic and inelastic collisions.

116 | Essential Physics

5.1 Introduction

The term 'work' has a different meaning in science than as we understood in everyday life. If a person
reads a book for hours or a coolie carries a load on his back for hours waiting a train in a station, then
there is no any work in scientific aspect. In this chapter, we shall have a look into the scientific aspect of
the term 'work'. Here, we also go into details of the term 'energy', forms of energy and work, laws
associated with energy etc.

5.2 Work Done by a Constant Force

In physics, the term 'work' is defined as, "work is said to be done by a force when the force produces a
displacement in the body on which it acts in any direction except the direction perpendicular to the
direction of force." So, for work to be done, a force must be applied on a body and the applied force
must produce a displacement in any direction except perpendicular to the direction of the force.

θ
F Fcosθ

SS
(i) (ii)

[Fig. 5.1, Work done by a force.]

→→

Suppose a force F applied on a body produces a displacement S in the direction of force as shown in

figure 5.1 (i). Then, the work done W is,

W = FS . . . (1)

However, the displacement is not always necessary to take place in the direction of force. Suppose, a

→→ →→
constant force F produces a displacement S in the body in such a way that S is inclined to F at an angle

θ as shown in figure 5.1 (ii), then work done W is,

W = component of force in the direction of displacement × magnitude of displacement

or, W = (Fcosθ)S

or, W = FScosθ . . . (2)
. . . (3)
→→

∴ W=F.S

Therefore, the work is defined as the dot product between force and displacement. Being dot product

of two vectors, it is a scalar quantity.

Special Cases

→→

i. When F and S are directed in same direction, then θ = 0°. So, W = FScos0° = FS

Also, W is positive when θ is acute i.e. 0° < θ < 90°. For example, when a load is lifted; the lifting

force and displacement act in the same direction. Therefore, the work done by the lifting force is

positive.

→→

ii. When F and S are directed in opposite directions, then θ = 180°. So, W = FScos180° = –FS

Also, W is negative when θ is obtuse i.e. 90° < θ < 180°. For example: When a load is lifted, the

gravitational force acts vertically downwards while the displacement is in the vertically upward

direction (θ = 180°). Therefore, the work done by the gravitational force during upward motion is

negative.

→→

iii. When F and S are perpendicular to each other, then θ = 90°. In this case, W = FScos90° = 0

Work and Energy | 117

So, work done by a force is zero if a body is displaced in a direction perpendicular to the direction of
applied force. For example, if a body is sliding over a smooth horizontal surface, then the work done by
the force of gravity and the reaction of the surface will be zero. This is because both the force of gravity
and the reaction are perpendicular to the displacement. Similarly, if a body is moving in a circle with
constant speed, then the centripetal force and displacement are mutually perpendicular at every point.
Therefore, the work done by the centripetal force is zero.

From above special cases, we conclude that the work done is maximum if force and displacement are
directed in same direction, minimum in opposite direction and zero if they are perpendicular to each
other.

Units and Dimensions of Work

We have, W = FS

In SI units, unit of work is Nm or Joules (J). Work done is said to be 1 Joule if a force of 1 Newton
displaces a body through 1 meter in the direction of the force.

In cgs units, unit of work is dyne cm or Erg. The work done is said to be 1 erg if a force of 1 dyne
displaces a body through 1 centimeter in the direction of the force.

Also, 1 Joule = 1 Nm = 1N × 1m = 105 dynes × 100cm = 107 dynes. cm
∴ 1 Joule = 107 erg
In terms of fundamental units, Joule = Nm = (kgms–2). m = kgm2s–2
The dimensional formula of work is [ML2T–2]

5.3 Work Done by a Variable Force

A variable force varies both in magnitude and direction. The work done by a variable force can be

calculated as follows.

Suppose a variable force displaces a body from point 'a' to point 'b' along a curved path 'ab' as shown in

figure 5.2. We can divide the path 'ab' into a large number of infinitesimally small displacements. Each

displacement should be so small that the force may F

be regarded as constant during that displacement. Let b
dW be the small amount of work done in an

→→ θ

infinitesimally small displacement dS and F be the
corresponding force . Then

→→ . . . (1) ds

dW = F .dS

The total work done Wab from a to b can be obtained
by adding the elements of the infinitesimally small
work done between a and b.

Mathematically, this can be done by integrating both a

sides of above equation (1), from a to b,

∫ ∫b b [Fig. 5.2. Workdone by variable force]

∴ dW = →→
F .dS
aa

∫b

or, Wab = a FdS cosθ

∫b . . . (2)

or, Wab = a (Fcosθ) dS

Equation (2) is the required expression for work done by a variable force.

118 | Essential Physics

Graphical Method:

F(x)

PQ b

a Fi

O x1 SR x2 X
dxi

[Fig 5.3 , Graphical method for measurement of work done by variable force.]

Let F(x) be variable force, which acts in x–direction only. Here, we have to calculate the work done by F

on moving a body from points x1 to x2. The total displacement x1x2 can be divided into a large number of
intervals as shown in figure 5.3. On each interval, force can be considered approximately constant. If dW

be small work done in small displacement dxi , then

dW = Fi.dxi . . . (3)

or, dW = PS.SR

or, dW = Area of strip PQRS . . . (4)

From equation (3) and (4), we get

dW = Fidxi = Area of strip PQRS

The total work done on moving the body from point x1 to x2 is,

x2 . . . (5)

ΣW = x1 Fidxi

Σx2 . . . (6)

Also, x1 Fidxi = Area of 'x1x2ba'

From equations (5) and (6), we get

x2 . . . (7)

ΣW = x1 Fidxi = Area of 'x1x2ba'

This shows that the area of 'x1x2ba' i.e. area under force versus displacement curve represents the

total work done.

Now, if we divide the interval between x1 to x2 into infinite number of intervals then Σ sign should be

replaced by integration sign ( ∫ ). Then equation (5) reduces to,

Work and Energy | 119

∫ x2 . . . (8)

W = F dx
x1

∫ x2

Here, x1 F dx is the integral of F with respect to x between the limits x1 to x2.
If F acts in three dimensions then, equation (8) can be written as,

∫W = r2 → → . . . (9)
r1 F .d r

→^^ ^

In turns of components; F = Fxi + Fyj + Fzk

→^ ^ ^
d r = dxi + dyj + dzk

∴ W= ∫ r2 (Fxdx + Fydy + Fzdz) . . . (10)
r1

5.4 Energy

The ability or capacity of doing work is called energy. It is a scalar quantity. It has same units and
dimensions as that of work. The dimensional formula is [ML2T–2] and SI unit is Joule. Energy of a body
is equal to the total amount of work that the body can do. It doesn't depend on the time taken to do work.
On the other hand, power depends on the time required to do work.

There are different types of energy such as mechanical energy, sound energy, heat energy, light energy,
atomic energy etc. In this chapter, we discuss mechanical energy only.

The energy possessed by a body by virtue of its motion or position is called mechanical energy. The
mechanical energy is of two types- kinetic energy and potential energy.

5.5 Kinetic Energy

The energy possessed by a body by virtue of its motion is called kinetic energy. A body moving with
higher speed will possess more kinetic energy than body moving with lower speed. A bullet in motion, a
moving hammer, flowing water, air in motion etc. are few examples of bodies possessing kinetic energy.

u=0 v
F

S

mg
[Fig 5.4, Kinetic energy of a moving body]

To derive expression for kinetic energy, consider a body having mass m is lying on a smooth horizontal

→

surface as shown in figure 5.4. Let a constant force F displaces the body in its own direction through a

→

displacement S . Let the body is at rest initially (u = 0) and its velocity becomes v after travelling the

distance S. Then, the work done (W) by the force is,

→→ . . . (1)

W =F .S = FScosθ°

Here, force and displacement are in same direction, so, θ = 0°.

120 | Essential Physics

∴ W = FScos0° = FS . . . (2)
From Newton's second law of motion, F = ma . . . (3)

F
or, a = m
Using, v2 – u2 = 2aS

or, v2 – 02 = 2mF S [Using equation (3)]

or, FS = 1 mv2 . . . (4)
2

From equations (2) and (4), we get

W = 1 mv2 . . . (5)
2

→

This work done (W) becomes the kinetic energy (K.E.) of the body after displacement S .

∴ K.E. = 1 mv2 . . . (6)
2

Therefore, the kinetic energy of a body is equal to one-half the product of the mass of the body and square

of its speed. Also, we see from equation (6) that kinetic energy of a body is directly proportional to the
mass of the body and square of the speed of the body. Since both m and v2 are positive, so, K.E. is always

positive.

Relation between Kinetic Energy and Linear Momentum

Suppose a body having mass m is moving with velocity v. Then linear momentum (P) of the body is,

P = mv . . . (7)

But, the kinetic energy (Ek) of the body is, Ek = 1 mv2
2

(mv)2
or, Ek = 2m

P2 . . . (8) [Using equation (7)]
or, Ek = 2m

or, P = 2mEk . . . (9)

From above equations (8) and (9), we conclude that a body cannot have momentum without having

kinetic energy and vice–versa.

v1 E1 E2 v2
m1 m2

Light

Stationary Heavy

[Fig. 5.5, Kinetic energy due to explosive force.]

Further, suppose a stationary mass suddenly explodes into two fragments, one heavy and another light as

shown in figure 5.5. Let the light fragment has mass m1, speed v1, momentum P1 and kinetic energy E1
after explosion. Let m2, v2, P2 and E2 be the corresponding parameters for the heavy fragment after
explosion. Then, from linear momentum conservation principle,

m1v1 + m2v2 = 0 . . . (10)
or, P1 + P2 = 0
∴ P1 = –P2

Work and Energy | 121

The negative sign suggests that the two fragments move in opposite direction after the explosion. This is

the reason for recoil of gun during firing a bullet.

Now, kinetic energy of light fragment is, E1 = 1 m1v12 = (m1v1)2
2 2m1

or, E1 = P12 . . . (11) [since, P1 = m1v1]
2m1

Similarly the kinetic energy of the heavy fragment is , E2 = P22 . . . (12)
2m2

Dividing equations (11) by (12), we get

E1 = P12 × 2m2 = (–p2)2 × 2m2 [using equation (10)]
E2 2m1 p22 2m1 p22

∴ E1 = m2 . . . (13)
E2 m1

This shows that the kinetic energy due to explosive force is inversely proportional to the mass of the

fragment. Therefore, if a stationary mass suddenly explodes into two fragments–one heavy and another

light, then the light fragment has greater kinetic energy than that of heavy fragment. This is the reason

that a bullet has greater kinetic energy than the kinetic energy of recoil of gun in case of firing.

