First mechanics final

Kinematics | 51

or, v2 – u2 = 2a(ut + 1 at2)
2

But ut + 1 at2 = S
2

∴ v2 – u2 = 2aS

Graphical treatment: From velocity–time graph, as shown in figure 3.6, the distance covered,

S = area of trapezium OABM

1
= 2 (OA + BM) × OM

1
= 2 (OA + BM) × AL

1 AL
= 2 (OA + BM) × BL × BL

BL
Now, OA = u, BM = v, BL = BM – LM = v – u and AL = slope of AB = a .

So that, AL = 1 . Thus, S = 1 (u + v) × 1 × (v – v2– u2 . Therefore, v2 – u2 = 2aS.
BL a 2 a u)= 2a

3.9 Distance Covered by a Uniform Accelerated Body in a
Particular Second

Suppose, a body has initial velocity u and uniform acceleration a. The distance covered by the body in nth
second of its motion is given by

Snth = Sn – Sn – 1 . . . (1)

where Sn and Sn – 1 represent distance covered in n and (n – 1) seconds. Since, distance covered in time t is

given by, S = ut + 1 at2 . . . (2)
2

Setting, t = n and n – 1, in equation (2), we have

Sn = un + 1 an2 . . . (3) and Sn – 1 = u(n – 1) + 1 a(n – 1)2 . . . (4)
2 2

Substituting for Sn and Sn – 1 in equation (1), we have

∴ Snth = [un + 1 an2] – [u(n – 1) + 1 a(n – 1)2]
2 2

= [un + 1 an2] – [u(n – 1) + 1 a(n2 – 2n + 1)]
2 2

= un + 1 an2 – un + u – 1 an2 + an – 1 a
2 2 2

1
= u + an – 2 a

∴ a . . . (5)
Snth = u + 2 (2n – 1)

3.10 Equation of Motion under Gravity

The equation of motion under gravity can be obtained by replacing 'a' by 'g' (i.e., g is acceleration due to
gravity). Accordingly, the equations of motion from above under gravity will become

v = u + gt

52 | Essential Physics

S = ut + 1 gt2
2

v2 – u2 = 2gS 1
and Snth = u + 2 g (2n – 1)

The value of g is 980 cms–2 or 9.8 ms–2. It is taken positive, when body falls towards the surface of earth

and negative when the body goes up against gravity.

3.11 Relative Velocity

Relative velocity is define as the time rate of change of position of one object with respect to the another
object. The velocity of a body is always measured with respect to a body at rest. The body at rest acts as
reference body. Let us now, determine the relative velocity of a body with respect to another, when both
of them are in motion. The relative velocity of a body with respect to another body when both are in
motion is the rate at which it changes its position with respect to the other body. To explain it, let us
consider two trains A and B moving along parallel railway lines in the following two cases:

1. When both the trains are moving in same direction with the same velocities: Suppose, the

trains A and B are moving with same velocity (say 50 A V
kmh–1) due east as shown in figure 3.7. To the observer in
East

train A, the observer in train B will appear at rest and

vice–versa. Therefore, we may say that the relative B V East
velocity of either train relative to the other is zero.

2. When both the trains are moving in same direction [Fig 3.7, velocity of moving trains in
same direction with same velocity]

with the different velocities: Now, let the train A moves

with the velocity (say 40 kmh–1) and the train B with the velocity (60 kmh–1) due east as shown in

figure 3.8. After one hour, the trains A and B will reach the points at distances of 40 km and 60 km

from their respective initial positions. Thus, the observer of train B will find himself ahead of the

observer of train A by 20 km after one hour i.e. observer of train B will find himself moving

forward with a velocity V = v – u = 20 kmh–1 with respect to the observer of train A. Thus, the

relative velocity of train B with respect to A is 20 kmh–1 due east. A V 40kmh–1 East
Similarly, it can be found that the relative velocity of train A with

respect to train B is 20 kmh–1 but due west (in opposite direction).

Let, us now develop the method to find the relative velocity of a B V –1 East

60kmh

body B moving with velocity v with respect to another body A [Fig 3.8, velocity of trains in same

moving with velocity u, when the two bodies are not moving along direction but unequal velocity.]

the same line. It is found in the following steps:

i. Bring the body A (the body with respect to which the relative velocity is to be determined)
at rest by providing it a velocity –u (equal and opposite to its own velocity).

ii. The process does not affect the relative velocity of body B, provide the same velocity –u to
body B also. Then,

iii. The resultant of the two velocities of body B; one its own velocity v and other the provided
velocity –u (which may be found by parallelogram law of velocities) gives the relative
velocity of body B with respect to A in both magnitude and direction.

If we apply this working method to the two cases as discussed above, we shall arrive at the same
result. Some examples of relative velocity are described as below:

Kinematics | 53

→→
i. Motion of satellite in equatorial plane: Let, V s and V e are the velocity of satellite moving in

equatorial plane and velocity of a point on the surface of earth relative to centre of the earth. The

→→→
relative velocity of satellite with respect to the surface of earth V r = V s – V e . Where satellite

moves from west to east in the direction of rotation of the earth on its axis but the satellite moves

from east to west opposite to the direction of rotation of the earth on its axis, the relative velocity

is given by Vr = Vs – (–Ve) ∴ Vr = Vs + Ve

The relative velocity of moving satellites with respect to earth as shown in figure 3.9.

Ve Vs Ve

Vs

[Fig 3.9, relative velocity of satellite with respect to earth]

ii. Relative velocity of rain with respect to man: When a man is walking on the road at raining

→→ N
time. Let, V m and V r represent the velocity of rain drop
man and rain respectively. The velocity of rain
falling vertically where man is walking

horizontally on the road. Therefore, the relative θ

velocity of rain with respect to man is given by – vm vm

→→ → W E
V = V r + –Vm O

θ

∴ → Vr2 + Vm2
|V| =

If a person is holding an umbrella to be tilted from vr

the vertical position by making angle θ. Then, S

Angle of umbrella is tanθ = Vm as shown in figure [Fig 3.10, Relative velocity between rain and
Vr man walking]

3.10.

→
iii. A swimmer crossing the river in the shortest time: Let, V s be the velocity of swimmer begins

to swim by making an angle θ with the perpendicular direction of flow of river. The velocity

component of swimmer is Vs cosθ.

∴ Vs cosθ = d
t

∴ d
t = Vscosθ . . . (1) where 'd' be the width of river as shown in figure 3.11 (a).

54 | Essential Physics

P P vr R

vs d vs vd
θ θ

Q Q
(a) (b)

[Fig 3.11, Crossing the river in short time]

At short time, θ = 0 so that, equation (1) will be

d . . . (2)
t = Vs

→→ →
Let, V , V s and V r be the resultant velocity of swimmer, velocity of swimmer and velocity of

→ Vs2 + Vr2 + 2VsVr cos90°
river respectively as shown in figure 3.11 (b). Then V =

or, →V = Vs2 + Vr2 and tanθ = Vr . We have from figure 3.11 (b), tanθ = PR
Vs PQ

or, PR = PQ tanθ = Vr PQ = Vr d. Where distance covered by swimmer PR.
Vs Vs

iv. Crossing the river along the shortest path: Let, RP
→→
V s and V r be the velocities of swimmer and river vs d
respectively when the swimmer cross the river θ
straight to the next bank along PQ = d as shown in
figure 3.12. Then, Q
[Fig 3.12, Crossing the river at shortest path]
V = Vs2 – Vr2 and

dd
crossing time t = v = Vs2 – Vr2

3.12 Projectile

"Any object thrown into space or atmosphere such that it moves under the effect of gravity alone is
called a projectile". It may be pointed out that for throwing the projectile into space, an engine or a fuel
of some kind may be used but after its firing, the projectile should move under the effect of gravity alone.
Thus, an aeroplane in flight is not a projectile but say a bomb dropped form the aeroplane is a projectile.
Examples of projectile are:

1. a bullet fired from the gun,
2. a shell fired from a canon,
3. a javelin or hammer thrown by an athletes,
4. a stone thrown from the top of hill or a tower etc.
The path followed by the projectile is called its trajectory. The projectiles moving in earth's atmosphere
are acted upon the resistance of air and the shape of the trajectory is greatly affected by air resistance.
Except for very light projectiles, the air resistance is negligible. While deriving expressions for the time
of flight, horizontal range etc. for a projectile, the air friction will be neglected. In the absence of air
resistance, the motion of projectile is taken to be the combination of following two independent motions:
1. Motion along horizontal path with uniform velocity and

2. Motion along vertical path under gravity (i.e., with uniform acceleration equal to g.)

Kinematics | 55

3.13 Projectile Fired Horizontally

Consider, a projectile fired from point O at a height h above the ground with a velocity v parallel to
horizontal. Consider point O is origin and the horizontal and vertical line OX and OY as the two axes.
Suppose at any time t, the projectile is at P. The position of the projectile at time t is given by x and y
coordinates of point P as shown figure in 3.13. The distance x along horizontal is covered with uniform
velocity v in time t. Therefore,

x = vt . . . (1) O v X
y
The distance y along vertical is covered with initial velocity
zero in time t and with uniform acceleration g. Since,

S = ut + 1 at2
2

We have, y = (0)t + 1 gt2 h P Vx
2 xβ

∴ y = 1 gt2 . . . (2) Vy V
2

This is an equation of trajectory. Again, from equation (1) we
get,

x
t=v

Substituting for t in equation (2) we get, A GROUND Q

y = 1 g xv2 YR
2 [Fig 3.13, ProFjeigct.il3e .f1ir6ed horizontally]

or, y = 2gv2 x2

∴ y = kx2 . . . (3) where k = g/2v2 is constant.

It is the equation of a parabola which is symmetrical about Y–axis. Hence, a projectile fired horizontally

moves along parabolic path.

Time of Flight (T)

It is defined as the 'total time for which the projectile remains in air". The total vertical distance covered
by a projectile when it hits on the ground is equal to height h of the tower. If T be the time of flight, then
using equation Sy = uy + 12gt2 we get,

h = 12gt2 where initial velocity uy = 0

∴ T= 2h
g . Since, t = T. Thus, time of flight is independent of mass of the projectile.

Horizontal Range (R)

The horizontal distance covered by a projectile during the time of flight is called horizontal range. We
have an equation

Sx = ux + axt2

R = ux T where Sx = R, t = T, ax = 0

∴ 2h 2h
R = u g . Since ux = u and T = g . This gives horizontal range of projectile.

56 | Essential Physics

Velocity of Projectile at an Instant

Let, V be the velocity of projectile at time t, when it is at point P. Then V is resultant of the two

velocities:

i. Vlocity Vx along horizontal which is equal to V and
ii. Velocity Vy along vertical which is given by Vy = 0 + gt

∴ Vy = gt
Therefore, the magnitude of velocity at point P is given by

V = Vx2 + Vy2 = (V)2 + (gt)2 = V2 + g2t2 . . . (1)

The resultant velocity V at point P acts along the tangent to the path at the point. If β is the angle it makes

with horizontal, then

tan β = Vy
Vx

or, tan β = gt
V

∴ β = tan–1 gVt . . . (2)

Note:
(1). Time (T) taken by the projectile to reach the ground (time of flight or time of descent) can be obtained from

2h
equation T = g . . . (1)
(2) Velocity of the particle both in magnitude and direction on reaching the ground can be obtained from above

2h
equations (2), by putting t = T or T = g Therefore, from equation above, velocity of projectile on

reaching ground at point Q is V = V2 + g2 t2 for t = T = 2h ∴ V= V2 + 2gh From equation above
g

g 2h
direction of velocity at point Q, tanβ =
g or, tanβ = 2gh
V V

(3) From above equation, it may be noted that if time t increases, tan β and hence value of β increases. In case

projectile is fired at greater height above the ground so that t is very large, the value of β approaches 90°

and the projectile will fall almost vertically downward.

3.14 Projectile Fired at an Angle with Horizontal

Consider a projectile fired from a point O with velocity v at an angle θ with the horizontal. Regard the
point O as origin, horizontal line OX as X–axis and the vertical line OY as Y–axis. Suppose at any time t,
the projectile is at point P. The position of projectile at time t is given by x and y coordinates of point P
as shown in figure 3.14.The initial velocity v of the projectile can be resolved into two components:

Y

VA

Vy

β

P Vx

vy=vsinθ v H
y

θ Q
x
O vx=vcosθ B X

R
[Fig 3.14, projectile Ffiirge.d3a.1t 6an angle with horizontal]

Kinematics | 57

1. vx = v cosθ along horizontal direction and 2. vy = v sinθ along vertical direction
The distance x along horizontal direction is covered with uniform velocity v cos θ in time t. Therefore,

x = (v cosθ)t . . . (1)

The distance y along vertical direction is covered in time t with initial velocity, u = v sin θ and

acceleration, a = –g (Since, g is directed downward i.e. opposite to direction of v sin θ)

Since, S = ut + 1 at2 . We have,
2

y = (v sinθ)t + 1 (–g)t2
2

∴ y = (v sinθ)t – 1 gt2 . . . (2)
2

Equation of trajectory : From equation (1), we have

x
t = v cosθ

Substituting for t in equation (2) we get,

y = v sin θ v x θ – 1 g  x 2
cos 2 v cos θ

∴ y = x tan θ − 2v2 g . x2 . . . (3)
cos2θ

As equation (3) is equation of parabola, it follows that a projectile fired at some angle with the horizontal

moves along a parabolic path.