5.6 Work–Energy Theorem

It states that, "the work done on a body by a resultant force is equal to the change in kinetic energy
of the body." This is also called Work–Energy Principle or Work Engery Theorem

uv

F

P Q
S

mg

[Fig. 5.6, Work done by a force.]

→

Suppose a body having mass 'm' is moving initially with velocity 'u' at point P. Let a constant force 'F ' is

→

applied to the body and the body covers a displacement 'S ' from point P to Q as shown in figure 5.6. Let
v be the final velocity at point Q. The work done by the force is,

→→

W = F .S = FScos0°

or, W = FS . . . (1)

From Newton's second law of motion, we have

F = ma

F . . . (2)
or, a = m . . . (3)
Using, v2 = u2 + 2aS
or, v2 – u2 = 2aS

From equations (2) and (3), we get

v2 – u2 = 2mF  S

122 | Essential Physics

or, 1 mv2 – 1 mu2 = F.S. . . . (4)
2 2

From equations (1) and (4), we get

1 mv2 – 1 mu2 = W
2 2

∴ ( K.E)f – (K.E)i = W . . . (5)

This equation shows that the work done by a force on moving a body is equal to change in kinetic energy

of the body. This proves work energy theorem.

Discussion of Work – Energy Theorem

i. If a particle is moving with a constant speed, then the change in kinetic energy is zero. So, the
work done by the resultant force is zero. For example, if a particle is moving with constant speed
in a circle, there is no change in kinetic energy of the particle. So, according to work energy
theorem, the work done by centripetal force is zero.

ii. If the kinetic energy of the body increases i.e. (K.E.)f > (K.E)i, then the work done (W) on the
body is positive. In such condition, the force and displacement are directed in same direction, i.e.
the work done by the applied force is converted in the form of kinetic energy of the body.

iii. If ( K.E)f < (K.E)i, then the work done on the body is negative. In such condition, the force and
displacement are directed in opposite direction, i.e. the applied force opposes the motion of the
body.

5.7 Potential Energy

The energy possessed by a body by virtue of its position or molecular arrangement (configuration) is
called potential energy. The potential energy is also called energy of configuration or mutual energy.
The potential energy is of the following three types:

i. gravitational potential energy

ii. elastic potential energy, and

iii. electrostatic potential energy

In this chapter, we discuss gravitational potential energy only. Water stored in a reservoir, lifted weight
from ground etc are few examples of bodies possessing P.E. due to position. Similarly, the coiled spring,
a stretched rubber, compressed gas etc are examples of bodies possessing P.E. due to configuration.

Gravitational Potential Energy m

Consider a body having mass 'm' is lying on the surface of earth and 'g' be h
the value of acceleration due to gravity at that place. The weight mg of the m
body is acting vertically downwards. A force 'mg' is required to lift the body
in upward direction. Let the body is lifted to a height 'h'. Then work done [Fig. 5.7. Gravitational potential energy.]
(W) is,

W = Force × displacement
or, W = F × h = mg × h
∴ W = mgh
This work done is stored in the body in the form of gravitational potential
energy. Therefore, Gravitational potential energy = mgh

Work and Energy | 123

The height 'h' is taken from the reference level (ground). The potential energy at the reference level is
understood to be zero. The potential energy at a point depends on the reference level. But, the difference
of potential energies between any two points is independent with the reference level.

Note : We can show the elastic potential energy stored in a stretched spring is, Ep = 1 kx2
2

Where k is spring constant of the spring and x is the extension produced.

5.8 Principle of Conservation of Energy

It states that, "energy can neither be created nor be destroyed but can be converted from one form to
another form." The total energy of an isolated system remains constant. The universe as a whole may be
treated as an isolated system. Therefore, the total energy of the universe is constant. When one part of the
universe loses energy, then another part of it must gain an equal amount of energy. During gain or loss of
energy, the form of energy may change, but total energy is conserved.

Principle of Conservation of Mechanical Energy

When a body falls freely from a height, its kinetic energy goes on increasing. The gain in kinetic energy is
at the expense of the gravitational potential energy, which is reduced. However, total mechanical energy
of the body remains conserved at each and every points of its path.

Suppose a body having mass 'm' is falling freely as shown in figure 5.8. Let h be the height of the body
above the reference level (ground) at initial position A.

At position A: A vA = u = 0
x
Velocity = vA = u = 0 [ body is at rest]
Height = h B

K.E = 1 mvA2 = 0
2

P.E. = mgh h h–x
vB
So, total mechanical energy (E) of the body at A is, Reference
C level
E = K.E + P.E = 0 + mgh

∴ E = mgh . . . (1)

At Position B : Let after falling through a distance x, the vC
body is at position B and vB be the velocity of the body at
position B.

Height at B = h – x

∴ P.E at B = mg (h – x) [Fig. 5.8, Freely falling body.]

Velocity at B = vB = ?
Using v2 = u2 + 2as

or, vB2 = 02 + 2gx = 2gx

∴ K.E. at B = 1 mvB2
2

1
= 2 m(2gx) = mgx

The total mechanical energy at B is

E = K.E + P.E = mgx + mg (h – x)

∴ E = mgh . . . (2)

124 | Essential Physics

At position C (ground): height from ground = 0

P.E. at C = mg × height = 0

Let vC be the velocity of the body when it strikes the ground.

Using, v2 = u2 + 2as

or, vC2 = 02 + 2gh = 2gh.

∴ K.E at C = 1 mvc2 = 1 m(2gh) = mgh
2 2

The total mechanical energy at C is, mgh (= K.E. + P.E.)
E = K.E + P.E = mgh + 0

∴ E = mgh . . . (3)

From equations (1), (2) and (3) it is clear that total mechanical h = 0 Height h
energy (sum of K.E. and P.E.) of a freely falling body remains
constant at all stages of its motion. Also, when the body falls (Above reference level)
freely, its P.E. goes on decreasing while its K.E. goes on
increasing, keeping total mechanical energy always constant. [Fig. 5.9, Variation of K.E. and P.E. with height]

The graph shown in figure 5.9 shows the variation of K.E and
P.E. with height from ground (reference level).

5.9 Conservative and Non–conservative Forces

A force is said to be conservative if work done by the force or against the force on moving a body

depends only on initial and final positions of the body and is independent on the nature of path followed

between the two positions. For example, gravitational force, elastic force in a stretched or compressed

spring, electrostatic force between two stationary W (A to B)

charges, magnetic force between two magnetic poles etc

are conservative forces.

Suppose a body is moved from point A to point B and A B

then again from point B to A. Here, if the force used is

conservative then, W (B to A)
W(A to B) = –W(B to A)
[Fig: 5.10, Work done around a closed path]
∴ W(A to B) + W(B to A) = 0

This shows that the work done by a conservative force on moving a body in a complete closed path is

always zero. The work done by a conservative force is completely reversible and the total mechanical

energy is always conserved.

Non–conservative Forces

A force is said to be non–conservative if work done by the force or against the force on moving a body is
dependent on the path followed between the initial and final positions. The frictional forces , the velocity
dependent forces such as air resistance, viscous force etc are non–conservative forces.

For a non–conservative force F, W(A to B) ≠ –W(B to A)
∴ W(A to B) + W (B to A) ≠ 0
This shows that the work done by a non–conservative force on moving a body in a complete closed
path is not zero. The work done against a non–conservative force may be dissipated as heat energy. The
work done by a non–conservative force is not completely reversible and the total mechanical energy is
not conserved.

Work and Energy | 125

5.10 Power

The time rate of doing work is called power. If W is the amount of work done in time t, then average

power (P), is,

W . . . (1)
P= t

Power is the ratio of two scalars –work and time. So, power is a scalar quantity. If time taken to complete

a given amount of work is less, then power is more and vice–versa.

→→

Also, W = F .S . . . (2)

From equations (1) and (2), we get

→→ → .→St 

F .S =F
P= t

→→ . . . (3)

∴ P = F .V
→
→S

Where, V = t = velocity

So, dot product between force and velocity is called instantaneous power.

→→

If θ be angle between F and V , than from equation (3), we get
→→
. . . (4)
P = F .V = FV cosθ

If θ = 0°, then, P = FV . . . (5)

Units and Dimensions of Power: From (1), we have

P = W = J s–1 or watt (S.I. unit) = Erg s–1 (C.G.S. unit)
t

1 watt = 1J s–1 = 107 Erg s–1

1 kilowatt = 103 watts

1 Megawatt = 103KW = 106 watts

The power of an automobile is expressed in horse power (HP) and

1 HP = 746 watts

The dimensional formula of power is,

Power = W = J = Nm = kgms–2m = kgm2s–3 = [ML2 T–3].
t S S s

5.11 Collision

A collision between the particles is defined as the mutual interaction between the particles for a short
interval of time during which the momentum and kinetic energy of the interacting particles should
change abruptly (immediately). A collision is said to take place when two bodies physically collide
against each other or the path of one body is changed by the influence of other body. Actual physical
contact between the particles is not necessary to be a collision. For example; the deflection of path of an
alpha particle approaching towards a nucleus is said to have undergone collision between the alpha
particle and the nucleus. The forces involved in a collision are the internal forces of the system i.e.
action– reaction forces. So, the total linear momentum is conserved. Also, the total energy is conserved.
There are two types of collisions: (i) Elastic collision and (ii) Inelastic collision.

Elastic Collision

A collision is said to be elastic collision if both the total linear momentum and total kinetic energy
remain conserved during the collision. The collision between gas molecules, sub–atomic particles etc are
elastic collisions. The following are the characteristics of the elastic collision:

126 | Essential Physics

i. The total linear momentum is conserved.
ii. Total kinetic energy is conserved.
iii. Total energy is conserved.
iv. The forces involved during the interaction are conservative forces.
v. Total mechanical energy is conserved.
A perfectly elastic collision is an extremely rare physical phenomenon. The collision between atomic or
sub–atomic particles, ideal gas molecules etc are the nearest approach to a perfectly elastic collision.

Elastic Collision in One Dimension [Head –on Elastic collision]

The elastic collision in which the colliding particles move along the same straight line path before and

after the collision is called one–dimensional elastic collision.

A A A
B B B

m1 m2 m1 m2 m1 m2

u1 u2 v1 v2

Before collision During collision After collision

[Fig. 5.11. One-dimensional elastic collission]

Consider two bodies A and B having respective masses m1 and m2 are moving with respective velocities
u1and u2 is a straight line such that u1 > u2 before collision as shown in figure 5.11. Let, the bodies collide
elastically and, v1 and v2 be their respective velocities after collision.

Applying law of conservation of linear momentum,

m1u1 + m2u2 = m1v1 + m2v2 . . . (1)

or, m1(u1 – v1) = m2(v2 – u2) . . . (2)

Since in an elastic collision, total kinetic energy also remains conserved. Therefore,

1 m1u12 + 1 m2u22 = 1 m1v12 + 1 m2v22 . . . (3)
2 2 2 2

or, m1(u12 – v12) = m2(v22 – u22) . . . (4)
or, m1(u1 – v1) (u1 + v1) = m2(v2 – u2) (v2 + u2)
Dividing equation (4) by equation (2), we get

u1 + v1 = v2 + u2

or, u1 – u2 = v2 –v1 . . . (5)

Here, (u1 – u2) is the relative velocity of approach before collision and (v2 – v1) is the relative velocity

of separation after collision. Thus, equation (5) shows that the relative velocity of approach before

collision is equal to the relative velocity of separation after collision in one–dimensional elastic collision.

From equation (5), we get

v2 = u1 – u2 + v1 . . . (6)

Putting value of v2 from equation (6) to equation (1), we get

m1u1 + m2u2 = m1v1 + m2(u1 – u2 + v1)

or, m1u1 + m2u2 = m1v1 + m2u1 – m2u2 + m2v1

or, m1u1 – m2u1 + m2u2 + m2u2 = m1v1 + m2v1

or, (m1 – m2)u1 + 2m2u2 = (m1 + m2)v1

∴ v1 = (m1 – m2)u1 + 2m2u2 . . . (7)
m1 + m2

Work and Energy | 127

Again from equation (5), we get

v1 = u2 – u1 + v2

Substituting this value of v1 in equation (1) and simplifying, we get

v2 = (m2 – m1)u2 + 2m1u1 . . . (8)
m1 + m2

From equations (7) and (8), the velocities of bodies A and B, after collision can be calculated.

Special Cases

a. When both the colliding bodies are of the equal masses then, m1 = m2 = m (say)
From (7), v1 = u2
From (8), v2 = u1

This shows that if two bodies having equal masses collide elastically in one–dimension, then they

simply interchange their velocities after the collision.

b. When target body B is at rest initially, then u2 = 0

From (7), v1 = (m1 – m2)u1 . . . (9)
m1 + m2

From (8), v2 = 2m1u1 . . . (10)
m1 + m2

Now, following three sub–cases arise.

i. Equal masses, i.e. m1 = m2 = m(say).

Then, from (9), v1 = 0

From (10), v2 = u1

This shows that when a body suffers an elastic collision with another body having equal masses at

rest, then the first body comes at rest whereas the second body starts moving with the same

velocity as that of the first body after collision.

ii. Massive projectile, i.e. m1 >> m2, then m2 → 0
From (9), v1 = u1
From (10), v2 = 2u1
This shows that if a massive body suffers an elastic collision with a light body at rest, then there is

no change in the velocity of the massive body but the light body acquires a velocity which is

nearly double the initial velocity of the massive body after the collision.

iii. Massive target, i.e. m1 << m2, then m1 → 0
From (9), v1 = – u1
From (10), v2 = 0
This shows that if a light body suffers an elastic collision with a massive body at rest then the

velocity of light body is reversed but the massive body remains practically at rest.

Inelastic Collision

A collision is said to be inelastic collision if total linear momentum remains conserved but total

kinetic energy doesn't remain conserved during the collision. Most of the collisions that we encounter

in our daily life are inelastic collisions. The collisions between two buses, the collision between bullet and

a target etc are inelastic collisions. The kinetic energy lost in an inelastic collision appears in the form of

sound energy, heat energy, light energy etc. Following are the characteristics of an inelastic collision:

i. Total linear momentum is conserved.

ii. Total kinetic energy is not conserved.

iii. The total energy is conserved.

iv. The forces involved during the interaction are non–conservative forces.

v. Total mechanical energy doesn't remain conserved.

128 | Essential Physics

When the two particles stick together after the collision then the collision is said to be perfectly inelastic
collision. Mud thrown on a wall, collision between proton and electron, a man jumping into a moving cart
etc are few examples of perfectly inelastic collision. Since, in an inelastic collision,

a. Total linear momentum remains conserved i.e.

m1u1 + m2u2 = m1v1 + m2 v2 . . . (1)
b. But, total K.E. doesn't remain conserved i.e.

1 m1u12 + 1 m2u22 ≠ 1 m1v12 + 1 m2v22 . . . (2)
2 2 2 2

Equations (1) and (2) are the equations of inelastic collision.

Loss of Kinetic Energy in Perfectly Inelastic Collision

A A
B
B
m1 u1 m2 u2 = 0 m1 m2 v

Before collision After collision

[Fig. 5.12, Perfectly inelastic collision]

Consider two bodies A and B collide each other and the collision is perfectly inelastic as shown in figure
5.12. Let body B is initially at rest and two bodies stick together after collision and move with common
velocity v.

From linear momentum conservation principle, m1u1 + m2u2 = (m1 + m2) v

∴ v = m1u1 . . . (1) [since, u2 = 0]
m1 + m2

K.E. before collision = 1 m1u12 + 1 m2u22 = 1 m1u12 . . . (2)
2 2 2

K.E. after collision = 1 (m1 + m2) v2 . . . (3)
2

Dividing equation (3) by equation (2), we get

K.E. after collision 1 + m2)v2 (m1 + m2)
K.E. before collision 2(m1 m1 u12
= = × v2
12m1u12

= (m1 + m2) × mm1 +1um1 22 [ using equation (1)]
m1u12

= (m1 + m2) × m12u12
m1u12 (m1 + m2)2

= m1 < 1.
m1 + m2

K.E after collision
or, K.E before collision < 1

∴ K.E. after collision < K.E. before collision

Thus, there is always loss of kinetic energy in an inelastic collision.

Work and Energy | 129

Coefficient of Restitution

The ratio of relative velocity of separation after collision to the relative velocity of approach before

collision is a constant and is called coefficient of restitution or coefficient of resilience. It is denoted by
'e' and it is a measure of degree of elasticity of collision. Value of 'e' depends on the nature of colliding

bodies. Mathematically,

e = v2 – v1
u1 – u2

a. For a perfectly elastic collision, v2 – v1 = u1 – u2. So, e = 1
b. For perfectly inelastic collision, v2 – v1 = 0. So, e = 0

c. In general, collisions are neither perfectly elastic nor perfectly inelastic. So, 0<e<1

d. For super elastic collision, e > 1.

Boost for Objectives

• Work done by centripetal force in displacing a particle along a circular path is zero.

• When a person-carrying load on his head moves over a horizontal road, work done against the gravitational

force is zero.

• Flowing water has both K.E. and P.E.

• Potential energy exists only for conservative forces, which are central forces. It does not exist for non–

conservative forces like forces of friction etc.

• Work is necessarily done if energy of the body gets changed.

• When a body is in static or dynamic equilibrium, then work done is zero.

• The potential energy of a system in the state of unstable equilibrium is maximum.

• P.E. decreases on the rise of an air bubble in water because work is done by upthrust.

• Oscillations of a body take place about stable equilibrium position only.

• dW = tanθ.
The slope of work–time curve at any instant gives us the power at the instant, as P = dt

• Work done by gravity is +ve if body moves downwards and is –ve if body moves upwards.

• Weight of hydrogen balloon in air is negative. Hence gravitational P.E of it decreases with altitude.

• When a man rowing a boat upstream is at rest with respect to shore, he is doing no work, though he is

applying a force. However, when he stops rowing and moves down with the stream, work is done on him.

• When two like charges are brought together P.E. increases (proton–proton, electron–electron) and when

unlike charges are brought together, P.E. decreases (proton–electron)

• When air bubble rises to the surface of water from below, PE decreases.

• Power is dissipated only by the tangential component of force and not by normal/radial component of force.

Thus power dissipated by centripetal force is zero.

• Mgh
Work done to take out all the liquid out of a full tank, W = 2

• A car of mass m1 and a bus of mass m2 moving with same velocity (v). If they are brought to rest by the
application of brakes, which provide equal retardation, then both will stop at the same distance.

[Hint: 1 mv2 = F.s or, 1 mv2 = ma.s
2 2

∴ s = v2 which is independent of mass]
2a

• If a shell fired from cannon explodes, in mid air then its total kinetic energy increases but total momentum is

conserved.

• Only momentum not kinetic energy is conserved in inelastic collision while both momentum and K.E. are

conserved in elastic collision.

• A bullet is fired from a rifle. If rifle recoils freely, the K.E of the rifle is less than that of bullet.

• A bullet of mass M hits a block of mass M'. The transfer of energy is maximum when M' = M

130 | Essential Physics

• A body of mass m1 moving with velocity u1 strikes head on (elasticity) with another body of mass m2 initially

at rest. Then the fraction of energy lost (or transferred) by the moving body in the collision itself is

4m1m2
(m1 + m2)2

• A body of mass m1 moving with velocity u1 strikes another body of mass m2 initially at rest, inelastically.

Then the fraction of energy lost by moving body in the collision itself is m2
m1 + m2

• When a particle (mass m and velocity v) strikes normally with a wall and returns with same velocity, the

change in momentum is 2mv.

• Elastic potential energy (U) = 1 kx2 where ( k = force constant of spring)
2

Short Questions with Answers

1. A car speeds up while the engine delivers constant power. Is the acceleration greater at the
Ans:
beginning of this process or at the end? Explain. [HSEB sample question]
2.
Ans: Acceleration is the rate of change of velocity. At the beginning, the car speeds up from rest and hence
3.
the rate of change of velocity is high. Therefore, the acceleration is greater at the beginning, and then
Ans:
the acceleration becomes zero at the end for constant power delivered. For constant power [P = F.v], v
4.
Ans: remains constant and hence acceleration becomes zero at the end of the process.

5. A compressed spring is clamped in its compressed position and then is dissolved in acid. What
Ans:
happens of its potential energy? [HSEB Sample question]

The elastic potential energy of the compressed spring is converted into internal kinetic energy of the

acid when dissolved. This internal kinetic energy, then raises the temperature of the acid.

When an object moves away from the earth, the gravitational potential energy increases; when it

moves closer, the potential energy decreases. But the potential energy for a spring increases both

when the spring is stretched and when it is compressed. Explain the reason for the difference in

behavior of these two potential energies.

In case of gravitational potential energy, the work is done to put a body from earth's surface to a certain

height against the gravitational force. So, P.E. increases. But, when the body moves closer to earth's

surface, the work is done by the gravitational force. So, P.E. decreases.

In case of a spring, whether it is compressed or stretched, the work has to be done against internal

restoring force of the spring. Therefore, P.E. of the spring increases in both cases.

A nonzero net force acts on an object. Is it possible for any of the following quantities to be

constant: (i) the particle's speed; (ii) the particle's velocity, and (iii) the particle's kinetic energy?

i. The particle's speed may be constant. As for example, in uniform circular motion, the speed

remains the same and a non–zero centripetal force is acting.

ii. The particle's velocity cannot be constant because a non–zero net force always produces

acceleration and to be acceleration, there should be change in velocity.

iii. The particle's kinetic energy may be constant, as in uniform circular motion.