Velocity of Projectile at any Instant

Let, V be velocity of projectile at time t, when it is at point P. Then, V is resultant of two velocities:

i. Velocity Vx along horizontal which always remains equal to v cos θ.
ii. Velocity Vy along vertical which is given by

Vy = v sinθ + (–g)t = v sinθ – gt
Therefore, magnitude of velocity at point P is given by

V = Vx2 + Vy2

= (v cos θ)2 + (v sin θ – gt)2

= v2 cos2θ + v2 sin2θ + g2t2 – 2vgt sinθ

∴ V = v2 + g2t2– 2vgt sinθ . . . (4)

The resultant velocity V at point P acts along the tangent to the curved path at the point. If β is the

angle, it makes with horizontal then,

tan β = Vy
Vx

or, tan β = v sin θ – gt
v cos θ

∴ β = tan- 1 v sin θ – gt  . . .(5)
 v cos θ 

Time of Flight (T)

It is the time taken by the projectile to return to the ground or the time for which the projectile remains in
air. It is denoted by T. As motion from O to A and A to Q are symmetrical, the time of ascent (for journey

T
from O to A) and time of descent (for journey from A to Q) will be each equal to 2 .

58 | Essential Physics

Initial velocity along vertical at point O = v sinθ, Final velocity along vertical at height point A = 0.
T

Acceleration = –g and time = 2 . Using the equation, v = u + at, we have

0 = v sin θ + (–g) T
2

∴ T = 2v sin θ . . . (6)
g

Maximum Height Attained (H)

It is the greatest height to which a projectile rises above the point of projection. It is denoted by H. Initial

velocity along vertical at O = v sin θ. Final velocity along vertical at highest point A = 0. Acceleration,
a = –g . Height attained =H. Then, using the equation v2 – u2 = 2as we get,

or, (0)2 – (v sinθ)2 = 2(–g) H (Since, S = H, a = – g)

∴ v2 sin2θ . . . (7)
H = 2g

Horizontal Range (R)

It is the distance covered by projectile along horizontal between the point of projection to the point on

ground, where the projectile returns again. It is denoted by R. Obviously, range R is the horizontal

distance covered by the projectile with uniform velocity v cos θ in time equal to time of flight. Therefore,

Horizontal range = horizontal velocity × time of flight

R = v cosθ × 2v sin θ = v2 (2 sin θ cos θ)
g g

v2 sin2 θ . . . (8)
R= g

Angle of Projection of Maximum Range

For a given velocity of projection, the horizontal range will be maximum when sin2θ = 1

or, 2θ = 90°

∴ θ = 45°

Two angles projection for the same horizontal range

Horizontal range is same for angle of projection θ and (90° – θ).

For angle of projection θ, horizontal range . . . (1)
v2 sin2θ

R1 = g

Let, R2 be horizontal range for angle of projection (90 – θ). Then,

v2 sin 2(90° – θ) v2 sin (180° – 2θ)
R2 = g =
g
v
As sin (180° – 2θ) = sin2θ, we have . . . (2). v
v2 sin2θ
[Fig. 3.15, projecting angle of projectile]
R2 = g

Therefore, from equation (1) and (2) we get,

R1 = R2
Hence, the horizontal range is same for angle of projection θ and

(90° – θ) as shown in figure 3.15.

Kinematics | 59

Note:
1. Different mass of two objects are dropped from same vertical height at same time, they reach the ground

at same time by neglecting air resistance.
2. Time of flight independent on the path followed but dependent on the vertical height.

Boost for Objectives

• If a body is in uniform motion (velocity = constant) or body has acceleration and velocity acting at 0° or

180°, then path followed is straight line and motion is rectilinear motion.

Uniform motion

• Velocity is constant and acceleration is zero.

• Body covers equal displacement in equal time interval.

• Equation for uniform motion s = ut.

• Time-displacement graph is straight line inclined to time axis.

• Time-velocity graph is straight line parallel to time axis

• Time-acceleration graph is straight line coincident to time axis.

Variable motion at constant acceleration

• Velocity changes at constant rate and acceleration is constant.

• Body may cover equal distances in equal time intervals but never covers equal displacements in equal time

intervals (a and u at 180°).

Distance and Displacement

• The length of actual path followed by a moving body is called distance. It is a scalar quantity.

• The shortest distance between initial and final position with direction for a moving body is called

displacement. It is a vector quantity.

• Displacement covered by a moving body in certain time interval may be zero, negative or positive but

distance can never be zero or negative. Distance > Displacement ∴ Distance
Displacement > 1

• For a moving body the distance travelled always increases with time but displacement may increase or

decrease with time.

• A body is said to be moving with uniform speed if it covers equal distances in equal intervals of time.

• A body is said to be moving with a variable speed if it covers equal distance in unequal intervals of time or

unequal distances in equal intervals of time.

• Instantaneous speed is limiting value of average speed as time interval tends to zero.

lim ∆x dx
vi = x → 0 ∆t = dt . It is defined for a point of time.

• Average velocity for a moving body in certain time interval may be zero, +ve or –ve but average speed can

never to zero or –ve.

• Average speed
Average speed > Average velocity. Therefore, Average velocity > 1 .

• An object is said to be moving with uniform acceleration if its velocity changes by equal amounts in equal

intervals of time.

• An object is said to be moving with a variable acceleration if its velocity changes by unequal amounts in

equal intervals of time.

Motion in plane and projectile motion

• When initial velocity and acceleration on a body act on any other angle except 0° and 180° then path followed

by the body is parabolic.

• For projectile motion, the acceleration is constant but velocity of the projectile changes in both magnitude and

direction.

Variations of few quantities throughout the path of the projectile

• Velocity, speed, K.E. and momentum increase gradually.

• Potential energy gradually decreases.

• Mechanical energy remains constant.

• Acceleration remains constant.

60 | Essential Physics

• Angle between velocity and acceleration decreases gradually.
• For maximum horizontal range, range of projection should be 45°, and maximum range Rmax is u2/g.

• For least possible speed to attain certain range R, angle of projection should be 45° and least speed is umin =

gR .

• Two angles of projection to obtain same range with particular speed are θ and (90° – θ) or 45° – α and 45° +

α.

Variations of different quantities throughout the path of oblique projectile.

• Speed, Velocity, Momentum and K.E. first decreases, become minimum at highest point and then increase.

• P.E. first increases, becomes maximum at highest point and then decreases.

• Mechanical energy remains constant throughout the path.

• Acceleration always remains constant.

• Angle between acceleration and velocity gradually decreases from (90° + θ) to (90° – θ) while acceleration

and velocity are at right angle at highest point.

If air resistance is not neglected, then

• Maximum vertical height, horizontal range, speed, momentum and K.E. of a projectile at striking point are

each less than those when air resistance is neglected.

• Time of flight will be more when air resistance is neglected.

• Angle at which projectile strikes the ground will be more than angle of projection.

• In a projectile motion, horizontal component of velocity remains constant but vertical component first

decreases to zero and then increases to initial value.

• If a passenger drops a body from a uniformly moving vehicle, the trajectory would appear.

i. a straight line for the passenger. ii. a parabola to a person on ground.

Short Questions with Answers

1. Is it possible that the displacement is zero but not the distance?
Ans:
2. Yes, when a body moves from one point to another and back to its original position, the displacement is

Ans: zero. For example, a body thrown upward returns after attaining height; 'h'. Displacement of the body

3. will be zero but the distance travelled will be twice the height i.e. '2h'.
Ans:
Can a body have zero velocity and finite acceleration? [HSEB 2051]

OR

Give with an example a case where the velocity of object is zero but its acceleration is not zero.

[HSEB 2071 S]

OR

Can you have a zero acceleration but not zero velocity? Explain with the help of graph.

Yes, a body can have zero velocity and finite acceleration. For example, if a Y

body is thrown vertically upward at the highest point, the body has zero

velocity but its acceleration is equal to the acceleration due to gravity, 'g'. P Q
Similarly, in case of a body moving with simple harmonic motion, it has

V

zero velocity and maximum acceleration at the extreme point. In the given (v

- t) graph, the velocity through out the portion OP is increasing where is non-

zero acceleration where as through the region PQ, the velocity is constant or O t X

uniform and acceleration is zero.

A projectile fired at an angle of 18° has certain horizontal range. State another angle of projection

for the horizontal range. [HSEB 2052]

Let, R1 be the horizontal range for angle of projection α, . . . (1)
u2 sin2α

R1 = g

Let, R2 be the horizontal range for angle of projection (90° – α),
u2 sin 2(90° – α) u2 sin (180° – 2α)
R2 = =
g g

Kinematics | 61

u2sin 2α . . . (2)
or, R2 = g

Thus, the horizontal range is same for angle of projections α and (90° – α)

Here, α = 18°

∴ 90° – α = 90° – 18° = 72°

Hence, required another angle of projection is 72°.

4. A car is moving on the road when the rain is falling vertically downwards. Why does the from
Ans:
wind-screen get wet? [HSEB 2053]
5.
Ans: →
Suppose, rain is falling vertically downward with a velocity v and a vehicle is moving in horizontal
6.
Ans: →
direction with velocity u . To find the relative velocity of rain
7.
Ans: with respect to the vehicle Let, the resultant R is represented

8. is magnitude and direction which makes an angle θ with –u u
Ans:
9. vertical. From the figure, we have tanθ = u and R2 = u2 + v2. Therefore, rain strikes the frθont wind vehicle making an angle
Ans: v
screen

R

Give reasons why a man getting out of a moving bus must

run in the same direction for a certain distance.

It is dangerous to jump out of a moving bus. The person Rain
inside the moving bus has inertia of motion as he moves with

the velocity of the bus. When he jumps out of the moving bus, his feet come to rest when they touch the

ground but, the upper portion of his body moves forward in the direction of moving of bus due to inertia

of motion. Therefore, the man falls in the direction of motion of the bus. So, a man getting out of a

moving bus must run in the same direction for a certain distance.

Can a body have a constant velocity but a varying speed?

The velocity of a body is the rate of change of displacement with time and the speed is the rate of

covering the distance of a particle with time. Hence, as displacement has both magnitude and direction,

the velocity also has the same. But, as the distance is a scalar quantity, the speed is also the scalar

quantity. Hence, if the speed varies, its velocity also varies. As a result, it is impossible to remain the

constant velocity with varying speed.

Can an object have eastward velocity while experiencing a westward acceleration? [HSEB 2056]

Yes, on the application of brakes on a moving vehicle, the direction of acceleration of it is opposite to

that of its motion. Similarly, for a body executing, simple harmonic motion if we consider moment

when the body is moving towards east from its mean position, the direction of its velocity will be

eastward since the direction of acceleration is always directed towards the mean position; i.e. towards

westward. Hence, the object can gave eastward velocity and the westward acceleration.

Can the direction of velocity of an object change keeping the acceleration constant? [HSEB 2069]

Yes, it is possible in case of projectile motion. When a body is projected upwards, its direction of

velocity changes after reaching at the highest point of the path but acceleration due to gravity 'g'

remains constant.

The acceleration of falling body is measured in an elevator travelling at a constant speed of 9.8

ms–1. What result is obtained? [HSEB 2050]

Here, speed of elevator, V0 = 9.8ms–1. If v is the speed of the falling body with respect to ground, V0 is

the velocity of the moving elevator with respect to ground and v' is the velocity of the body with respect

to moving frame, then,

v = v' + v0 [ v = u + gt = 0 + gt = gt]
or, v' = v – v0
or, v' = gt – 9.8

dv'
or, dt = g

Hence, the acceleration of the falling body measured from a uniformly moving elevator is the same as

measured from the ground.

62 | Essential Physics

10. A small ball and another big ball are dropped from the same height. Which one will reach the

Ans: ground first? Or two stones P and Q of different masses m and 2m respectively are dropped

11. simultaneously from the top of a tower and reach the ground with different energies. Which one
Ans:
12. is faster? [NEB 2074]
Ans:
13. The different masses of the bodies are falling from the same hight at a same time. Both will reach the
Ans:
14. 2h
Ans: ground at the same time because time of descend, t = g is falling body.here h and g are

15. independent with mass.
Ans:
Can an object with constant acceleration reverse its direction? [Model Question]

Yes, when a body is thrown vertically upward, it is moving with constant acceleration. But after

reaching certain height, it can reverse its direction.

A projectile moves in a parabolic path without air resistances. Is there any point at which

acceleration is parallel to velocity? [Model Question]

No, there is not any point at which acceleration is parallel to velocity. This is because acceleration acts

in vertical direction and velocity in horizontal direction.