Explain why the force of gravity due to the earth does not pull the moon in closer and closer on an

inward spiral until it hits the earth's surface? [HSEB 2070]

The moon moves around the earth in circular path. The gravitational force of S Moon
attraction between the earth and moon provides the necessary centripetal force for

the moon to keep on moving around the earth in circular path. The displacement of F

moon and gravitational force are always perpendicular (i.e. θ = 90°). In this case, the

work done by the gravitational force (F) is, Earth

W = →F .→S = FS cos θ = FS cos 90° = 0

Hence, the work done by the gravitational force is zero. So, the moon doesn't move closer and closer to

hit the earth's surface.

Work and Energy | 131

6. How does the kinetic energy of a body change if its momentum is halved? [HSEB 2069]
Ans: P2

7. We have, the relation between kinetic energy (E) and linear momentum (P) is, E = 2m
Ans:
P
8. If its momentum is halved i.e., P1 = 2, then new kinetic energy (E1) of the body becomes,
Ans:
E1 = P12 = 1 P22= 1 2Pm2 
9. 2m 2m 4
Ans:
∴ E
10. E1 = 4
Ans:
So, when momentum is halved, then K.E. becomes one–fourth of its initial K.E.

Two bodies of different masses are moving with the same kinetic energy of translation. Which one

has more momentum? [HSEB 2064]

P2
The relation between linear momentum (P) and kinetic energy (E) is, E = 2m

Let two bodies having masses m1 and m2 are moving with linear momenta P1 and P2 respectively. Since,

they are moving with equal kinetic energies, then E1 = E2

or, P12 = P22
2m1 2m2

or, P1 = m1
P2 m2

∴ P∝ m

Thus, for same kinetic energy, the heavier body has more linear momentum than that of lighter body.

"The earth moving round the sun in an orbit is acted upon by a force, hence the work must be

done on the earth by this force." Do you agree with this statement? [HSEB 2063]

The gravitational force of attraction between the sun and the earth provides the necessary centripetal
force for the earth to keep on moving around the earth in circular orbit. The direction of displacement
(path of orbit) and centripetal force are perpendicular (i.e. θ = 90°). Hence, the work done by this force
is, W = →F .→S = F S cos θ = F S cos 90° = 0,

Hence, the work done must be zero. So, we can't agree with the given statement.

How does the K.E. of an object change if its momentum is doubled? [HSEB 2060]

P2
We have, the relation between linear momentum (P) and kinetic energy (E) is, E = 2m

When its momentum is doubled, i.e.

P1 = 2P, then its new K.E. (E1) becomes, E1 = P12 = (2P)2 = 42Pm2 
2m 2m

∴ E1 = 4E
Hence, K.E. is increased by 4 times when momentum of an object is doubled.

Distinguish between conservative and non–conservative forces. [HSEB 2069, 2066, 2057]

A force is said to be conservative if work done by the force or against the force on moving a body

depends only on initial and final positions of the body and is independent on the nature of path followed
between the two positions. The work done by a conservative force on moving a body in a complete
closed path is always zero. For example; gravitational force. The work done by a conservative force is
completely reversible and the total mechanical energy is always conserved.

A force is said to be non–conservative if work done by the force or against the force on moving a body
is dependent on the path followed between the initial and final positions. The work done by a non–
conservative force on moving a body in a complete closed path is not zero. For example; frictional
force. The work done by a non–conservative force is not completely reversible and the total mechanical
energy is not conserved.

132 | Essential Physics

11. What are elastic and inelastic collisions? Give examples of each. [HSEB 2068]
Ans:
A collision is said to be elastic collision if both the total linear momentum and total kinetic energy
12. remain conserved during the collision. For example; the collision between gas molecules, sub–atomic
Ans: particles, etc. It is an ideal collision. The forces involved during the interaction are conservative forces.

A collision is said to be inelastic collision if total linear momentum remains conserved but total
kinetic energy doesn't remain conserved during the collision. For example: the collision between two

buses. The forces involved during the interaction are non–conservative forces.

In a syphon, water is lifted above its original level during its

flow from one container to another. Where does it get the

needed potential energy from? [HSEB 2050]

Syphon is an arrangement of two vessels kept at different V1 A B
heights in which water (liquid) can be transferred from a vessel V2
V1 at higher level to another vessel V2 at lower level without
disturbing the positions of two vessels as shown in figure. Two h
open ends of a bent tube are dipped into the liquid of both C
vessels V1 and V2. Then the liquid transfers from the vessel V1
at higher level to vessel V2 at lower level due to the pressure V2
difference exerted by the liquids in two vessels. The liquid gets
the needed potential energy from the pressure difference.

13. Differentiate between conservation of kinetic energy and
Ans:
conservation of linear momentum. [HSEB 2055]
14.
Ans: The conservation of kinetic energy means the total K.E. of a system remains constant during collision.
When K.E. remains conserved, then the K.E. doesn't transfer into other forms of energy. In an elastic
15. collision, the total K.E. of the system remains constant. But, in an inelastic collision, the total K.E. of
Ans: the system doesn't remains constant. The K.E. conservation equation is,

1 m1u12 + 1 m2u22 = 1 m1v12 + 1 m2v22
2 2 2 2

The conservation of linear momentum means total linear momentum of a system remains constant
during a collision. In both elastic and inelastic collision, total linear momentum remains conserved. The
linear momentum conservation equation is,

m1u1 + m2u2 = m1v1 + m2 v2

Are conservative and non–conservative forces related with the path of a force involved? Explain

in brief. [HSEB 2069]

A conservative force is not related with the path of force involved. The work done by a conservative
force on moving a body between two points is independent with the path followed and only depends on
initial and final positions of the body. The work done by a conservative force on moving a body in a
complete closed path is always zero.
But, a non–conservative force is related to the path of force involved. The work done by a non–
conservative force on moving a body between two points depends on path followed by the body. The
work done by s a non–conservative force on moving a body in a complete closed path is not equal to
zero.
What is the work done by a coolie walking on a horizontal platform with a load on his head?
In order to balance the load on his head, the coolie has to apply a force on it in upward direction equal
to weight of the load. The displacement is along the horizontal platform. Thus the angle between force
→F and displacement →S is 90°. Hence, the work done W is,
W = →F .→S = F S cos θ = F S cos 90° = 0,
So, the work done by the coolie is zero.

Work and Energy | 133

16. A stationary mass suddenly explodes into two fragments; one heavy and another light. Which one
Ans:
has greater kinetic energy and why?
17.
Ans: We know, the K.E. due to explosive force is inversely proportional to mass of the fragment; i.e.

18. E1 = m2
Ans: E2 m1
19.
Ans: ∴ K.E. ∝ 1
mass

Hence, the lighter fragment has greater kinetic energy than that of heavy fragment.

A car is moving with uniform velocity. Is the engine of the car doing any work under this

condition?

The frictional force is acting against the motion of the car. The force applied by the engine balances the

frictional force and maintains the uniform velocity. The frictional force is acting in a direction opposite

to the motion of car. So, there is displacement of the car against frictional force. Since the velocity is

uniform, so, applied force by engine = frictional force. The work done by the engine of the car is equal

to product of frictional force and displacement of car.

A cricket ball sometimes rebounds from the ground with greater velocity than with which it was

bowled. How can it be possible?

A cricket ball may have spinning motion besides being in translational motion. In some cases, the

spinning energy may be added to the energy of translation. In such condition, the cricket ball may

rebound from ground with greater velocity than with which it was bowled.

Can a body have momentum without kinetic energy? Explain.

We have, kinetic energy of a body of mass m and moving with velocity v is,

E = 1 mv2 = (mv)2 = P2
2 2m 2m

and P = 2mE

20. When K.E. is zero, then P = 2mE must be zero. Hence, a body cannot have momentum without
Ans: kinetic energy.

21. Can a body have energy without momentum? Explain.
Ans:
If momentum is zero, then the body is at rest. Then K.E. must be zero. But, the body at rest may have
22. potential energy like a body placed at some height from the ground. Thus, a body can have energy
Ans: without momentum.
23.
Ans: A particle is moving in a circular path of a given radius with number of rotations per second (i)
constant, (ii) increasing, (iii) decreasing. What happens to work done in three cases?

The work done is equal to the change in kinetic energy of the particle.

(i) For constant speed, no change in kinetic energy and hence work done is zero.
(ii) For increasing speed, K.E. of the particle increases and hence work is done on the particle by the

existing force.
(iii) For decreasing speed, K.E. of the particle decreases and hence work is done by the particle

against the existing force.

Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due
to the sun is not normal to the comet's velocity in general. Yet the work done by the gravitational
force over every complete orbit of the comet is zero, why?
We know, gravitational force is a conservative force. Therefore, the work done by the gravitational
force of the sun over every complete orbit of the comet should be zero.
Can linear momentum of a system be changed without changing its kinetic energy? Explain.
Linear momentum is a vector quantity and kinetic energy is a scalar quantity. In uniform circular
motion, K.E. remains constant (speed is constant) but linear momentum of the particle changes at every
points due to change in direction of motion. Hence, linear momentum of a system can be changed
without changing its kinetic energy.

134 | Essential Physics

24. Is collision possible even without actual contact of colliding particles? Explain.
Ans: Yes. For example, when two positively charged particles, say protons, approach each other, then they
are scattered to different directions. This interaction is also a collision, where there is no physical
25. contact of colliding particles.
Ans: When a book slides along a table top, the force of friction does negative work on it. Can friction
ever do positive work? Explain.
When a book slides along a table top, the force of friction does negative work on it because the force of
friction and displacement are directed in opposite direction [W = →F .→S = F S cos180° = – FS]. Yes, the
friction can do positive work. For example; when a man walks on ground, the friction acts in the
direction of the displacement and hence the work done by frictional force is positive
[W =→F .→S = FS cos 0° = FS].

Numerical Examples

1. A box of mass 15 kg placed on horizontal floor is pulled by a horizontal force. What will be the work

done by the force if the coefficient of sliding friction between the box and the surface of the floor is

0.3 and body moved at unit distance. [NEB 2074]

Solution:

Mass of box (m) = 15 kg. R = mg
Coefficient of sliding friction (µ) = 0.3

Distance moved (S) = 1m v
g = 10ms-2.