Can a body have a constant speed but changing velocity? Explain with example. [HSEB 2068]

Yes, when body is moving in a circular path. In circular path the speed is at each point is constant but

velocity is changing at each point

Can a body be considered to be at rest and motion at the same time? [HSEB 2060]

This is definitely possible. The feeling of rest and motion depends upon the condition of the observer.

For one observer, a body might be in motion and for some, it might be at rest, but if the observer

himself/herself is in motion with same velocity as the body being observed, the body seems to be rest.

Similarly, even if the body is at rest for one person, it might seem to be in motion for the other.

A swimmer wants to reach to a point just opposite on the other bank of the river. How should he

swim and why? [HSEB 2065] B river (vr)

Let us consider, river is flowing with velocity vr as shown in figure.
A swimmer wants to reach to a point B just opposite on the other

bank of the river from the point A. Then he has to move with the

direction making an angle θ with the AB as shown in figure. θ

16. A ball is projected horizontally from the top of a building and A

another is dropped gently from the same point at the same time, which one will hit the ground

first? [HSEB 2057, 2062]

OR

Two stones are projected simultaneously from a height, one falls freely while the other is

projected horizontally which one reaches the ground first? Explain. [HSEB 2059]

Ans: Both will hit the ground at the same time. Because, time of descent of a freely falling body or a

2h
horizontally launched projectile is T = g . It means, the time of descent depends only on the height
of fall and the acceleration due to gravity and nothing else. So, both of them will reach the ground at
the same time.

17. If initial velocity of a projectile is doubled what happens to its maximum range?
OR

What would be the effect on the maximum range in doubling the velocity of a projectile?
[HSEB 2058]

Kinematics | 63

Ans: The horizontal range 'R' for body projected with velocity 'u' making an

v2sin 2θ y
g.
angle 'θ' with horizontal is, R = If the velocity of firing is

doubled, Vn = 2v. So, other quantities remaining constant. The new range
Vn2sin2θ (2V)2 sin 2θ 4V2 sin2θ
will be, Rn = g = g= g= 4R. So, the u

new range will be four times the old one. g

18. O x
Ans:
A bomb is to be dropped from a moving helicopter on a target on the

ground. Explain how it can hit the target. [HSEB 2071]

When the bomb is dropped from helicopter, it moves in parabolic path as shown in figure. So, to hit the

target by bomb it should drop before reaching above the target as shown in figure.

19. Find the angle of projection at which the horizontal range and the maximum height of a projectile
Ans:
are equal. [HSEB 2061]
20.
Ans: The horizontal range and the vertical height for the angle of projection 'θ' and the velocity of projection

21. V2sin2θ V2sin2θ
Ans: 'v' are given by , H = 2g , and R = g

According to the given condition,

∴ V2sin2θ V2sin2θ
2g = g

or, V2sin2θ V22sinθcosθ
2g =
g

or, sinθ = 2 cosθ
2

or, sinθ = 4
cosθ

or, tanθ = 4
or, θ = tan–1(4)

∴ θ = 75.96°

Hence, the required angle of projection is 75.96°.

Because of air resistance, two objects of unequal mass do not fall at precisely the same rate. If two

bodies of identical shape but unequal mass be dropped simultaneously from the same height,

which one reaches the ground first? [HSEB 2050]

As the bodies have identical shape, they would displace equal volume of air. Hence, the buoyant force

(B) on bodies due to the upthrust of air is same. Let, m1 and m2(m1 > m2) be masses of two bodies.
For first body, we have

m1 g – B = m1a1 . . . (1)
B

or, a1 = g – m1

For second body, we have

m2g – B = m2a2 . . . (2) Since m1 > m2.
B

or, a2 = g – m2

From (1) and (2) we conclude, that a1 > a2 .
Hence, the body having a large mass falls more rapidly.

An athlete runs some distance before taking long jump. Why?

The magnitude of long jump is equal to the range obtained by the person during the jump, given by, R =
v2sin2θ

g . If the jump is to be longer, higher range should be obtained, which requires that the velocity of

launching 'v' should be high if a person jumps simply, the only velocity obtained is the vertical velocity

64 | Essential Physics

'Vv'. If he/she runs some distance, there will be some horizontal velocity 'Vh' as well. This will give the
resultant velocity as VR2 = Vh2 + Vv2 ∴ VR = Vh2 + Vv2 .
This shows that, VR > Vh, i.e. the person can get higher launching speed and make the range high.

22. A projectile moves in a parabolic path without air resistance. Is there y
Ans: O
any point at which its acceleration is perpendicular to the velocity?

Explain. [HSEB 2070]

When a projectile moves in a parabolic path without air resistance, at the

maximum height velocity and acceleration are perpendicular to each u
x
other as shown figure, where velocity is horizontal and acceleration is g

vertical.

Numerical Examples

1. An object is dropped from the top of the tower of height 156.8 m and at the same time another object
is thrown vertically upward with the velocity of 78.4ms–1 from the foot of the tower, when and where

the two objects meet? [HESB 2063]

Solution:

Here, Height of the tower (h) = 156.8m

Let, there are two objects, one dropped from the top of the tower and another thrown vertically upward

from the foot of the tower meet at a distance x from the top after time t.

S = ut + 1 at2 Top
2

or, x = 1 gt2 . . . (1). Since u = 0.
2

For the projected body, we have, x

S = ut + 1 at2 h
2

or, h – x = 78.4t + –21gt2 . . . (2)

Substituting value of x from equation (1) in equation (2) we get,

h – 1 gt2 = 78.4t – 1 gt2
2 2

or, 156.8 = 78.4t

∴ t = 2 seconds. ground

From (1) we get, x = 1 10 × 22 = 20m
2

Hence, they meet at a distance of 20m from the top and after 2 seconds.

2. A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m.

Find the height of tower. [HSEB 2069]

Solution:

Distance covered in last second = 25m. We know that,

1
or, u + 2 g(2t – 1) = 25 [since, u = 0]

50
or, 2t – 1 = 10

or, 2t = 6

∴ t = 3 seconds

Now, we know that, h = ut + 1 gt2 = 1 × 10 × 32 [ since,u = 0]
2 2

∴ h = 45m. Hence, height of the tower is 45m.

Kinematics | 65

3. A man wishes to swim across a river 600m wide. If he can swim at the rate of 4 km/h in still water

and the river flows at 2 km/h. Then in what direction must he swim to reach a point exactly opposite

to the starting point and when will he reach it? [HSEB 2069]

Solution: M v MW
Here, width of river (d) = 600m = 0.6km

Velocity of man with respect to water, VMW = 4 km/hr
represented by O→M

Velocity of water with respect to ground VWG = 2 km/hr O
→ θ

represented by ON . N v WG

Let, the man should swim making angle θ with the P v MG

water current to reach the exactly opposite end.

→→ → →
Then, V MW + V WG = V MG , represented by OP that

points exactly opposite end.

From the figure. In ∆OMP,
OM2 = OP2 + PM2

or, 42 = OP2 + ON2 [ PM = ON] 600 m

or, OP2 = 42 – 22
or, OP2 = 12

or, OP = 2 3 km/hr.

Then, ⊇MOP = tan–1POMP  = tan–12 2  = tan–1  1 = 30°
3  3

Then, θ = ⊇ PON + ⊇MOP = 90° + 30° = 120°

So, the man should swim at 120° with water current. Also, the time to cross the river,

d 0.6
t = VMG = 2 = 0.173 = 0.173 × 60min = 10.39 minutes.
3

4. A projectile is fired from ground level with a velocity 600 m/s at 30° to the horizontal. Find the
horizontal range, the greater vertical height to which it rises and the time to reach the greatest

height. [like NEB 2074]

Solution: Y
Velocity (v) = 600ms–1

Angle of projection with horizontal (θ) = 30° v sin θ
v = 600 ms–1
For horizontal range we have,
v2sin2θ (600)2 × sin(2 × 30)

R = g = 10 = 31176.91m

For maximum vertical height, 30° H X
v2sin2θ (600)2 × sin2 300 O R

R = 2g = 2 × 10 =4500m

Now, time taken for reaching the maximum vertical height, T' = T = 2vsinθ = vsinθ . Where, T is time
2 2g g

taken by projectile to reach the ground. (Time of flight)

Then, T' = 600 × sinθ = 30 seconds
10

∴ T' = 30 seconds.

Hence, the time to reach greatest height is 30 seconds.

V2sin2θ
Again, greatest vertical height H = 2g

∴ (600)2sin230°
H = 2 × 10 = 4500m

Hence, maximum height attained is 4500m.

66 | Essential Physics

5. A projectile is fired from the ground level with a velocity 600m/s at 30° to the horizon. Find its

horizontal range. What is the least with which it could be projected in order to achieve the same

horizontal range? (g = 10m/s2) [HSEB like 2065, 2068]

Solution: Y

Given, velocity (v) = 600 m/s; Angle with horizontal (θ) = 30° v

Horizontal range (R) = ?

We have, horizontal range, 30° R X
v2 sin 2θ (600)2 × sin(2 × 30) 360000 3 O

R = g = 10 = 20 = 31176.91m

Hence, the horizontal range is 31176.91m.

Now, for the maximum horizontal range, θ should be 45°. If Vl is the least speed to project the

substance to the same horizontal range,

R = Vl2 sin2θ
g

or, 31176.91 = Vl × sin 90°
10

or, Vl2 = 311769.1
∴ Vl = 588.36 ms–1

Hence, the least speed for the same horizontal range is 558.36m/s.

6 . A body is projected horizontally from the top of a tower 100m high with a velocity of 9.8ms–1. Find

the velocity with which it hits the ground. [HSEB 2060]

Solution:

Given, Height from which body is projected, H = 100m
Horizontal velocity, u = 9.8 ms–1

We know, v = u + at

∴ vx = ux + axt

(Since, ax = 0 i.e. no force acts in the horizontal direction then no change in velocity during the motion).
Thus, vx = ux = u = 9.8 ms–1

Now, vy = uy + ayt u

∴ vy = gt [since, uy = 0]

For vertical motion:

We have, H = uy + 1 gt2
2

t2 = 2H [since, uy = 0]
g
Vx
or, t2 = 2 × 100 θ
9.8

∴ t = 4.47 second V
Then, vy = gt = 10 × 4.47 = 44.7 ms–1 Vy
We know, v2 = vx2 + vy2 = (9.8)2 + (44.7)2
∴ v = 45.78 ms–1

If it makes θ with horizontal, then,

tanθ = vy
vx

∴ θ = tan–1  vvyx = 77.64°.

Hence, the body hits the ground with velocity of 45.78 ms–1 making an angle of 77.640 with the

horizontal (ground).

Kinematics | 67

7. A base ball is thrown towards a player with an initial speed of 20ms–1 and 45° angle with horizontal.

At the moment, the ball is thrown, the player is 50m from the thrower. At what speed and in what

direction must he run to catch the ball at the same height at which it is released? [HSEB 2069]

Solution: Y

Given, Velocity of ball (v) = 20 ms–1

We have, horizontal range, v
v2 sin 2 θ 202 × sin 900

R = g = 10

∴ R = 40m

Distance from the thrower to player = 50m 45°

Distance run to catch the ball = (50 – 40)m = 10m O 40 m 10 m X

Now, time of flight (T) = 2 v sinθ = 2 × 20 × sin45° = 2.83 seconds
g 10

The player has to travel 10m in 2.83 seconds.

Now, speed of the player (V) = d = 10 = 3.5 ms–1
t 2.83

Hence, the speed of the player is 3.5 ms–1 towards the thrower.

8. A car travelling with a speed of 15m/s is braked and it slows down with uniform retardation. It covers

a distance of 88m as its velocity reduces to 7m/s. If the car continue to slow down with the same rate,

after what further distance will it be brought to rest? [HSEB 2070 S]

Solution:

Given, Case I : Case II :

Initial velocity (u) = 15m/sec Initial velocity (u) = 7m/sec
Distance covered (s) = 88m Retardation (a) = 1m/sec2

Final velocity (v) = 7m/sec Final velocity (v) = 0m/sec

Retardation (a) = ? Distance covered (S) = ?

We have, the relation Again we have,
v2 = u2 + 2aS
v2 = u2 + 2aS v2 -u2 02 - 72
v2 - u2 72 152
or, a = 2s = 2 – 88 = -1m/sec2 or, s = 2a = 2 × (-1) = 24.5m
×

9. A vehicle having a mass of 500 kg moving with a speed of 10m/s. Sand is dropped into it at the rate of

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

Solution:

Given, Mass of a vehicle (m) = 500kg;

Velocity (v) = 10 m/s

We know, the rate of sand dropped ddmt  = 10kg/min = 10 kg s–1 = 1 kgs–1
60 6

Force required (F) = ?

By Newton's second law of motion, we get,

dm(v – u)
F = dt .Since, Speed is uniform,

dm 1
or, F = dt (v – u) = 6 (10 – 0)

∴ F = 1.67N

Hence, the required force to keep the vehicle moving with uniform speed in 1.67N.
10. A projectile is launched with an initial velocity of 30 ms–1 at an angle of 600 above the horizontal.

Calculate the magnitude and direction of its velocity 5 second after launch.

Solution:

Given initial velocity (u) = 30 m/s; Angle of projection (θ) = 60°

Time taken (t) = 5 sec;

68 | Essential Physics

Velocity (v) = ? Y

Now, after 5 second,

Vx = u cosθ = 30 × cos60° = 15 m/s θ
Vy = uy – gt

= u sinθ – gt = 30 sin 60° – 10 × 5 = – 24 m/s

Again, V = Vx2 + Vy2 = 152 + 242 vx = u cos θ
∴ V = 28.3 m/s

Also, tan θ = Vy = 15 = – 5 60° = θ X
Vx –24 8 O ux = u cos θ

∴ θ = tan–1 VVxy  = tan–1 –58 = 57°.

Hence, the magnitude and direction of projectile is 28.3 m/s and 57° respectively.

11. A swimmer's speed along the river (down stream) is 20 km/hr and can swim up – stream at 8 km/hr.

Calculate the velocity of the stream and the swimmer's possible speed in still water. [HSEB 2069]

Solution:

Speed of swimmer in downstream (VR) = 20km/hr; Speed of swimmer in upstream (V'R) = 8 km/hr
Let, Vs = velocity of swimmer and Vr be the velocity of river with respect to still water. Then,
For downstream, the resultant velocities,

VR = Vs + Vr . . . (1)
or, 20 = Vs + Vr
For upward stream, the resultant velocity is,

V'R = Vs – Vr . . . (2)
or, 8 = Vs – Vr
Adding (1) and (2) we get,

28 = 2 Vs
or, Vs = 14 km/hr and from (1), Vr = 6 km/hr
Hence, the velocity of swimmer in stream and velocity of swimmer in still water are 14km/hr and

6km/hr respectively.

12. A bullet is fired with a velocity of 100m/s from the ground at an angle of 60° with the horizontal.

Calculate the horizontal range covered by the bullet. Also, calculate the maximum height attained

[HSEB 2071]

Solution:

Given, Initial velocity of projectile (u) = 100m/s; Angle (θ) = 60° with horizontal.

Horizontal range (R) = ?; Maximum height (H) = ?

u2 sin2θ (100)2 × sin 1200
We have, R = g =
10

∴ R = 866 m

Again, maximum height

u2 sin2θ (100)2 × (sin 60°)2
H = 2g =
2 × 10

∴ H = 375 m.

Hence, Horizontal range is 866m and maximum height is 375m

13. An airplane is flying with a velocity of 90.0 m/s at an angle of 23.0° above the horizontal. When the
plane is 114m directly above a dog that is standing on level ground, a suitcase drops out luggage
compartment. How far from the dog will the suitcase land? You can ignore air resistance.
[HSEB 2072]

Solution:
Given, velocity (v) = 90.0m/s; Angle (θ) = 23.0°
Height of plane (h) = 114m; Distance of suitcase from dog (R) = ?

Kinematics | 69

We have a relation,

1 at2
h = ut + 2

or, 114 = –v sinθt + 1 gt2
2

or, 1 × 10 × t2
114 = –90 sin23°t + 2

or, 114 = – 76.16t + 5t2

or, 5t2 – 76.16t – 114 = 0

Solving this equation we get, t = 9.4,2

But, time (t) cannot be negative, so t = 9.4 sec

Now, R = vxt = vx = v cosθ × t = vxcos23° × t = 90 × cos23 × 9.4 = 778.7m from the dog.
Hence, the suitcase land 778.7 m far from the dog.

Important Numerical Problems

1. A person travelling towards east at the rate of 2ms–1. Finds that the wind seems to blow directly from

the north, on doubling his speed, it appears to come from the north east. Find the magnitude and

direction of the wind.

Solution: From figure (i) N
Velocity of person (P') = –2 ms–1; Velocity of wind (Q) = V(suppose)
p' = 2 m/s 2 m/s
Angle between resultant and speed of person (β) = 90° W
E

Then, Tanβ = Q sinθ Q =V
P' + Qcosθ

or, Tan 90° = 2 V sinθ S
+ Vcosθ Fig (i)

or, 1 = 2 V sinθ
0 + Vcosθ

or, 2 + Vcosθ = 0 . . . (1) p' = 4 m/s 4 m/s
b =45° E
or, Vcosθ = –2
Q =V
From figure (ii)
Velocity of person, (P]) = 4ms–1

Let, velocity of wind (Q) = V

Angle between resultant and speed of person (β) = 45°

Then, Tanβ = Vsinθ Fig (ii)
+ Vcosθ
4

or, 4 + Vcosθ = V sinθ . . . (2)
or, 4 – 2 = Vsinθ

V sinθ = 2
Solving (1) and (2), we get,

Tanθ = –1
∴ θ = 135°
Squaring and adding,

V2 cos2θ + V2 sin2θ = (–2)2 + 22
or, V2 = 8

∴ V=2 2
Hence, the magnitude of wind is 2 2 ms–1 towards south - east.
2. A stone on the edge of a vertical cliff is kicked to that its initial velocity is 9 ms–1 horizontally. If the

cliff is zoom height, calculate (i) time taken by stone to reach the ground.

(ii) How far from cliff the stone will hit the ground? [T.U. 2058,2065]

70 | Essential Physics

Solution:

Given, Height (h) = 200m

Velocity of projectile (v) = 4 m/s t
The time taken by the projectile to reach the ground is given by, h

2h 2 × 200
T= g = 10 = 6.3 sec.

Also, the distance of the stone and the foot of the cliff when it hits the ground of given byR

x = velocity × time taken

∴ x = 40 × 6.32 = 25.30m.

Hence, time taken by the projectile to reach the ground and the distance of the stone and foot of the cliff

are 6.3 sec and 22.30 m respectively.
3. A projectile is fired with a velocity to 320 ms–1 at an angle of 30° to the horizontal. Find (i) the time to

reach its greatest height (ii) its horizontal range. (iii) with the same velocity what is the maximum

possible range?

Solution:
Here, Velocity of projection (u) = 320 ms–1; Angle of projection, (θ) = 30°

Time to reach greatest height (t) = ? ; Horizontal range (R) = ?

(i) We know, t = u sinθ = 320 × sin30° = 16 second.
g 10

Hence, the time to reach greatest height is 16 second.
u2 sin2θ (320)2sin2 × 300

(ii) Horizontal range, R = g = 10 = 8868m

(iii) The maximum possible range will be attained by the ball is thrown at an angle θ1 = 45° with the

horizontal.

Then, maximum possible range = u2sin2θ1 = 10240m
g

4. A stone is projected horizontally with 20m/s from top of a tall building. Calculate is position and

velocity after 3 sec neglecting the air resistance. u
Solution:

Given, initial velocity, u = 20m/sec

Time taken, t = 3 sec y
Let, the stone be at point P.

Now, P
x vx
vx = u = 20m/sec
vy = uy + gt = 0 + 10 × 3 = 30m/sec v
v = vx2 + vy2 = 202 + 302
Vertical distance, y is given by

y = uyt + 21gt2 = 0 + 1 × 10 × 32 = 45m vy
2

Horizontal distance x = u.t = 20 × 3 = 60m.

Angle is given by, θ = tan-1vvxy = tan-1 3200 = 56.3°

So, the particle lies 45m below the top of tower with an angle 56° with horizontal.

Practice Short Questions

1. Write the differences between speed and velocity.
2. Distinguish between average velocity and Instantaneous velocity. If the velocity does not change from instant

to instant, will the average velocities be different for the different interval?
3. If the displacement of a body is proportional to square of time, state whether the body is moving with uniform

velocity or uniform acceleration.
4. Can you have zero acceleration but non-zero velocity? Explain with help of a graph.
5. Can the direction of the velocity of a body changed when its acceleration is constant?

Kinematics | 71

6. Can a body have constant speed but changing velocity?
7. Can an object with constant acceleration reverse its direction? Explain.
8. Rain drops hitting the side windows of a car in motion often leave diagonal streaks. Why?
9. A ball is dropped gently from the top of a tower and another bull is thrown horizontally at the same time

which ball hit the ground earlier? Explain.
10. A projectile fired at an angle of 18° has certain horizontal range. State another angle of projection for the

same horizontal range.
11. What would be the effect on maximum range in doubling the initial velocity of a projectile.
12. A bomb is to be dropped from a moving helicopter on a target on the ground. Explain how it can hit the

target.
13. A projectile moves in a parabolic path without air resistance. Is there any point at which its acceleration

perpendicular to the velocity?
14. A swimmer wants to reach to a point just oppositive on the other bank of the river. How should be swim and

why?
15. From a higher tower, one ball is dropped first rest and second ball is simultaneous horizontally. Neglecting air

resistance,

Practice Long Questions

1. What is relative velocity? Explain with example.

2. Define projectile projected horizontally from the top of a tower is parabolic. Calculate the time of flight and

horizontal range travelled by the projectile.

3. Show that the path followed by a projectile fired at an angle of θ with a horizontal is parabola and derive

general expression for the time of flight and horizontal range. [HSEB 2069S]

4. Calculate the horizontal range and maximum height travelled by the projectile when fired at an angle of θ

with a horizontal. [HSEB 2067S]

5. A stone is projected with a velocity in a direction making an angle θ with horizontal. Derive expression for (i)

maximum height (ii) time of flight (iii) Horizontal range. Also find the condition for maximum horizontal

range. [HSEB 2064]

6. A projectile is projected with a velocity V making an angle θ with the horizontal. Derive relation for its

trajectory also show the components of velocity at any point on the trajectory in a diagram [NEB 2074]

Practice Numerical Questions

1. A subway train starts from rest at a station and accelerates at a rate of 1.60 m/s2 for 14.0S. It runs at constant
speed for 70.0S and slows down at a rate of 3.50 m/s2 until it stops at next station. Find the total distance

covered. [Ref. 1] [Ans: 156.8 m, 1568m, 1796.48m]

2. A jet plane is flying at a constant altitude. At time t1 = 0 it has components of velocity Vx = 90 m/s, Vy =

110m/s, Vy = 110m/s. At time t2 = 30.0s, the components are Vx = 170m/s, Vy = 40m/s. For this time interval,

calculate (a) the components of the average acceleration (b) the magnitude ad direction of average
[Ref. 1] [Ans: (a) –8.67ms–2, –2.33ms–2 (b) 8.98ms–2, 195°]
acceleration.

3. A military helicopter on a training mission is flying horizontally at a speed of 60.0m/s and accidentally drops

a bomb (fortunately not an arm) at an elevation of 300 m. You can ignore air resistance. (a) How much time

is required for the bomb to reach the earth? (b) How far doe sit travel horizontally while falling? (c) Find the

horizontal and vertical components of its velocity just before it strikes the earth. (d) If the velocity of the

helicopter remains constant, where is the helicopter when bomb hits the ground?
[Ref. 1] [Ans: 7.825, 469m, 76.6ms–1]

4. An airplane is flying with a velocity of 90m/s at an angle of 23° above the horizontal. When the plane is

114m directly above a dog that is standing on ground level, a suitcase drops out of the luggage compartments,

How far from the dog will the suitcase land? You can ignore air resistance. [Ref. 1][Ans: 795m]

5. A tennis ball rolls of edge of table top 0.76m above the floor and strikes the floor at a point 1.40m

horizontally from the edge of the table. You can ignore air resistance. (a) Find the time of flight (b) Find the

magnitude of initial velocity (c) Find the magnitude and direction of the velocity of the ball just before is
[Ans: 0.785, 4.2ms––1, 7.7 ms–1 = 8.8ms–1
strikes the floor.

6. An airplane pilot wishes to fly due west. A wind of velocity 8000 km/hr is blowing towards the south.

(a) If the air speeds of the plane (its speed in still air is 32 km/hr, in which direction should the pilot

head? (b) What is the speed of the plane over the ground? [Ref. 1] [Ans: 14.5°, 310 km/s]
7. A car moving with a velocity of 10ms–1 accelerates uniformly at 1ms–2 until it reaches a velocity of 15ms–1.

Calculate (a) the time taken (b) the distance travelled during the acceleration (c) the velocity reaches 100m

from the place where the acceleration began. [Ref. 2] [Ans: (a) 5r (b) 62.5m (c) 17.3 m/s]

72 | Essential Physics

8. A ball is thrown vertically upward with an initial speed of 20ms–1. Calculate (a) the time taken to return to the

thrower, (b) the maximum height reached. [g = 10m/s] [Ref. 2][Ans: (a) 4 r (b) 20m]

9. A projectile is fired with a velocity of 320ms–1 at an angle of 30° to the horizontal. Find (a) the time to reach

its greatest height (b) its horizontal range (c) with the same velocity, what is the maximum possible range?