Work done (W) = ? m F = Ff
Horizontal force (F) = Ff Ff

or, F = µR [∴µ = Ff ] mg
R

or, F = µ mg = 0.3 × 15 × 10 =45N

Now, work done W is,

W = FS

or, W = 45 × 1

∴ W = 45 J.
2. An explosive of mass M placed at a point explodes into one-third and two-third parts. If the initial

kinetic energy of the smaller part is 1000 J. What will be the initial K.E. of t he bigger part ?
[NEB 2074]

Solution:

Stationary explosive mass = M

M
Mass of light (smaller) part (m1) = 3

2M
Mass of heavy( bigger) part (m2) = 3

Initial K.E. of smaller part (E1) = 1000 J
Initial K.E. of bigger part ( E2) = ?
We know, K.E. due to explosive force is inversely proportional to the mass of the fragments (parts).

∴ E2 = m1
E1 m2

Work and Energy | 135

or, E2 = M/3 = 1
1000 2M/3 2

1
or, E2 = 2 × 1000 = 500 J.

Therefore, the initial K.E. of bigger part is 500 J.

3. A stationary mass explodes into two parts of mass 4 units and 40 units respectively. If the larger mass

has an initial K.E. 10J, what is the initial kinetic energy of the smaller mass? [NEB 2074]

Solution:

Mass of smaller mass (m1) = 4 units

Mass of larger mass (m2) = 410 units

Initial K. E. of larger mass (E2) = 10J.

Initial K.E. of smaller mass (E1) = ?

Since, E1 = m2
E2 m1

or, E1 40
10 =4

∴ E1 = 100 J.

So, initial K.E. of smaller mass is 100 J.

4. A block of mass 50 kg rests on a horizontal surface. The coefficient of static and sliding friction
between the block and the surface are 0.40 and 0.20 respectively. Calculate the minimum force,
which will start the block, and the minimum force which will keep the block in motion once it has
been started. Find the horse power of the engine which pulls it with a constant speed of 20 ms–1.
[T.U. 2051]

Solution:
Mass of the block (m) = 50 kg
Coefficient of static friction (µs) = 0.40. Coefficient of kinetic friction (µk) = 0.20
Minimum force, which will start the block in motion, is limiting friction (Fs) and it is given by
Fs = µsR = µs mg = 0.40 × 50 ×10 = 200N
And, the minimum force which will keep the block in motion is the kinetic friction (Fk) and it is given
by
Fk = µkR = µkmg = 0.20×50×10 = 100N
Now, the power required to pull the block with constant speed of 20 ms–1 is
2000
P = F×V = Fk × v = 100 × 20 = 2000W = 750 H.P. = 2.67 H.P.

5. A block of weight 150N is pulled 20m along a horizontal surface at constant velocity. Calculate the

work done by the pulling force if the coefficient of kinetic friction is 0.20 and pulling force makes an

angle of 60° with the vertical. [T.U. 2057]

Solution: Fcos60°
Given, weight of block (W) = mg = 150N
F

µk = 0.20 R

Since, velocity is constant, 60°
∴ Ff = Fsin60°
. . . (1) m Fsin60°
and, R + Fcos60° = mg Ff

or, R = mg – Fcos60° . . . (2) 20m

Now, µk = Ff = Fsin60° mg
R mg – Fcos60°

or, Fsin60° = µk mg – µk Fcos60°
or, F[sin60° + µk cos60°] = µk mg

136 | Essential Physics

∴ F = sin60° µkmg 0.20 × 150 = 31.06N
+ µk cos60° = 0.866 + 0.20 × 0.50

Work done, W = Fsin60° × 20 = 31.06 × 0.866 × 20 = 537.90J

6. You throw a 20N rock vertically into the air from ground level. You observe that when it is 15m

above the ground, it is travelling at 25 m/s upward. Use the work–energy theorem to find (i) its speed

as it left the ground and (ii) its maximum height. [HSEB 2072 D]

Solution:

Given, Weight of rock (mg) = 20N

20 20 Top
Mass of rock (m) = g = 10 = 2 kg

Height (h) = 15m

Final velocity at 15m height (v) = 25 m/s.

(i) Initial speed (u) = ? 25ms–1

(ii) Maximum height (hmax) = ?
Now, (i) From Work Energy Theorem,

Work done = change in K.E. 15m
u
→ . → = 1 mu2 – 1 mv2
2 2
F S

or, FS cos180° = 1 mu2– 1 mv2 m
2 2

or, 1 mu2 – 1 mv2 Ground
mgh (–1) = 2 2

or, –2gh = u2 – v2

or, u2= v2 – 2gh = 252 – 2(–10).15 = 625 + 300

∴ u = 925 = 30.41 ms–1

(ii) At the maximum height, v = 0, S = hmax
Also, Work done = change in k.E.

or, FS cos180° = 1 mu2 – 1 mv2
2 2

or, 1 mu2 – 1 m .02
mghmax (–1) = 2 2

or, –2ghmkx = u2
u2 925
∴ hmax = – 2g = – 2(–10) = 46.25 m

7. A 650 kW power engine of a vehicle of mass 1.5 × 105 kg is rising on an inclined plane of inclination

1 in 100 with a constant speed of 60 km/hr. Find the frictional force between the wheels of the

vehicle and the plane. [HSEB 2071 C]

Solution: R vF

Power of engine (P) = 650 KW = 650,000W
Mass of vehicle (m) = 1.5×105kg

sinθ = 1 m
100 Ff θ

60×1000 50 mg
Constant speed (v) = 60 km/hr = 60×60 m/s = 3 m/s.
θ
Frictional Force (Ff) = ? O
Since, Power = Force × velocity

or, P = F.v.

or, P = (Ff + mgsinθ) v

or, P – mgsinθ = 650000 × 3 – 1.5 × 105 × 10 × 1 = 39,000 – 15,000
Ff = v 50 100

∴ Ff = 24,000N

Work and Energy | 137

8. A stationary mass explodes into two parts of mass 4 kg and 40kg. The initial kinetic energy of larger

mass is 10J. Find the velocity of the smaller mass. [HSEB 2071 D]

Solution:

Smaller mass (m1) = 4 kg; Larger mass (m2) = 40kg;
Velocity of smaller mass (v1) = ?
K.E. of larger mass (E2) = 10J;

Let, E1 be the k.E. of smaller mass. We know,

E1 = m2
E2 m1

or, E1 40
10 =4

∴ E1 = 100J

1 m1v12
Also, E1 = 2

or, 1 × 4 × v12
100 = 2

or, v12 = 50

∴ v1 = 50 = 7.07 ms–1

9. The constant force resisting the motion of a car of mass 1500 kg is equal to one fifteenth of its weight

if, when travelling at 48 km/hr, the car is brought to rest in a distance of 50m by applying the brakes,

find the additional retarding force due to the brakes (assumed constant) and heat developed in the

brakes. [HSB 2070 B]

Solution:

Given, mass of car (m) = 1500 kg

Weight of car (mg) = 1500 × 10 = 15000N

1
Frictional force (Ff) = 15 × 15000 = 1000N

48×1000 40
Initial velocity of car (u) = 48 km/hr = 60×60 m/s = 3 m/s

Final velocity of car (v) = 0;

Distance covered (s) = 50m.

Additional force due to brakes (Fb) = ?
Using, v2 = u2 + 2as

or, 02 = 4302 + 2.a. 50

∴ a=– 1600 = – 16 ms–2
9 × 100 9

So, total retarding force F is,

F = ma = 1500 × – 196 = – 2666.67N

Thus, magnitude of total retarding force is, F = 2666.67N.

Now, F = Ff + Fb
or, 2666.67 = 1000 + Fb
∴ Fb = 2666.67 – 1000 = 1666.67N
Again, heat developed in brakes = work done by Fb = Fb× S = 1666.67 × 50 = 83,333.50J
So, additional retarding force due to brakes (Fb) and heat developed in the brakes are 1666.67 N and
83,333.50 J respectively.

10. A stationary mass explodes into two parts of mass 4 units and 40 units respectively,. If the larger

mass has an initial K.E. of 100J, what is the initial K.E. of the smaller mass? [HSEB 2069 A, 2068]

Solution:

Smaller mass (m1) = 4 units; Larger mass (m2) = 40 units.
K.E. of larger mass (E2) = 100J K.E. of smaller mass (E1) = ?

138 | Essential Physics

We know, E1 = m2
E2 m1

or, E1 40
100 =4

∴ E1 = 10 × 100 = 1,000J
11. A 0.15 kg glider is moving to the right on a frictionless horizontal air track with a speed of 0.80 ms–1.

It has a head on collision with a 0.300 kg glider that is moving to the left with a speed of 2.2 ms–1.

Find the final velocity (magnitude and direction) of each glider if the collision is elastic.

[HSEB 2069 B]

Solution:

Given, mass of first glider (m1) = 0.15 kg.; Mass of second glider (m2) = 0.300 kg
After collision,
Before collision
Velocity of first glider (u1) = 0.80 ms–1 Final velocity of first glider (v1) = ?
Velocity of second glider (u2) = – 2.2 ms–1 Final velocity of second glider (v2) = ?

We know, in an one dimensional elastic collision,

v1 = (m1– m2) u1+ 2m2u2 = (0.15 – 0.300) 0.80 + 2 × 0.300 × (–2.2)
m1 + m2 0.15 + 0.300

– 0.15 × 0.80 – 1.32 1.44 = – 3.2 ms–1
= 0.45 = – 0.45

Also, v2 = (m2 – m1) u2 + 2m1u1
m1 + m2

(0.300 – 0.15) × (–2.2) + 2×0.15 × 0.80 – 0.33 + 0.24 0.09 = – 0.2 ms–1
= 0.15 + 0.300 = 0.45 = – 0.45

So, the first and second gliders move with respective velocities 3.2 ms–1 and 0.2 ms–1 towards left after

collision.

12. A typical car weighs about 1200N. If the coefficient of rolling friction is µr= 0.015, what horizontal
force is needed to make the car move with constant speed of 72 km/h on a level road? Also, calculate

the power developed by the engine to maintain this speed. [HSEB 2067]

Solution:

Given, weight of car (mg) = 1200N. R = mg

1200 1200 m v
Mass of car (m) = g = 10 = 120kg Ff F = Ff

Coefficient of rolling friction (µr) = 0.015
Constant speed (v) = 72 km/hr = 20 m/s.
Horizontal force (F) = ?
Power developed by engine (P) = ?

Since the speed is constant,

∴ Horizontal force = Frictional force mg
or, F = Ff
or, F = µrR = µr mg = 0.015 × 1200
∴ F = 18N

Also, power (P) = F × v = 18 × 20 = 260 Watt.

13. A car of mass 1000kg moves at a constant speed of 25 m/s along a horizontal road where frictional

force is 200N. Calculate the power developed by the engine. [HSEB 2066]

Solution:

Given, mass of car (m) = 1000kg; Constant speed (v) = 25 m/s.

Frictional force (Ff) = 200N; Power developed by engine (P) = ?
Since, speed is constant,

∴ Horizontal force = Frictional force

or, F = Ff = 200N
Now, power developed (P) is, P = F.v = Ff . v = 200 × 25 = 5000 Watts = 5 KW.

Work and Energy | 139

14. A train of mass 2 × 105 kg moves at a constant speed of 72 kmh–1 up a straight incline against a
frictional force of 1.28 × 104N. The incline is such that the train rises vertically 1.0m for every 100m

travelled along the incline. Calculate the necessary power developed by the train. [HSEB 2061]

Solution:

Mass of train (m) = 2×105kg R vF

Constant speed (v) = 72kmh–1 = 20ms–1 m
Ff θ
Frictional force (Ff) = 1.28×104N

sinθ = 1
100

Power developed by train (P) =?

P = F × v = (Ff + mgsinθ) × v

= 1.28×104+2×105 × 10 × 1100 × 20 θ
O
= (1.28 × 104 + 2 × 104) × 20 = 3.28 × 104 × 20
= 65.6 × 104 = 656 × 103 Watts = 656 KW. mg

15. A ball of mass 4 kg moving with a velocity 10ms–1 collides with another body of mass 16 kg moving

with 4 ms–1 from the opposite direction and then coalesces into a single body. Compute the loss of

energy on impact. [HSEB 2056]

Solution:

u1= 10ms–1 u2= –4ms–1 v=?

m1 = 4 kg m2 = 16 kg m1 + m2
Before collision After collision

From principle of conservation of linear momentum,

(m1+m2) v = m1u1 + m2u2
or, (4 + 16) v = 4×10 + 16 × (–4)

∴ 40 – 64 24 = –1.2 ms–1
v = 20 = – 20

The negative sign suggests that they move together in the initial direction of mass m2 after impact.
Now,

Energy loss on impact = Total k.E. before collision-Total k.E. after collision

= 21 m1u12 + 1 m2u22 – 1 (m1 + m2) v2
2 2

= 21 × 4 × 102 + 1 × 16 × (–4)2 – 1 (4 + 16) × (–1.2)2 = 328 – 14.4 = 313.6J
2 2

16. A bullet of mass 20g travelling at 100ms–1 embeds itself in the centre of a block of wood of mass 1kg,

which is suspended by light vertical string 1m in length. Calculate the maximum inclination of the

string to the vertical. S [HSEB 2053]
Solution:

Given, length of string (l) = 1m θ
Mass of bullet (m1) = 20g = 20 × 10–3kg
Initial velocity of bullet (u1) = 100ms–1 l–h l
Mass of wooden block (m2) = 1kg
Initial velocity of wooden block (u2) = 0 C v=0
Let u be the common velocity of block and bullet after impact. h B

From liner momentum conservation principle, u
A
m1u1 + m2u2 = (m1 + m2) u
or, 20×10–3 × 100 + 1 ×0 = (20×10–3 + 1) u

140 | Essential Physics

∴ u = 2 = 1.96 ms–1
1.02

Let the maximum inclination of the string to the vertical be θ and h is the vertical height through which

the block rises as shown in figure. Here,

Gain in P.E. = Loss sin k.E.

1 mu2 – 1 mv2
mgh = 2 2

or, 2gh = u2 – v2
u2 – v2 (1.96)2 – 02
or, h = 2g = 2 × 10 = 0.19m

From figure, cosθ = l – h 1 – 0.19
l = 1 = 0.81

∴ θ = cos–1 (0.81) = 35.9°
17. A bullet of mass 10g is fired from a gun of mass 1kg with a velocity of 100ms–1. Calculate the ratio of

kinetic energy of the bullet and the gun. [HSEB 2051]

Solution:

Given, Mass of bullet (m1) = 10gm = 0.01 kg.; Mass of gun (m2) = 1kg
Velocity of bullet (v1) = 100 ms–1;

Let, K.E. of bullet = E1 and K.E. of gun =E2

E1 =?
E2

Since, firing is a type of collision. So, we have, E1 = m2 1
E2 m1 = 0.01

∴ E1 100 = 100:1
E2 =1

18. Find the power of an engine in kilowatts, which pull a train of mass 600 tones up an incline of 1 in

100 at the rate of 60 km/hr. The weight of the engine is 200 tonnes and the resistance due to friction

is 50 Newton's per tonne. [HSEB 2050]

Solution:

Power of engine (P) =? R vF
Mass of train = 600 tonnes; mass of engine = 200 tonnes

Total mass of engine and train (m) = 600 + 200 = 800 tonnes = 800000 kg

Constant velocity (v) = 60 60 × 1000 = 50 ms–1 m
km/hr = 60×60 3 Ff θ

sinθ = 1
100

Frictional force per tonne = 50N θ

Total frictional force (Ff) = 50×800 =40,000N. O
The power developed (P) is,
mg
= F×V = (Ff + mgsinθ) × V= 40‚000 + 8‚00‚000 × 10 × 1100 50
P × 3

50 50
= (40,000 + 80,000) × 3 = 1,20,000 × 3 = 20,00,000 Watts = 2,000 KW

Important Numerical Problems

1. Sand drops vertically at the rate 2kgs–1 on to a conveyer belt moving horizontally with a velocity of

0.1 ms–1. Calculate the extra power needed to keep the belt velocity constant. [HSEB 2063 S]

Solution:

Given, Mass of sand falling per second mt  = 2kgs–1

Work and Energy | 141

Velocity of conveyer belt (v) = 0.1 ms–1

Extra power needed (P) = ?

v–u m [Since, u = 0]
Since, P = F×v = ma×v = m × t × v = t (v – u) × v = 2(0.1 – 0) × (0.1)

= 0.02 Watt.
2. A bullet of mass 10g is fired vertically with a velocity of 100 ms–1 into a block of wood of mass 190g

suspended by a long string above the gun. If the bullet comes to rest in the block, through what

height does the block move?

Solution:

Given, mass of bullet (m1) = 10g; mass of block (m2) = 190g
Initial velocity of bullet (u1) = 100ms–1; initial velocity of block (u2) = 0
Common velocity after impact (u) =?

Using linear momentum conservation principle,

(m1 + m2) u = m1u1 + m2u2
or, (10 + 190) u = 10 × 100 + 190× 0

∴ 1000 = 5 ms–1 v=0
u = 200

Second case: hu
Initial velocity of block and bullet (u) = 5ms–1 u2 = 0
m2 = 0
Final velocity of block and bullet (v) = 0

Total mass of block and bullet (m) = m1 + m2 = 10g + 190g = 200g = 0.2 kg.
Height moved by block (h) = ?

Therefore, Work done = Loss in K.E.

or, → . → = 1 mu2 – 1 mv2 u1
2 2 m1
F S

or, FS cos180° = 1 mu2 [since, v = 0]
2

or, mgh (–1) = 1 mu2
2

∴ u2 52 25
h = – 2g = – 2×(–10) = 20 = 1.25 m

3. A man cycles up a hill whose slope is 1 in 20 with a velocity of 3.2 kmh–1 along the hill. The weight of

the man and the cycle is 100kg. What work per minute is he doing? What is his horse power? Take

g = 10ms–2. R
Solution: v

Given, sinθ = 1
20

Velocity (v) = 3.2 kmh–1 = 3.2× 1000 = 8 ms–1 θ
60×60 9

Mass (m) = 100kg

Time (t) = 1 min = 60s θ
Work done per minute (W) = ? mg
Power (P) = ?

P = F × V = mgsinθ × V 18 44.44 H.P. = 0.059 H.P.
= 100 × 10 × 20 × 9 = 44.44 Watt = 750

And, W = Power × time = 44.44 × 60 = 2666.4J.

4. How many joules of energy does a 100 Watt light bulb use per hour? How fast would a 70 kg person

have to run to have that amount of energy?

Solution:

Given, Power of bulb (P) = 100 Watt; Time of use (t) = 1 hr = 3600 s

Energy consumed (E) =?

142 | Essential Physics

Therefoe, E = P× t = 100 × 3600 = 3.6 × 105J.

Mass of person (m) = 70kg

Velocity of person (v) =?

1 mv2 = E
From energy conservation principle, 2

or, 1 × 70 × v2 = 3.6 × 105
2

or, v2 = 3.6 × 105 ×2
70

∴ v = 101.4 ms–1

5 The system in figure is released from rest with the 12 kg block 3 m

above the floor. Use the principle of conservation of energy to find

the velocity with which the block strikes the floor. Neglect friction

and inertia of the pulley.

Solution: 12 kg

For upward motion of 4kg block, the equation of motion is,

T – m1g = m1a 4 kg 3 m
or, T – 4×10 = 4a

or, T – 40 = 4a . . . (1)

For downward motion of 12 kg block, the equation of motion is

m2g – T = m2a
or, 12×10 – T = 12a

or, 120 – T = 12a ….(2)
Adding equations (1) and (2), we get
T

120 – 40 = 4a + 12a T m2 a
a m2g
or, 80 = 16a
m1
∴ 80 = 5 ms–2
a = 16

Now, for 12 kg block m1g
u = 0; v =?; a = 5 ms–2; s = h = 3m

Using, v2 = u2 + 2as
or, v2 = 02 + 2 × 5 × 3 = 30

∴ v = 30 = 5.48 ms–1
6. The turbine pits at the Niagra falls are 50m deep. The average power developed is 3.75 × 106 Watt. If

the efficiency of the generator is 85%, how much water passes through the turbines per minute?
Take g = 10ms–2.

Solution: output power (Pout) = 3.75 × 106 Watts
Given, Height of water fall (h) = 50m;
Efficiency (η) = 85%; time (t) = 1 min. = 60s
Mass of water (m) = ? , g = 10 ms–2

Input power = Pin (say)

Then, η = Pout × 100%
Pin

3.75 × 106
or, 85% = Pin × 100%

∴ 3.75 × 106 × 100% = 4.41 × 106 Watts.
Pin = 85%

Work and Energy | 143

The total work done (W) by generator in 1 min is,

W = Pin × t = 4.41 × 106 × 60 = 2.65 × 108J.

Also, work done by generator = Gain in P.E. by water

or, 2.65 × 108 = mgh

or, 2.65 × 108 = m × 10 × 50
2.65 × 108
∴ m = 500 = 5.29 × 105 kg

7. A dock worker applies a constant horizontal force of 80N to a block of ice on a smooth horizontal

floor. The frictional force is negligible. The block starts from rest and moves 11m in 5 sec. (i) What is

the mass of the block of ice? (ii) If the worker stops pushing at the end of 5 sec, how far does the

block move in the next 5 sec?

Solution:

Given, horizontal Force (F) = 80N; frictional Force (Ff) = 0,

Initial velocity of block (u) = 0; distance moved (s) = 11m

Time taken (t) = 5 sec

(i) Mass of block of ice (m) = ? (ii) Distance moved (s) = ?