[Ref. 2] [Ans: (a) 16rr (b) 8868m (c) 10240m]

10. A small smooth object slides from rest down a smooth inclined plane inclined at 30° to the horizontal. What

is (a) the acceleration down the plane (b) to reach the bottom if the plane is 5m long? The object is now

thrown up the plane with an initial velocity of 15m/s. (c) How long does the object take to come to rest? (d)
[Ref. 2] [Ans: (a) 5m/s2 (b) 1.41 r (c) 3s (d) 12.5m
How far up the plane has the object then travelled?

11. Two ships A and B are 4km apart. A is due west of B. If A moves with a uniform velocity of 8kmr–1 due east
and B moves with a uniform velocity of 6 kmhr–1 due south, calculate (a) the magnitude of the velocity of A

relative to B, and (b) the closest distance apart of A and B. [Ref. 2][Ans: (a) 36.87° (b) 2.4km]

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

1. An aeroplane files 400m north and 300m east and then files 1200m upwards then net displacement is

a. 1200m b. 1300m c. 1400m d. 1500m

2. A person turns as right by 90° after travelling each 20m in st. line. What is his maximum displacement
after three successive turns?

a. 20 5 m b. 40 5 m c. 40 2 m d. 80m

3. The length of a second hand in a watch is l cm. The change in velocity in 15 sec is

a. π cm/s b. π cm/s c. 0 d. π 2
30 2 30 30 cm/s

4. A car travelling along a circular track at constant speed 20m/s has completed half revolution on the track.
Its average velocity will be

a. 10πm/s 20 c. 20m/s 40
b. π m/s d. π ms

5. A vehicle climbs a hill at a speed of 40km/hr and returns to same place at speed 60 km/hr. What is the
average speed for whole journey?

a. 48km\hr b. 50km\hr c. 32km\hr d. 52km\hr

6. A particle is thrown vertically upward with speed 100m/s.The time to reach the back on earth

a. 10s b. 20s c. 15s d. 5c

7. A vehicle moving along a straight road travels for half time at 40 km/hr and then travels for half time at
60km/hr the average velocity for entire journey is

a. 50 km/hr b. 48 km/hr c. 54 km/hr d. 45 km/hr

8. If the displacement of a body is proportional to square of time. Then, body has

a. Constant velocity b. constant acceleration

c. Increasing acceleration d. decreasing acceleration

9. If the displacement for a particle moving along x- axis at any instant of time t is given as x = t2 -20 in

meter

i) What is its average velocity for first four second of its motion?

a. 4m/s b. 6m/s c. 8m/s d. 10m/s

ii) What is its velocity at 4 sec?

a. 6m/s b. 4m/s c. 8m/s d. 10m/s

10. The displacement of the particle is given as x= 6 t3– 3t2 +4t +5. The motion of the particle is

a. Constant (uniform) b. Uniformly accelerated

c. Decelerated d. Accelerated

11. The instantaneous acceleration of a particle at any instant of time t is a = 2t (in m/s2) which starts from

rest from origin, what is its velocity at 5 sec?

a. 10m/s b. 25m/s c. 5m/s d. 30m/s

Kinematics | 73

12. A body moving initially due east at velocity 10m/s has constant acceleration 4m/s due west for 4sec. What
is distance covered by the body in 4 sec?

a. 8m b. 12m c. 17m d. 18m

13. A particle is moving at speed of 5m/s towards east. After 10s its velocity changes and becomes 5m/s toward
north. What is the average acceleration during this interval?

a. 1 m/s2 due north- west b. 2 m/s2 due west-east c. 1 m/s2 due east- west d. 2 m/s2 due east- west
22

14. A person is throwing ball into the air one after the others. He is throwing 2nd ball when 1st ball is at highest

point. If he is throwing two balls every second, how high do they rise?

a. 5m b. 3.19m c. 2.5m d. 1.25m

15. A ball is dropped from height of 45m. Then,

i) Distance covered by it in last second of its motion will be

a. 25m b. 15. 6m c. 16.2m d. 16.5m

16. If stone is released from top of tower, The ratio of distant cover by it in 2nd and 3rd second is

a. 3:5 b. 2:4 c. 1:2 d. 4:3

17. If stone is released from top of tower of height h, it reaches the ground in t sec. The position of stone after

t
3 sec from the ground will be

3 h 8h h
a. 4 h b. g c. g d. 4

18. A body is dropped from a height 'h' with initial velocity 0 m/s and on reaching the ground it attains a

velocity of 3m/s. If an clothes block is dropped from the same height with initial velocity 4m/s , then what is

velocity attained by the body on striking the ground ?

a. 4m/s b. 5m/s c. 6m/s d. 8m/s

19. A stone is dropped from the top of tower. If it reaches the earth in 6 sec, the height of tower is nearly

a. 200m/s b. 196m/s c. 206m/s d. 0m/s

20. A stone is dropped from a building of height 'h' and it reaches after t sec on earth .From the same building

if two stones are thrown one upwards and other downwards with same speed and they reach the earth

surface after t1 and t2 seconds respectively, then

a. t = t1 – t2 b. t = t1+t2 c. t = t1t2 d. t = t12t22
2

21. The time taken by a mass projected vertical upwards to reach the maximum height (with air resistance not

neglected) is 10 sec, the time of descent of the mass from same height will be

a. 6.2 sec b. 8 sec c. 10sec d. 12.3 sec

22. When a ball is thrown upward with air resistance not neglected it takes 10 sec to reach a height. If then

returns to ground in time.

a. 8.2sec b. 6sec c. 10sec d. 12.4secs

23. A body is thrown vertically upward attains a maximum height H while moving upwards if it covers first

3H H
4 distance from the ground in time T them time taken to cover remaining distance 4 will be

a. T T T 3T
b. 2 c. 3 d. 4

24. A stone is dropped from the top of tower of height h. After 1 second another stone is dropped from 20m

below the top. Both reach the bottom simultaneously. What is the value to?

a. 3125m b. 312.5m c. 31.25m d. 25.31m

25. A body falls freely from rest and has velocity v after it falls through a height h. The distance it has to fall

further for its velocity to become double is

a. h b. 2h c. 3h d. 4h

26. The equation of the trajectory of an oblique projectile is

y= 3 x – 1 x2 here x and y are in meter and g is ms-2? Then, the angle of projectile is
gt2

a. 0° b. 90° c. 45° d. tan–1 3

27. Two projectile are fired from the same point with same speed at angles of projection 60° and 30°

respectively. Which of the following is true?

a. Their range will be same b. their maximum height will be the same

c. Their time of flight will be same d. their landing velocity will be same the same

74 | Essential Physics

28. The range of projectile when projected at an angle θ is R. If the angle of projection is 2θ but the range

remains the same, the angle will be

a. 10° b. 20° c. 30° d. 40°

29. A projectile time of flight T is related to horizontal range by equation gT2 = 2R, The angle of projection in

degree is

a. 45° b. 90° c. 30° d. 60°

30. A ball is thrown horizontal from the top of a tower .Horizontal component of its velocity will

a. Increase b. decrease c. remains same d. first increase then decrease

31. A cricket ball is struck with K.E. of 'k' at an angle of 45° with the horizontal. The K.E. at maximum height

will be

a. 0 b. k/2 c. k d. 2k

32. R is the horizontally range for projectile at an angle of projection of 15°. For the same range another

angle of projection will be

a. 30° b. 75° c. 60° d. 45°

33. Two projectile A and B are thrown at a same angle with velocity uB = 2uA then what is the relation between
their range

a. RA= RB b. 2RA= RB c. RB = 4RA d. 2RB= RA

34. If mass and velocity of body in a projectile motion is double then lines momentum becomes.

a. 2 times b. 4 times 1 d. constant
c. 2 times

35. The maximum range of gun on a horizontal train is 16km. If g is 10m/s2 , the muzzle velocity of the shell

will be

a. 400m/s b. 200m/s c. 800m/s d. 256m/s

36. For what angle of projection with horizontal, the horizontal range of a projectile is equal to its maximum

vertical height b. tan–1(2) c. tan–1(3) d. tan–1(4)
a. tan–1 (1)

37. A projectile is projected from ground after 1 second of projection; its velocity makes an angle of 45° with

horizontal. Further one second it moves horizontally, then find angle of projection

a. tan–1(2) b. tan–1(3) c. tan–1(4) d. tan–1(1)

38. A ball is projected upwards from the top of towers with a velocity 50m/s at 30° to horizontal. If the height

of tower is 70m, after how many seconds will the ball but the ground?

a. 2s b. 5s c. 7s d. 9s

39. A ball thrown by one player reaches the other in 2 seconds. The maximum height attained by the ball

above the point of projection will be

a. 2.5m b. 5m c. 7.5m d. 10m

40. A man can throw a stone 80m away. The maximum height to which he can throw the stone is

a. 10m b. 20m c. 40m d. 80m

41. The time of flight and horizontal range of a projectile are related as g T2 = 2R, The angle of projection will

be

a. 30° b. 45° c. 60° d. 90°

42. The time of flight for a projectile is 4s. What is its maximum vertical height ? (g = 10m/s2)

a. 5m b. 10m c. 20m d. 25m

43. A stone is projected twice from the same point, each time with same speed but at different angles and the

range obtained is same. If the maximum height obtained by the stone in the two throws were 9m and 4m.

Find the range.

a. 10m b. 36m c. 6m d. 24

44. A projectile is projected with K.E. K. If it has maximum range , than its K.E. at the highest point will be

a. 6.25k b. 0.5k c. 0.75k d. k

Answer Key

1. b 2. c 3. b 4. d 5. a 6. b 7.d 8. b 9. (i)a(ii)c 10. b
19. b 20. c
11.d 12. c 13. a 14. a 15.a 16.a 17.c 18. b 29. a 30. c
39. a 40. c
21. a 22. a 23. a 24. c 25. c 26. a 27. b 28. d

31. a 32. a 33. b 34. b 35. c 36. b 37. a 38. d

41. b 42. c 43. b 44. c

Ch apter

4 Laws of Motion

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

Syllabus:

Laws of Motion-Newton's law of motion; Inertia, force, linear momentum, impulse, Conservation of linear
momentum; Free-body diagrams; Solid frictions: Laws of solid friction and their verifications; Application
of Newton's laws: Particle in equilibrium, dynamics of particle.

Objectives:

The objectives of this sub unit are to make the students be able to:
– understand Newton's law of motion and use them in practice.
– understand the law of conservation of linear momentum as the fundamental law of conservation,

and
– understand the cause and the effect of the friction.

Activities (Micro syllabus):

1. Give the concept of force. Review the concept of the state of rest and state of motion of a body.
2. State and explain Newton's first, second and third law of motion. Give concept of equilibrium.
3. Define inertia and appreciate that the mass is a measure of inertia.
4. Define impulse and establish a relation with momentum.
5. State and prove the law of conservation of linear momentum.
6. Explain the necessicity of free body diagram and use them to solve varieties of problems in

mechanics.
7. Explain, (i) sliding friction and rolling friction (ii) static and dynamic friction and (iii) the cause of

friction.
8. State and verify the laws friction.
9. Define the coefficient of friction, angle of repose and angle of friction.
10. Discuss the method of reducing friction.
11. Numerical problems: Focused numerical problems given:

(a) As exercise in the University Physics (Ref. 1) and
(b) In advanced level Physics (Ref.2) and also in Physics for XI (Ref .3).

76 | Essential Physics

4.1 Introduction

The study of motion of objects by considering the causes of motion (force and inertia) is called kinetics
(dynamics). In this chapter, we study motion by considering its cause as well. We define force as pull or
push which is responsible to change the state of a body. An externally applied force can bring a body
from rest to motion or from motion to rest.

Motion is the simplest and most important phenomenon in nature. It is easily observable in case of
macroscopic objects. In case of atomic or even nuclear level, it is not directly observable but motion is
very much there. Motion may be as simple as a stone throwing up into air or may be as complicated as the
launching of a satellite into its orbit. Whatever may be the type of motion, we have to study forces which
generate the motion. In this chapter, we study the roles of forces in causing motion.

4.2 Newton's First Law of Motion

It states "Every body in this universe continues in its state of rest or uniform motion in a straight
line unless compelled by an external force to change that state."

It follows from first law that an external force must act on a body to change in the state of rest or of
uniform motion in straight line. This leads to the following definition of force.

"Force is that push or pull on a body which changes or tends to change the state of rest or uniform

motion in a straight line."

According to this law, a ball set in motion will continue to be in uniform motion along a straight line for

ever if we can eliminate two external forces completely - force of friction and air resistance. This leads to

the following fundamental property of matter, "A material body, by itself, is unable to change its state

of rest or of uniform motion in a straight line." This property is called inertia. Therefore, Newton's

first law of motion is also called "law of inertia." There are three types of inertia.