1 at2 Force (F) = 0. So, acceleration (a) = 0
Since, s = ut + 2 Velocity (v) = 4.4ms–1

or, 11 = 0 × 5 + 1 × a ×52 Time (t) = 5 sec.
2 Here, the velocity remains uniform. So, using,

25a S = vt
or, 11 = 2 or, S = 4.4 × 5 = 22m
∴ S = 22m
∴ 22 ms–2
a = 25

Final velocity,

22 22 = 4.4ms–1
v = u + at = 0 + 25 × 5 = 5

Also, F = ma
22

or, 80 = m × 25

∴ 80×25
m = 22 = 90.91 kg

8. A 6.0 kg box moving at 3.0 m/s on a horizontal, frictionless surface runs into a light spring of force

constant 75 N/cm. Use the work –energy theorem to find the maximum compression of the spring.

Solution:

Given, Mass of box (m) = 6.0kg; velocity of box (v)_ = 3.0 m/s.

Force constant of spring (K) = 75N/cm = 7500 N/m; maximum compression (x) = ?

From work-energy theorem, the K.E. of the box is converted into the P.E. on the spring while

compressing it. That is,

1 mv2 = 1 Kx2
2 2

or, x2 = mv2 6 × 32 = 0.0072
K = 7500

∴ x = 0.0072 = 0.085 m = 8.5 cm.

144 | Essential Physics

Practice Short Questions

1. The moon is accelerating towards the earth. Why isn't getting closer to us? [HSEB 2070]

2. If you are given two objects one is heavier and another is lighter, both have the same kinetic energy. Which

one has more linear momentum? [HSEB 2066]

3. What is meant by elastic and non–elastic collision? [HSEB 2054]

4. A bullet is fired from a rifle. If the rifle recoils freely, determine whether the K.E. of the rifle is greater than

or equal to or less than that of the bullet.

5. A heavy body and a light body possess equal momentum. Which one possesses larger kinetic energy?

6. A light body and heavy body have the same kinetic energy. Which one will have greater momentum?

7. A porter moving vertically up the stairs with a suitcase on his head does work, why?

8. The momentum of a body is increased by 100%, what is the percentage increases in its kinetic energy?
9. Does the kinetic energy of a car change more when it speeds up from 10 to 15 ms–1 or from 15 to 20 ms–1?

Explain.

10. Can the overall energy of a body be negative?

11. Two boys are carrying equal loads and run to reach same place from same place. If one has reached faster

than another, who has done greater work? What would be the power?

12. Is it necessary that work done in the motion of a body over a closed loop is zero for every force in nature?

Why?

13. What happens to the potential energy when

(i) two protons are brought close together,

(ii) one proton and one electron are brought close together?

14. No energy is consumed in planetory motion. Explain, why?

15. If a rocket explodes in mid–air, what happens to its total momentum and total K.E.?

16. Friction is non–conservative force. Explain.

17. Does the work done depend on how fast or slow a body is moved?

Practice Long Questions

1. Define perfectly inelastic collision. Show that there is always loss of K.E. during perfectly inelastic collision.

2. Write expressions for work done by a constant and a variable force. Show that the work done by the resultant

force on a particle is equal to the change in kinetic energy of the particle. [HSEB 2072 C]

3. What do you mean by variable force? Derive an expression for work done by a variable force. [HSEB 2071 S]

4. What is elastic collision? Prove that the colliding objects exchange their velocities in one-dimensional elastic

collision. [HSEB 2071 C]

5. What is elastic collision? Show that in an elastic collision between two particles, the relative velocity of

separation after collision is equal to the relative velocity of approach before collision. [HSEB 2070 C]

6. Define work. Derive an expression to calculate the work done by a variable force. [HSEB 2070 D]

7. What is the principle of conservation of energy? Show that total mechanical energy of a body is conserved

when it moves under the action of gravitational field. [HSEB 2069 B]

8. What do you mean by elastic and inelastic collisions? Write with one example of each. Show that the sum of

kinetic energy and potential energy of a freely falling body remains constant at any instant. [HSEB 2068 Old]

9. What are conservative and non–conservative forces? Illustrate your answer by reference to the energy

changes occurring in a body whilst falling freely under gravity. [HSEB 2066]

10. State and prove conservation of the mechanical energy. [HSEB 2065]

11. What are elastic and inelastic collision? Give an example of each. Write the energy and momentum equations

for an inelastic collision. [HSEB 2059]

Practice Numerical Questions

1. A car of mass 1000 kg moves at a constant speed of 20 ms–1 along a horizontal road where the frictional force

is 200 N. Calculate the power developed by the engine. [HSEB 2058][ Ans: 4 KW]

2. A stationary radioactive nucleus of mass 210 units disintegrates into an alpha particle of mass 4 units and a

residual nucleus of mass 206 units. If the kinetic energy of the alpha particle is E, calculate the kinetic energy

of the residual nucleus. 2E
[ Ans: 103]

Work and Energy | 145

3. A 10 kg ball and 20 kg ball approach each other with velocities 20 ms–1 and 10 ms–1 respectively. What are

their velocities after collision if the collision is perfectly elastic? [Ans: –20 ms–1, 10 ms–1 ]

4. You push your physics book 1.50 m along a horizontal tabletop with a horizontal force of 2.40 N. The

opposing force is 0.600 N. (a) How much work does your 2.40 N force do on the book ? (b) What is the work

done on the book by the friction force? (c) What is the total work done on the book?

[Ans: 3.60 J, –0.900 J, 2.70J]

5. A fisherman reels in 12.0 m of line while pulling in a fish that exerts a constant resisting force of 25.0 N. If

the fish is pulled in at constant velocity, how much work is done on it by the tension in the line? Ans: 300J
6. Two tugboats pull a disable supertanker. Each tug exerts a constant force of 1.80 × 106 N, one 14° west of

north and the other 14° east of north, as they pull the tanker 0.75 km towards the north. What is the total work

they do on the supermarket? [Ans: 2.62 × 109 J]

7. A sled with mass 8.00 kg moves in a straight line on a frictionless horizontal surface. At one point in its path,

its sped is 4.00 m/s.; after it has traveled 2.5 m beyond this point, its speed is 6.00 m/s. Use the work–energy

theorem to find the force acting on the sled, assuming that this force is constant and that it acts in the

direction of the sled's motion. [Ans: 32N]

8. A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downwards at an angle of

36.9° below the horizontal. If the block of ice starts from rest, what is its final speed?: You can ignore

friction. [Ans: 2.97 m s–1]

9. A tandem (two–person) bicycle team must overcome a force of 165N to maintain a speed of 9.00 m/s. Find

the power required per rider, assuming that each contributes equally. Express your answer in watts and in

horsepower. [Ans: 473 W, 0.99 HP]

10. An object A moving horizontally with kinetic energy of 800 J experiences a constant horizontal opposing

force of 100 N while moving from a place X to another place Y, where XY is 2 m. What is the energy of A at

Y? In what further distance will A come to rest if this opposing force continues to act on it? [Ans: 600 J, 6m]
11. A ball mass 0.1 kg is thrown vertically upwards with an initial speed of 20 ms–1. Calculate (i) the time taken

to return to the thrower, (ii) the maximum height reached, and (iii) the kinetic and potential energies of the

ball half–way up. [Ans: (i) 4 sec (ii) 20m (iii) 10 J, 10 J]

12. A 4 kg ball moving with a velocity of 10.0 ms–1 collides with a 16 kg ball moving with a velocity of 4.0 ms–1.

(i) in the same direction and (ii) in the opposite direction. Calculate the velocity of the balls in each case if

they coalesce on impact, and the loss of energy resulting from the impact.
[Ans: (i) 5.2 ms–1 57.6 J (ii) –1.2 ms–1, 313.6J]

13. A ball of mass 0.1 kg is thrown vertically upwards with a velocity of 20 ms–1. What is the potential energy at

the maximum height? What is the potential energy of the ball when it reaches three–quarters of the maximum

height while moving upwards? [Ans: 20J, 15 J]

14. A car of mass 1000kg moves at a constant speed of 20 ms–1 along a horizontal road where the fraction force is

200 N. Calculate the power developed by the engine. If the car now moves up an incline at the same constant

speed, calculate the new power developed by the engine. Assume that the frictional force is still 200 N and

that sinθ = 1 , where θ is the angle of the incline to the horizontal. [Ans: 4 kW, 14 kW]
20

15. Sand falls at a rate of 0.15 kg s–1 on to a conveyor belt moving horizontally at a constant speed of 2 ms–1.

Calculate (i)( the extra force necessary to maintain this speed, (ii) the rate at which work is done by this force,

(iii) the change in kinetic energy per second of the sand on the belt. [Ans: (i) 0.3 N (ii) 0.6 W (iii) 0.3W]
16. A railway truck of mass 4×10 kg moving at a velocity of 3 ms–1 collides with another truck of mass 2 × 104

kg which is at rest. The coupling joins and the trucks move off together. What fraction of the fist truck's

initial kinetic energy remains as kinetic energy of the two trucks after the collision? 2
[Ans: 3 ]

17. A train of mass 2.0 × 105 kg moves at a constant speed of 72 km h–1 up a straight incline against a frictional
force of 1.28 × 104 N. The incline is such that the train rises vertically 1.0 m for every 100 m traveled along

the incline. Calculate (i) the rate of increase per second of the potential energy of the train and (ii) the

necessary power developed by the train. [Ans: (i) 400 kW (ii) 656 kW]

18. A vehicle of mass 2000 kg traveling at 10 ms–1 on a horizontal surface is brought to rest in a distance of 12.5

m by the action of its brakes. Calculate the average retarding force. What power must the engine develop in
order to take the vehicle up an incline of 1 in 10 at a constant speed of 10 m s–1 , if the frictional resistance is

equal to 200 N? [Ans: 8000 N, 22 KW.]