1. Inertia of rest 2. Inertia of motion 3. Inertia of direction

1 Inertia of rest: It is that property by virtue of which a body in the state of rest tends to remain at

rest. Examples;

i. The passengers in a bus experience jerk in the backward direction when the bus suddenly

starts moving. It is because, when the bus suddenly starts moving, the lower part of our

body shares the motion but upper part tends to remain in the state of rest due to inertia of

rest. Similar arguments can be applied for the following cases.

ii. The leaves and fruits fall down when a tree is shaken.

iii. When a blanket is given sudden jerk, the dust particles fall off.

iv. A coin placed on a cardboard over a glass falls into the glass when the cardboard is given a

quick jerk.

v. When a bullet is fired into a tightly fitted glass pane from a reasonably close distance, it

makes a clear circular hole in the glass pane. It is because; the particles of glass around the

hole tend to remain at rest due to inertia of rest. Therefore, they are unable to share the fast

motion of the bullet.

2. Inertia of motion: It is that property by virtue of which a body in the state of uniform motion

tends to maintain its uniform motion.

Examples:

i. The passengers standing in a moving bus jerk forward when the bus stops suddenly. It is

because, the lower part of the body comes to rest along with the bus but the upper part of

the body remains in a state of motion due to inertia of motion. Similar arguments can be

applied for the following cases:

ii. An object thrown upward in a moving bus continues to move with the bus.

iii. An athlete runs for some distance before taking a long jump.

Laws of Motion | 77

3. Inertia of direction: It is that property of a body by virtue of which it tends to maintain its
direction.

Examples;

i. Mud sticking to a wheel flies off tangentially when the wheel rotates at a high speed.

ii. The passengers in a moving bus experience a jerk in the outward direction when the bus
suddenly takes a turn.

iii. A stone tied to one end of a string and being whirled in a horizontal circle tends to fly off
tangentially along a straight line when the string breaks.

The lighter objects require small force to change their state of rest or state of motion in comparison to
heavier ones. This suggests that the lighter objects have small inertia and heavier objects have large
inertia. Therefore, the inertial property of a matter depends on its mass and mass is the measure of
inertia.

4.3 Linear Momentum

Consider a light body A and a massive body B are at rest. Let a certain force applied on the light body A

increases its velocity from zero to a certain value v in time t. Now in order to increase the velocity of the
massive body B from zero to the same velocity v in the same time interval t, comparatively larger force

will be required. Similarly, if they are moving with equal speeds and have to be brought to rest in a given

time interval, then a greater force will be required to stop the massive body B than the light body A. This

is the reason to why the brakes of a train have to be more powerful than the brakes of a bicycle. Further, if
a tennis ball hits us at a speed of 5km h–1, we may not show much concern. But, imagine what happens if
a train hits us at a speed of 5kmh–1. Therefore, it is not only the velocity but also the mass of moving
object, which has to be taken into account while measuring the quantity of motion of a body.

The total quantity of motion possessed by a moving body is called linear momentum of the body

→

and it is measured as the product of mass and velocity of the body. It is denoted by p . The linear

→

momentum of a body having mass m and moving with velocity v is given by

→→

p = mv

→→

In magnitude, |p | = m|v |

or, p = mv.
The S.I. unit of momentum is kgms–1 and its dimensional formula is [MLT–1]. Being the product of a

→

vector (velocity) and a scalar (mass), the linear momentum is a vector quantity. p has same direction as

→

that of v .

Newton's Second Law of Motion

It 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."

→

Let, a constant external force F is applied to a body and the linear momentum of the body changes from

→
dp
→→ →
p to p + dp in time interval dt. Then, time rate of change of linear momentum = dt

According to Newton's second law,

→
dp
dt →

∝F

78 | Essential Physics

→
dp
or, → ∝ dt

F

→

→ dp
or, F = K dt

Where, K is a proportionality constant. The value of K depends on the units chosen for the measurement

of force. In both CGS and SI system of units, the units are so chosen that K = 1.

→
→ dp
→→
or, F = dt . [Since, p = mv ]

→d →
or, F = dt (mv )

Since, mass m of a body is constant in Newtonian mechanics.

→

→ dv
or, F = m dt .

→

dv →
But, dt = a = acceleration of the body.

→→

∴ F = ma

This is the mathematical form of Newton's second law of motion. The second law of motion gives a

measure of force as the product of mass and acceleration of a body. The S.I. unit of force is Newton (N)
and its dimensional formula is [MLT–2]. Force is a vector quantity.

In magnitude, F = ma
If m = 1 kg and a = 1 ms–2, then
F = 1 kg × 1ms–2 = 1kgms–2 = 1N

Therefore, the force required to produce acceleration of 1ms–2 on a body of mass 1kg is called 1 Newton.
In CGS system F = ma = 1g × 1cms–2 = 1gcms–2 = 1 dyne

Also, 1 N = 1 kg × 1ms–2 = 1000g × 100cms–2 = 105 gcms–2 = 105 dynes.

4.4 Newton's Third Law of Motion

It states that, "To every action, there is always an equal (in magnitude) and opposite (in direction)

→

reaction."Suppose the interaction (action and reaction) between two bodies A and B. Let FAB be force
→

exerted by B on A and FBA be the force exerted by A on B during the interaction. Then, from Newton's

third law of motion,

→→ FAB AB FBA

FAB = –FBA [Fig: 4.1, Newton's Third law of motion]
It is clear from this equation that these two forces of action
and reaction are equal in magnitude but opposite in direction

and always act along the line joining the centres of the two bodies. The action and reaction forces

constitute a mutual simultaneous interaction between the two bodies. Therefore, the forces always exist

in a pair because of action and reaction. A single isolated force does not exist in nature. Newton's third

law of motion suggests that forces originate as a result of interaction between the bodies and hence this

law is also called 'law of interaction'. Whenever we take a single force, we are considering one aspect of

mutual interaction.

The action and reaction forces given by Newton's third law of motion are always equal and opposite, but
they never cancel to eachother because they always act on two different bodies. The two equal and

Laws of Motion | 79

opposite forces cancel each other only if they act on same body. This law is valid when bodies are at rest
or in motion or in contact or at a distance from each other.
Examples:
i. Let a body having weight W is resting on a horizontal surface. The body exerts a force (action)

equal to its weight W on the ground and the ground exerts a reaction R on the body in upward
direction as shown in figure 4.2 (a) such that

R = W.
→→

In vector notation, R = –W .

R

RV

Hθ

W

(a) Body resting on a horizontal surface (b) Walk of a man F

[Fig. 4.2, Examples to illustrate Newton's Third Law of Motion ]

ii. Consider a man walking on a ground as shown in 4.2 (b). When the man walks, he pushes the

ground slantingly backwards with a force F (action) and the ground exerts a force R(reaction) to

→→

the person in the opposite direction such that R = F. In vector notation. R = –F

The reactions R can be resolved in to two rectangular components; horizontal component H(=

Rcosθ) and vertical component V (=Rsinθ). The horizontal component H helps the man to walk

forward and vertical component V balances the weight W of the man.

iii. When a rubber ball is struck against a ground, the ball bounces upward direction due to reaction of

the ground.

iv. During swimming, the swimmer pushes the water backwards with his hands with a force (action)

and the water offers a reaction R to the swimmer in forward direction. As a result, the swimmer is

pushed forward.

v. When a man jumps out of a boat, then the boat is pushed in the backward direction due to reaction.

4.5 Impulse of a Force F
force
While producing motion, the effectiveness of a force depends not
only upon the magnitude of the force, but also on duration of time Fav
for which the force acts. The total effect of a force is called
impulse of the force and it is measured as the product of force
and time of impact of the force. That is,

Impulse = Force × Time of impact of the force

or, Impulse = F × t . . . (1) t2 t

Also, from Newton's second law of motion, O t1

F = ma t1– t2 = ∆t
[Fig. 4.3, Impulse is the product of everage
m(v – u) v–u
or, F = t , [since,a = t ] force and time of impact of the force. ]

or, F× t = mv – mu. . . . (2)

80 | Essential Physics

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

Impulse = mv – mu . . . (3)

Therefore, the impulse may also be defined as the total momentum change produced by a force. The
impulse may be positive or negative or zero according as the momentum change is positive or negative or

zero. The impulse is a vector quantity and its direction is same as that of force.
In S.I. system, the unit of impulse is NS or kgms–1. The dimensional formula of impulse is [MLT–1].

When a variable force acts for short interval of time, then the impulse is measured as taking the product
of average force and time of impact of the force. So,

Impulse = Average force × time of impact = Fav × ∆t

A force having very short time of impact is called an impulsive force. The magnitude of impulsive force

changes with time, first zero to a maximum and then maximum to zero as shown in figure 4.3.

Applications of Impulse

These are based on the fact that the force is large if the total change in momentum takes place in a very
short interval of time and vice-versa. When two forces F1 and F2 act on a body to produce the same
impulse, then their times of application t1 and t2 must be such that

F1 × t1 = F2 × t2
Followings are some practical applications of impulse.

1. A cricket player lowers his hands while catching a fast moving cricket ball. While lowering hands,
the time of impact to catch the ball increases and the player has to apply less average force to
catch the ball. Consequently, the ball will also apply less force (reaction) on his hands and the
player will not get much hurt on his hands.

2. A person falling from a certain height receives more injuries when he falls on a cemented floor
than when he falls on a heap of sand. When the person falls on a cemented floor, then the person is
stopped abruptly and a large change in momentum takes place in a very short interval of time. So,
a large force is exerted by the floor on the person. But, when the person falls on a heap of sand,
then the person is stopped comparatively in longer time interval. The increase in time interval
reduces the force exerted by the floor on the person.

Similar explanations holds for the following cases:

3. When a moving vehicle strikes against a wall, a large amount of force acts on the vehicle.

4. Buffers are provided between the bogies of a train.

5. Automobiles are provided with spring system.

6. Striking a nail with a hammer.

7. China plates are wrapped in paper or straw pieces while packing.

4.6 Principle of Conservation of Linear Momentum

It states that, "If the vector sum of external forces acting on a system is zero, then its total linear
momentum remains conserved or constant."

The forces that the particles of the system exert on each other are called internal forces. The internal
forces can change the momenta of individual particles in the system but not the total linear momentum of
the system.

Derivation from Newton's Second law of Motion

From Newton's second law, Applied force = Rate of change of linear momentum

Laws of Motion | 81

dp
or, F = dt

In the absence of external force F = 0

or, dp ∴ P = constant
dt = 0

Therefore, in the absence of external force, the total linear momentum of the system is conserved.

Derivation from Newton's Third Law of Motion

A A A B
B B

m1 m2 m1 m2 m1 m2
u1 u2 FAB
FBA v1 v2

Before collision During collision After collision

[Fig. 4.4. One-dimensional collision]

Let two particles A and B having respective masses m1 and m2 are moving with respective velocities u1
and u2 in a straight line before collision such that u1 > u2 as shown in figure 4.4.

Let v1 and v2 be their respective velocities after collision.

Before collision, After collision,

Momentum of A = m1u1 Momentum of A = m1v1
Momentum of B = m2u2 Momentum of B = m2v2
Total momentum of the system = m1u1 + m2u2 Total momentum of the system = m1v1 + m2v2
Therefore,

Change in momentum of A = m1v1 – m1u1
Change in momentum of B = m2v2 – m2u2
Let, FAB = Force exerted by B on A during collision.
FBA = Force exerted by A on B during collision.
∆t = Small time of collision.

From Newton's third law of motion.

FAB = –FBA
change in momentum of A change in momentum of B
or, time of collision = – time of collision

or, m1v1 – m1u1 = – m2v2 – m2u2
∆t ∆t

or, m1v1 – m1u1 = –m2v2 + m2u2

∴ m1u1 + m2u2 = m1v1 + m2v2
Therefore, the total linear momentum of the system before collision is equal to the total linear momentum
of the system after collision.

The internal forces i.e. action and reaction cannot change the total momentum of a system. The momenta
of individual particles may change due to internal forces. But, the total momentum of a system can be
changed only by the external forces acting on the system. The recoil of a gun, firing of bullets by machine
gun, explosion of a bomb, the working of a rocket or a jet plane etc are based on the law of conservation
of linear momentum.

82 | Essential Physics

4.7 Newton's Second law is the Real law of Motion:

We can show first law and third law as the special cases of Newton's second law of motion. So, second
law is also called real law of motion.

First Law as a Special Case of Second Law

From Newton's second law of motion, F = ma
Here, F is the force applied on a body of mass 'm' and 'a' is the resulting acceleration. In the absence of
external force,

F=0
or, ma = 0
∴ a = 0 [ since, m ≠ 0]
This implies that a body at rest will remain at rest and a body in uniform motion in a straight line will
continue to do so in the absence of external force. This is the first law of motion. Hence, first law is
contained in the second law of motion.

Third Law as a Special Case of Second Law

Consider a collision (mutual interaction) between two bodies A and B. Let,

FBA = Force (action) exerted by A on B during interaction.

FAB = Force (reaction) exerted by B on A during interaction

dPA = Rate of change of momentum of A.
dt

dPB = Rate of change of momentum of B.
dt

From Newton's second law of motion.

FBA = dPB and FAB = dPA
dt dt

Adding, FBA + FAB = dPB + dPA
dt dt

d . . . (1)
or, FBA + FAB = dt (PB + PA)

According to Newton's second law of motion, the external force F acting on the bodies is,

d . . . (2)
F = dt (PB + PA)

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

FBA + FAB = F . . . (3)
In the absence of external force, F = 0

or, FBA + FAB = 0

∴ FAB = – FBA

This is Newton's third law of motion. Hence, third law is contained in second law of motion.