146 | Essential Physics

MBBS, B.E., B.Sc. Nursing Entrance Preparation Questions

1. Two identical balls A and B are moving towards each other with velocity +5 m/s and –3 m/s, undergoes head on
collision. There velocities after collision will be

a. + 5 and – 3 b. – 3 and + 5 c. + 3 and – 5 d. – 5 and + 3

2. A train is moving with velocity V1 and another train is moving with velocity V2 in the same direction in the same

track. If the distance between the two trains is 'a' then minimum retardation of train so as to avoid collision.
– V2)2 b. r < (V1 – V2)2 V2)2 d. r < (V1 – V2)2
a. r > (V1 2a 2a c. r < (V1 – 2
a

3. A body of mass 'm' moving with velocity V splits in two parts such that m comes to rest instantly. What is the K.E.
4

of other mass ?

a. 1 mv2 b. 2 mv2 c. 4 mv2 d. mv2
2 3 3

4. If 2 bodies A and B, being A lighter, having same linear momentum are moving, then the K.E. of A is

a. More than B b. Less than B c. Same as B d.None

5. If K.E. of a body is increased by 2 times then it's momentum increases by:

a. 2 times 1 c. 2 times d. 2 2 times
b. 2 times

6. A force F = (10 + 0.5 x) acts on a particle in x – direction where F is in Newton and x is in meter. How much

work is done by the force for a displacement from x = 0 to x = 2m?

a. 4 Joule b. 8 Joule c. 21 Joule d. 34 Joule

7. A bullet of 0.01kg is fired from a rifle of mass 20kg with a speed of 100m/s. The recoil velocity of the rifle must

be

a. 0.01 m/s b. 0.05 m/s c.1 m/s d.20 m/s

8. A body of mass 10kg at a height of 500m from the earth has energy of 850 KJ. It is moving with velocity nearly

equal to

a. 200 m/s b. 300 m/s c. 400 m/s d. 500 m/s

9. Kinetic energy of a body of mass m is E. Its linear momentum will be equal to:
b.(2mE)1/2 c. (mE/2)1/2
a. 2 mE d. 2 E/m

10. A bullet of mass 10 gm is fired with velocity 63 ms–1 on a block and wood of mass 200 gm at rest and the bullet

sticks inside the wood. The common velocity of bullet and wood is

a. 6 m/s b. 9 m/s c. 3 m/s d. 1 ms–1

11. Two bodies having different mass have same momentum. Which of them has greater kinetic energy?

a. Heavier one b. Lighter one c. Both same d. Description is incomplete

12. A substance is moved by a distance S using force F. The work done is

a. FS cos θ b. FS sin θ c. FS tan θ d. FS sec θ

13. Which of the following is always conserved in collision?

a. Kinetic energy b. Angular momentum c. Linear momentum d. Torque

14. Two masses 'm' and '4m' are moving with equal kinetic energies, the ratio of magnitudes of their momentum is:

a. 4:1 b. 1:2 c. 2 : 1 d. 6 : 1

15. In an inelastic collision

a.The initial K.E. is equal to the final kinetic energy. b.The final K.E is less than the initial kinetic energy.

c. The K.E remains constant. d. The K.E first increase and then decreases.

16. The work done by the external forces on a system equals the change in

a. Total energy b. Kinetic energy c. Potential energy d.None of these

17. A body initially at rest explodes into to masses M1 and M2 and they move apart with speed V1 and V2 respectively.

What is the ration of V1/V2 c . M21/M22 d. M22/M21

a. M1/M2 b. M2/M1

18. In perfectly elastic collision, K.E. after collision

a. greater than that of initial b. smaller than that of initial c. is same as that of initial d. is zero

Answer Key

1. b 2. a 3. b 4. a 5. c 6. c 7. b 8. c 9. b 10. c

11. b 12.a 13. c 14. b 15. b 16. B 17. B 18. C

Ch apter

6 Circular Motion

Teaching Manual Physics Grade – XI, Higher Secondary Education Board
Curriculum and Training Division Sanothimi, Bhaktapur

Syllabus:

Circular Motion–Angular displacement, velocity and acceleration; Relation between
angular and linear velocity and acceleration; Centripetal acceleration, centripetal force;
Conical pendulum ; Motion in a vertical circle; Motion of cars and cyclist round a
banked track.

Objectives:

The objective of this subunit is to introduce the concept of circular motion.

Activities (micro syllabus):

1. Define various quantities to describe circular motion; angular displacement,
angular velocity, angular acceleration, time period and frequency

2. Find the relation between linear and angular velocity, and acceleration.
3. Define and derive centripetal acceleration and centripetal force.
4. Define centrifugal force and mention its application.
5. Explain the conical pendulum.
6. Discuss the motion of a body in a vertical circle and derive the relations for

maximum and minimum tensions.
7. Discuss the motion of cars and cyclist round a banked track.

148 | Essential Physics

6.1 Introduction

When displacement, velocity and acceleration of a moving particle lie along a same straight line, then the
motion is called rectilinear motion. If displacement, velocity and acceleration of the moving particle
don't lie along a same straight line, then the motion is called curvilinear motion.

Circular motion is a special type of curvilinear motion in which a particle moves along a circle.
Motion of electrons around the nucleus in different orbits is an example. Circular motion is a two
dimensional motion in which the magnitudes of both velocity and acceleration remains constant but their
directions are changing continuously. In uniform circular motion, the speed of the particle is constant. In
non–uniform circular motion, the magnitude of speed is also changing.

6.2 Characteristics of Uniform Circular Motion

i. Angular Displacement : Consider a Y

particle is moving along a circle in anti– B
clockwise direction as shown in figure
6.1 (i). Let the particle is at A at time t = ∆θ

0 and at B after time t such that ∠BOA θ A O θ1 θ2 X
= θ. The angle 'θ' is called angular O

displacement of the particle in time

→→
interval 't'. In figure, OA and OB are the

position vectors (or radius vectors) of (i) (ii)
the particle at time 't' = 0 and 't' = t [Fig. 6.1, Motion in a circle]
respectively. Thus, the angular

displacement (θ) is defined as the angle traced out by the position vectors of the particle in a
given interval.

Mathematically, θ = Arc length A{B . . . (1)
Radius of arc
= OA

The SI unit of angular displacement is radian and it is a dimensionless quantity. A finite angular

displacement (θ or ∆θ) is not a vector quantity but an infinitesimally small angular displacement

(dθ) is a vector quantity. Angular displacement is positive in the counter – clockwise direction and

negative in the clockwise direction.

ii. Angular Velocity: The rate of change of angular displacement is called angular velocity and it

is denoted by ω.

__

a. Average angular velocity (ω or ωav): It is defined as the ratio of the total angular

displacement to the total time taken by the particle to undergo this displacement.

Suppose a particle is moving along a circle as shown in figure 6.1 (ii). Let θ1 and θ2 be the

angular displacements at time t1 and t2 respectively. Then, the average angular velocity

(ωav) in the time interval ∆t = t2 – t1 is,

ωav = Total angular displacement = θ2 – θ1 = ∆θ . . . (2)
Total time taken t2 – t1 ∆t

b. Instantaneous angular velocity (ω): In non–uniform circular motion, the magnitude and

direction of the velocity of particle is changing at every instant of its motion. The velocity

of the particle measured at a particular instant of time is called instantaneous angular

Circular Motion | 149

velocity at that instant. It is the limiting value of the average angular velocity in a small

time interval as the time interval approaches zero. That is, ω = lim 0 ∆θ = dθ . . . (3)
∆t → ∆t dt

For uniform circular motion, ωav= ω. ω
The S.I. unit of ω is rads–1 and its

dimensional formula is [M°L°T–1]

The angular velocity (ω) is a vector

quantity. It is represented by an arrow

drawn along the axis of rotation. The

direction of the arrow (i.e. ω) depends ω
on the sense of rotation and is given by

right hand thumb rule. When the (a) (b)
fingers of right hand are curled in the

sense of revolution of the particle, then
the thumb held perpendicular to the [Fig. 6.2. Direction of angular velocity (ω)]

curved fingers points in the direction of angular velocity as shown in figure 6.2(a) and (b).

The length of arrow gives the magnitude of ω.

c. Uniform angular velocity: The angular velocity of a particle is said to be uniform if the

particle describes equal angular displacement in equal interval of time.

iii. Time Period (T): The time taken by the particle to make one complete revolution is called time

period of circular motion.

When the particle completes one revolution then θ = 2π and t = T. The angular velocity (ω) is

ω = θ = 2π . . . (4)
t T

∴ T = 2π . . . (5)
ω

iv. Frequency (f or n): The number of revolutions completed by the particle in 1 second is called

1
frequency of the particle in circular motion. Therefore, frequency = time period

1 . . . (6)
or, f = T

The S.I. unit of frequency is revolutions per second or Hertz (Hz) and its dimensional formula is
[M0L0T–1].

It is a scalar quantity. From equations (4) and (6), we get

ω = 2πf . . . (7)

v. Angular Acceleration (α) : If angular velocity of a particle is non–uniform, then the particle has

angular acceleration.

a. Average angular acceleration (αav): It is defined as the ratio of change in angular velocity
of the particle to the time taken by the particle to undergo this change in angular velocity.

If ω1 and ω2 be the angular velocities of the particle at times t1 and t2 respectively, then the

average angular acceleration (αav) in the interval t1 – t2 = ∆t is given by

αav = ω2 – ω1 = ∆ω . . . (8)
t2 – t1 ∆t

b. Instantaneous angular acceleration (α)

It is defined as the limiting value of the average angular acceleration as the time interval ∆t

approaches zero. That is,

α = lim 0 ∆ω = dω . . . (9)
∆t → ∆t dt

150 | Essential Physics

The S.I. unit of α is rads–2 and dimensional formula is [M0L0T–2]. It is a vector quantity.

From equations (3) and (9), we get

α = d dθ = d2θ . . . (10)
dt  dt dt2

vi. Relation between linear velocity and angular velocity: Let a particle having mass 'm' is moving

along a circle of radius 'r' with constant speed 'v' as shown in figure 6.3. Let 'ds' be small linear

displacement from point A to B in small time 'dt' and 'dθ' be the small angular displacement at the

centre O of the circle. Then, B

{dθ AB ds ds
OA r
= = dθ A
r
O

or, ds = rdθ

Also, ds = vdt [ ds
∴ vdt = rdθ v = dt ]
[Fig 6.3, Motion in a circle.]

dθ
or, v = r dt

∴ v = rω . . . (11)

Where, ω = dθ is the angular velocity of the particle. Thus,
dt

linear velocity = radius of circular path × angular velocity

vii. Relation between linear Acceleration and Angular Acceleration:

From equation (11), v = rω

Differentiating both sides with respect to time 't', we get

dv = d (rω) = r dω
dt dt dt

∴ a = rα . . . (12),

where α = dω = angular acceleration.
dt

Thus, Linear acceleration = radius of circular path × angular acceleration.

6.3 Centripetal Force and Centripetal Acceleration

When a particle is moving in a circle with constant speed then the particle is said to be moving with

uniform circular motion. In this case, the velocity changes only in direction but not in magnitude. An

external force must be acting on the particle to change direction of velocity continuously in a uniform

circular motion. Also, to change direction of velocity only, the component of external force along the

direction of velocity must be zero. This is possible only if the direction of the external force is

perpendicular to the direction of velocity. Since, velocity is along the tangent to the circle, therefore, the

direction of external force must be along the normal (i.e. radius) of the circle at every point. In other

words, the external force must always be directed towards the centre of the circle at every point and it is

called "centripetal force". An inward force necessary for a particle to keep on moving along a

circular path is called centripetal force. If m be the mass of the particle then the centripetal force F is

F = ma . . . (1)