4.8 Applications of Newton's Laws of Motion

A. Apparent Weight

Consider a person having mass m is standing on a weighing machine in an elevator or a lift as
shown in figure 4.5. The true/actual weight of the person is mg which acts vertically downward
through the centre of gravity of the person. The weighing machine offers a reaction R to the
person, which is given by the reading of the weighing machine. The reaction R is the weight
experienced by the person. If the lift is in motion, then the weight experienced by the person R is
called apparent weight of the person, which may be different from the actual weight mg. Let us

Laws of Motion | 83

discuss the relation between apparent weight (R) and actual weight (mg) in the following

different situations:

v = constant
a=0

RR

v=0 a a g a>g
a=0

mg mg mg mg mg mg

(I) (II) (III) (IV) (V) (VI)

[Fig. (I) A man in the lift at rest, (II) A man moving upward at constant speed in a lift. (III) A man

moving upward in a lift at constant acceleration. (IV) A man moving downward in a lift at constant

acceleration. (V) A man in a lift falling freely. (VI) A man in a lift falling downward with an

acceleration greater than g(a > g)]

[Fig. 4.5. Apparent weight in a lift]

Case I : When the lift is at rest
In this case, the acceleration and hence the net force on the person are zero. So, from Newton's
second law of motion,

R – mg = ma
or, R – mg = 0 [since, a = 0]
∴ R = mg
So, the apparent weight (R) is equal to the actual weight (mg) of the person.
Here, R = mg
or, mg' = mg
∴ g' = g
Therefore, the effective value of acceleration due to gravity (g') is equal to actual value of g.
Case II : When the lift is moving upward or downward with uniform velocity
In this case, as velocity is uniform, a = 0 and F = 0.
Therefore, R – mg = ma
or, R – mg = 0 [since, a = 0]
∴ R = mg
Also, mg' = mg
∴ g' = g
So, this case is similar to case I.

Case III : When the lift is moving upward with uniform acceleration "a''
In this case, the resultant force F should act in upward direction, which is given by,

F = R – mg

Also, from Newton's second law of motion, F = ma

∴ ma = R – mg
or, R = mg + ma

In this case, the reaction (R) increases and person feels that his weight has increased. Therefore,
the apparent weight (R) of the person becomes more than the actual weight (mg) of the person.

Here, R = mg + ma

or, mg' = mg + ma

∴ g' = g + a
In this case, the effective value of acceleration due to gravity (g') becomes more than the actual
value of g.

84 | Essential Physics

Case IV: When the lift is moving downward with uniform acceleration "a"
In this case, the resultant force F should act in downward direction, which is given by

F = mg – R
Also, F = ma
∴ ma = mg – R
or, R = mg – ma
In this case, the reaction (R) decreases and person feels that his weight has decreased. Therefore,
the apparent weight (R) of the person becomes less than the actual weight (mg) of the person.
Here, R = mg – ma
or, mg' = mg – ma
∴ g' = g – a
In this case, the effective value of acceleration due to gravity (g') is less than the actual value of g.
Case V: Freely falling lift
If the cable supporting the lift breaks, then the lift starts to fall freely. The acceleration of the lift
will be equal to g and hence, R = mg – ma becomes,

R = m(g – a)
or, R = m (g – g) [since, a = g]
∴ R=0
In this case, the floor of the lift will not exert any reaction R to the person. The apparent weight of
the person (R) becomes zero and the person will be in a state of weightlessness.
Case VI: When the lift is moving downward with acceleration greater than g:
When a > g, then from equation, R = m(g – a), the reaction (R) will be negative. The person would
experience negative reaction (R) and the apparent weight is also negative. The person appears
moving in upward direction and will stick on the ceiling of the lift.

B. Horse and Cart Problem

Consider a horse having mass Mh is pulling a cart having mass Mc on a horizontal surface as
shown in figure 4.6. When the horse pulls the cart, then according to Newton's third law of motion,

the cart also pulls the horse with equal (in magnitude) but opposite force. Then, how does the cart

move?

Let, TCH = Forward pull of the horse on the cart.

THC = Backward pull (reaction)of RC
the cart on the horse. But, both

these forces are internal forces due Cart
to interaction (action and reaction)

within the horse and cart system.

Therefore, these internal forces THC TCH
(TCH and THC) cannot contribute
the motion of the system as a VR

whole.

The weight WC of the cart is f H
balanced by normal reaction RC of
the surface to the cart. Similarly,

the weight Wh of the horse is WC Wh
balanced by the vertical [Fig. 4.6. Horse and cart problem]
component V of the reaction R of

the surface to the horse. Now, there are only two external forces responsible for the motion of the

system as a whole;

i. horizontal component H of the reaction R of the surface to the horse, and

Laws of Motion | 85

ii. force of friction 'f' between the wheels of the cart and the ground surface.

If H > f, then the system moves forward with an acceleration 'a' under the influence of resultant

external force H – f. Here, resultant forward external force F is,

F=H–f . . . (1)

Applying Newton's second law of motion.

F = (MC + Mh) a . . .(2)

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

(MC + Mh)a = H – f

∴ H–f . . . (3)
a = Mc + Mh

4.9 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 various problems
in mechanics.

When a system consists of a number of bodies connected together by strings, ropes, rods etc then we can
consider each body at a time by taking account of all the forces acting on it. Then we can write equations
of motion for each body of the system by using Newton's laws of motion. We can solve the equations of
motion obtained for different bodies of the system to determine the values of unknown quantities. Let us
explain come important cases of free body diagrams in the following examples:

A. Problem of Block on a Smooth Horizontal R F
Surface a

1. Let a block having mass m is pulled by a horizontal force F. m
Then,
F = ma mg
F
or, a = m [Fig. 4.7. ]
Also, normal reaction R is,
R = mg

2. Let the block is pulled by applying a force F by making an R Fsinθ F
angle θ with the horizontal.Then, Fcosθ

Fcosθ = ma a
m
or, a = Fcosθ θ
m

Also, in the vertical direction,

R + Fsinθ = mg

∴ R = mg – Fsinθ mg
[Fig. 4.8. ]
In this case of pulling, the normal reaction R (also called
apparent weight) is less than the actual weight (mg) of the
body.

86 | Essential Physics

3. Let the block is pushed by applying a force F by marking an R
a
angle θ with the horizontal. Then,
m
Fcosθ = ma Fcosθ

Fcosθ θ F
or, a = m F F

Also, in the vertical direction, we have

R = mg + Fsinθ

So, in case of pushing, the normal reaction R (also called

apparent weight) is more than the actual weight (mg) of the mg

body. [Fig. 4.9. ]

During pulling, the apparent weight (R) is less R

than actual weight (mg). But, during pushing, the m m'

apparent weight (R) is more than the actual weight

(mg). Hence, it is easier to pull a body than to

push it.

4. Let the block is pulled by applying force F on a

horizontal smooth surface by the help of a string

having mass m'. Then,

F = (m + m') a

∴ F mg
a = m + m' [Fig. 4.10]

B. Problem of Two Bodies in Contact f

1.

m1 m2 m1

Ff f F

[Fig. 4.11. ] [Fig. 4.12. ]

Let two bodies having masses m1 and m2 are placed in contact with each other on a smooth
horizontal surface. Let an external force F is applied on body of mass m1 as shown in figure 4.11.
Then, common acceleration (a) of the two bodies is

F . . . (1)
a = m1 + m2

Let f be the force of contact between two bodies. Considering the free body diagram of mass m1

as shown in figure 4.12, the equation of motion of body m1 is,

F – f = m1a

or, f = F – m1a

or, f = F – m1m1 F m2 [Using equation (1)]
+

or, f = F1 – m1m+1m2

∴ f = m1m+2m2 F . . . (2)

2.

m1 m2 m2

f' f' F f' F

[Fig. 4.13. ] [Fig. 4.14. ]

Laws of Motion | 87

Let external force F is applied on body of mass m2 as shown in figure 4.13. Then, common
acceleration "a" is,

F
a = m1 + m2

Considering free body diagram of mass m2 as in figure 4.14, we have,
F – f ' = m2a

Where f ' is the force of contact between the two bodies.

or, f ' = F – m2a

or, f ' = F – m2 m1 F m2
+

or, f ' = F 1 – m1m+2m2

∴ f ' = m1m+1m2 F . . . (3)

From equations (2) and (3), we conclude that, during contact between two bodies,

Force of contact = Mass on which force is not directly applied × Applied force

Total mass of two bodies

C. Problems of Three Bodies in Contact m3

f'

m1 m2 m3 m2 f'
Ff f' f' f

f

[Fig. 4.15. ] m1
F
f

[Fig. 4.16. ]

Let three bodies of masses m1, m2 and m3 are placed in contact on a smooth frictionless surface as
shown in figure 4.15.

Let, F = Force applied on body of mass m1. . . . (1)
f = contact force between m1 and m2.
f ' = contact force between m2 and m3.
F

The common acceleration is, a = m1 + m2 + m3

Now, considering free body diagram of mass m3, f ' = m3a . . . (2)
. . . (3)
Considering free body diagram of mass m2, f – f ' = m2a . . . (4)

Considering free body diagram of a mass m1, F – f = m1a . . . (5)

From equations (1) and (2), we get, f ' = m1 m3F m3
+ m2 +

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

F – f = m1 m1 + F + 
m2 m3

88 | Essential Physics

or, f = F – m1 m1F m3
+ m2 +

or, f = F1 – m1 + m1 + m3
m2

∴ f = (m2 + m3)F . . . (6)
m1 + m2 + m3

From equations (1), (5) and (6) the common acceleration "a" and the forces of contacts "f "and

f ' can be calculated.

D. Problem of Tension in the String Connecting Two Bodies

1.

m1 T T m2 F m1 T
m2 F
T

[Fig. 4.17. ]

Let two bodies having masses m1 and m2 are connected by a string and they are placed over
smooth horizontal surface as shown in figure 4.17. Let 'T' be the tension on the string and 'a' be the

common acceleration. Now, considering the free body diagram of m1,

T = m1a . . . (1)

Also, considering the free body diagram of m2,

F – T = m2 a . . . (2)

Now, solving (1) and (2), we can find values of a and T.

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

F = (m1 + m2) a

∴ F . . . (3),
a = m1 + m2

From equations (1) and (3), we get

T = m1F . . . (4)
m1 + m2

2. Consider a body having mass m1 is placed on a smooth R
horizontal table and is connected to a string passing

over a frictionless pulley and carrying a mass m2 at its m1 T
another end as shown in figure 4.18. a

Considering free body diagram of mass m1,

T = m1a . . . (1) m1g T

Where, T = Tension on string a
and a = common acceleration m2

Again, considering free body diagram of mass m2,

m2g – T = m2a . . . (2) m2g

From equations (1) and (2), we get [Fig. 4.18. Motion on horizontal surface.]

m2g – m1a = m2 a

or, m2g = (m1 + m2)a

∴ a = m2g . . . (3)
m1 + m2

Laws of Motion | 89

From equations (1) and (3), we get

T = m1 m1m+2gm2

∴ T = mm1 1+mm2 2 g . . . (4)

3. Consider a body having mass m1 (lying on an inclined plane) is connected to one end of a string

which passes, over a frictionless pulley as shown in figure 4.19 (a). A mass m2 is connected to

other end of the string. Let 'a' be the common acceleration and 'T' be the tension on the string.

Now, considering free body diagram of mass m1.

T – m1gsinθ = m1a . . . (1)

Where θ = angle of inclination of the inclined plane with the horizontal.

Also, considering free body diagram of mass m2, . . . (2)
m2g – T = m2a

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

m2g – m1gsinθ = (m1 + m2) a

∴ a = (m2 – m1sinθ)g . . . (3)
m1 + m2

Now, from equations (2) and (3), we get

m2g – T = m2(m2 – m1sinθ)g T
m1 + m2 T

or, T = m2g – m2(m2 – m1sinθ)g a m2 a
m1 + m2 m1 m2g

or, T = m2g 1 – (m2 – m+ 1msi2nθ) θ
m1 m1g

or, T = m2gm1m+1 m+1msi2nθ θ

[Fig. 4.19. (a) Motion on an inclined plane]

∴ T = mm1 1+mm2 2 (1 + sinθ) g . . . (4)

4. Connected motion : Consider two masses m1 and m2 (m2 < m1) connected to the two ends of a
weightless and inextensible string which passes over a smooth

frictionless pulley as shown in figure 4.19 (b). Let 'a' be the

common acceleration produced in the masses and 'T' be the

tension in the string. Here, the heavier mass m1 moves with T
acceleration 'a' downwards and lighter mass m2 moves with the
same acceleration 'a' upwards.

Considering free body diagram of heavier mass m1, ... T m2 a
m1g – T = m1a m1 a m2g

(1)

Considering free body diagram of lighter mass m2,

T – m2g = m2 a . . . (2)

Solving equation (1) and (2), the tension 'T' on the string and m1g

common acceleration 'a' of the masses can be calculated. [Fig. 4.19(b). Connected motion]

Adding equations (1) and (2), we get
m1g – m2g = (m1 + m2)a

90 | Essential Physics

∴ a = mm11 – mm22 g . . . (3)
+ . . . (4)

Dividing equations (1) by (2), we get,

m1g – T = m1a
T – m2g m2a

or, m1(T – m2g) = m2(m1g – T)

or, (m1 + m2)T = 2m1m2g

∴ T = m21m+1mm22 g

E. Problem of Three Bodies Connected by Two Strings

m1 T2 T2 m2 T1 T1 m3 F m1 T2
T2 m2 T1
F
T1 m3

[Fig. 4.20 ] . . . (1)
. . . (2)
Common acceleration of the system is, . . . (3)
. . . (4)
F . . . (5)
a = m1 + m2 + m3
. . . (6)
Considering the free body diagram of m1,
T2 = m1a

Considering free body diagram of m2,
T1 – T2 = m2a

And, considering free body diagram of m3,
F – T1 = m3a

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

T2 = m1 m1F m3
+ m2 +

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

F – T1 = m3 m1 + F + m3 
m2

or, T1 = F – m1 m3F m3
+ m2 +

∴ T1 = (m1 + m2)F
m1 + m2 + m3

Laws of Motion | 91

4.10 Friction

Let, a block is projected with initial velocity v along a surface of ground. The block will finally come to
rest. This means while the block is moving it experiences an opposing force that points in a direction
opposite to its motion. This opposing force is called the force of friction. Hence, "Friction is an
opposing force which is developed between the surfaces in contact when a body actually moves or
tends to move over the surface of another body."

When the surface of one body slides over that of another, then each body exerts a frictional force on the
other. The frictional force acting on each body is in a direction opposite to its motion relative to the other
body. The frictional forces exist between surfaces even when no relative motion is actually present but
there is only a tendency for the relative motion.

4.11 Origin of Friction

We can explain origin or cause of friction on the basis of following two theories.

Classical Theory

According to this theory, the surfaces of bodies are never perfectly smooth but have irregularities called

projections and depressions. When a smooth surface is

minutely observed under a powerful microscope, it shows

irregularities. When the surfaces of two bodies come in contact, P Q
there is an interlocking of the irregular projections and R S
depressions of the two surfaces PQ and RS as shown in figure

4.21. When some external force is applied to move one body on

the surface of another body, then a force of interlock comes in

to play between the irregular projections and depressions of two [Fig. 4.21. Irregular projections of two

surfaces in contact. This force of interlock tries to resist the surfaces]

motion and acts opposite to the direction of applied force. This opposing force is called friction. Hence,

according to classical theory, the force of interlock between the irregular projections and depressions

of two surfaces in contact is the cause of friction.

Modern Theory

According to this theory, when surfaces of two bodies come in contact, then intermolecular force of
attraction, also called force of adhesion, between the molecules of two surfaces comes in to play which
tries to resist the relative motion between the bodies, This intermolecular force of attraction between
the molecules of two surfaces in contact is the cause of friction.

When the surfaces of two bodies come in contact, the actual microscopic area of contact is much less than
the apparent macroscopic area of contact. Since the actual area of contact is very small, therefore, the
pressures at contact points will be very large. Because of this large pressure, the contact points deform
plastically and many contact points actually become "cold-welded" together. The ''cold -welding"
between the molecules at the contact points creates a strong force of adhesion between the molecules of
two surfaces in contact. When one body is pulled across the surface of another body, then a frictional
opposition is associated with the breaking of these thousands of tiny cold-welding points.

92 | Essential Physics

4.12 Types of Solid Friction

A. Static friction B. Kinetic or Dynamic friction

R R

mg Applied
Fig (i) A block at rest Force
F
limiting friction
mg
Fig (ii) A block in motion

(Applied force)

(body at rest) (body at motion)

(iii) A graph between applied force and frictional force

[Fig. 4.22. Satic and kinetic friction.]

A. Static Friction

In the absence of applied force, the weight (mg) of the block is balanced by normal reaction (R) of
the ground on the block as shown in figure 4.22 (i). When no attempts are made to slide the block,
then there is no friction. When a small force is applied on the block then the block doesn't move.
This means, whenever an attempt is made to slide the block by applying a force, then an opposing
force is developed between the surfaces in contact, called force of friction. The block starts to
move only when the applied force exceeds a certain limiting value.

The force of friction, which is developed between the surfaces in contact before the actual
motion begins, is called static friction. Its value is zero when the applied force is zero and
becomes maximum when the block just begins to slide.
The static friction is a self-adjusting force because it adjusts its magnitude and direction so as to
become exactly equal and opposite to the applied force.

Limiting Friction: The maximum value of static friction at which a body just begins to slide
over the surface of another body is called limiting friction. For motion to take place, the
applied force must be greater than the limiting friction.

B. Kinetic or Dynamic Friction

The kinetic or dynamic friction comes into play when the two bodies in contact are in relative
motion. It acts opposite to the direction of instantaneous velocity. There are two types of kinetic or
dynamic friction.

Laws of Motion | 93

i. Sliding Friction: The force of friction which is developed between the surfaces in contact

when a body is sliding over the surface of another body is called sliding friction. It has a

constant value depending upon the nature of two surfaces in relative motion.

The sliding friction is always less than the limiting static friction.

ii. Rolling friction: The force of friction, which is developed between the surfaces in contact

when a body is actually rolling over the surface of another body, is called rolling friction.
The rolling body and the surfaces on which it rolls

are compressed by a small amount as shown in figure
4.23. As a result, the rolling body has to climb a hill

continuously. Beside this, the rolling body has to
detach continuously itself from the surface over

which it is rolling. This is opposed by the adhesive
forces between the molecules of two surfaces in

contact. Depression

Because of these factors, an opposing retarding force
is developed between the surfaces in contact, called [Fig. 4.23. Rolling frition on a wheel

rolling friction. For same magnitude of normal rolling over a surface.]

reaction, the rolling friction is much less than sliding friction. That is why, we prefer to

convert sliding friction into rolling friction. The ball and roller bearing on a rotating part of

a machine, make use of this principle.

4.13 Coefficient of Friction

Experimentally, it has been found that the force of limiting friction is directly proportional to normal

reaction in static friction, That is.

Limiting friction ∝ Normal reaction

or, F ∝ R

or, F = µsR . . . (1)

Where, µs is called coefficient of limiting friction for given two surfaces in constant,

∴ µs = F = Limiting friction . . . (2)
R Normal reaction

Therefore, the coefficient of limiting friction (µs) is defined as the ratio of limiting friction to normal
reaction in static friction. µs is unitless and dimensionless. The value of µs depends on the state of polish
between the two surfaces in contact. If the surfaces are smooth, then µs has small value.

When the body starts moving, then the coefficient of friction is called coefficient of kinetic friction. It is

denoted by µk. The coefficient of kinetic friction (µk) is defined as the ratio of kinetic friction to normal
reaction in kinetic friction. That is,

µk = Fk = Kinetic friction . . . (3)
R Normal reaction

Here, µk is also unitless and dimensionless.
Since, limiting friction > Kinetic friction

or, F > Fk

or, F > Fk
R R

or, µs > µk.
So, coefficient of static limiting friction is always greater than coefficient of kinetic friction for a

pair of surfaces. The actual values of µs and µk depend on the nature of two surfaces in contact. Both µs
and µk can exceed unity, although commonly they are less than one. The static friction is a self-adjusting
force but the kinetic friction is not a self-adjusting force.

94 | Essential Physics

4.14 Laws of Friction

A. Laws of limiting friction
I. The force of limiting friction is always opposite to that in which motion tends to take place.
II. It acts tangentially to the two surfaces in contact.
III. The magnitude of limiting friction is directly proportional to the normal reaction between
the two surfaces.
IV. The magnitude of limiting friction is independent of the shape or the area of the surfaces in
contact provided that the normal reaction remains the same.
V. It depends on the material, the nature of surfaces in contact and their state of polish.

B . Laws of sliding friction
I. The sliding friction opposes the applied force.
II. It has a constant value, depending upon the nature of two surfaces in relative motion.
III. The magnitude of sliding friction is directly proportional to the normal reaction.
IV. It is independent of the area of the surfaces in contact so long as the normal reaction
remains the same.
V. It doesn't depend on velocity, provided that velocity is neither too large nor too small.

4.15 Verification of Laws of Limiting Friction

B

AA BA

(i) (ii) (iii)

[Fig. 4.24. Verification of laws of limiting friction.]

Consider a wooden block A is placed on the surface of a table. It is attached to one end of a string which
passes over a frictionless pulley carrying a scale pan at the free end as shown in figure 4.24 (i). Go on
adding weights till the wooden block A just begins sliding. It is the evident that the force of friction was
opposing the motion of the block A. This verifies law I.

The force of friction F which is opposing the motion of the block A acts along the horizontal surface. This
verifies law II.

When the block A just begins sliding, the total weight added in the scale pan along with the weight of the
scale pan is equal to the limiting friction F. The weight (mg) of the block A is equal to the normal
reaction R. Now, put another block B (having same mass as that of block A) over block A as shown in
figure 2.24 (ii). The normal reaction R becomes 2mg. Adding weights on the scale pan, determine the
new value of limiting friction. It will be observed that the ratio of liming friction and normal reaction
is constant in the two cases. This verifies law III.
Now, place the blocks A and B over the table connected by a string as shown in figure 4.24 (iii). It will be
observed that the same load on the pan as before will just slide the blocks. Here, the area of contacts of
the blocks to the surface in two cases is different but the forces of limiting friction are same. This verifies
that the limiting friction is independent with the area of surfaces in contact i.e. law IV.

If the wooden block A is replaced by glass block of same weight mg, then it will be observed that the
limiting friction will be different in the two cases. This verifies that the limiting friction depends on
material, the nature of surfaces and the state of polish i.e. law V.

Laws of Motion | 95

4.16 Angle of Friction

The angle made by the resultant of the limiting friction F and normal reaction R with the normal reaction
R, is called the angle of friction.

Let a block having weight W is resting on a horizontal C A
surface as shown in figure 4.25. Let a force is applied to the
block and the block is about to move. Then, R
θ
OB = F = Limiting friction
OA = R = Normal reaction

OC = Resultant of limiting friction F and normal B F Applied force
reaction R. O

∠COA = θ = Angle made by resultant of F and R

with R, called angle of friction.

From ∆OAC, Tanθ = AC = OB
OA OA

or, Tanθ = F . . . (1) W
R [Fig. 4.25. Angle of Friection]

From definition of coefficient of limiting friction (µs),

µs = F . . . (2)
R

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

Tanθ = µs . . . (3)

So, the tangent of angle of friction is equal to the coefficient of limiting friction.

4.17 Angle of Repose or Angle of Sliding

It is the angle made by an inclined plane with the horizontal surface when a body placed over the inclined
plane just starts sliding down.

Let a block having weight mg is placed on an inclined plane OB. The R B
angle of inclination of the inclined plane OB with the horizontal F A
surface OA can be changed as shown in figure 4.26. Go on increasing
the angle of inclination very slowly till the block just begins to slide m

down the inclined plane. This particular value of angle of inclination α
is called angle of repose or angle of sliding. Let it be denoted by 'α'.

The value of 'α' depends on the material and nature of the two

surfaces in contact. mg

The weight mg of the block can be resolved into two rectangular α
components:
O

mgcosα → opposite to normal reaction R [Fig. 4.26. Angle of Repose]
and mgsinα → down the inclined plane.

The component mgcosα is balanced by normal reaction R and mgsinα component is equal to limiting
friction F. That is,

mg cosα = R . . . (1)

and mgsinα = F . . . (2)

Dividing equation (2) by equation (1), we get

Tanα = F . . . (3)
R

But, F = µs . . . (4)
R

96 | Essential Physics

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

Tanα = µs . . . (5)

Therefore, the tangent of angle of repose is equal to the coefficient of limiting friction.

Also, Tanθ = µs . . . (6)

Where θ is angle of friction. From (5) and (6), we get

Tanθ = Tanα

∴ θ=α . . . (7)

So, the angle of friction is equal to the angle of repose.

4.18 Friction is a Necessary Evil

Friction is Necessity (Advantages)

i. It will not be possible to walk without friction between ground and our feet. When the ground
becomes slippery after rain, it is made rough by spreading sand over it to increase friction.

ii. Tyres of vehicles are made rough to increase friction.
iii. The various parts of a machine are capable to rotate due to the friction between the belt and pulley.
iv. It would be impossible to write, to climb, to fix a nail, to drive etc. in the absence of friction.

Friction is an Evil (Disadvantages)

i. Wear and tear in different parts of a machine is due to friction.
ii. The friction between different parts of rotating machines produces heat and causes damage to

them.

iii. Extra power is necessary to apply to a machine to overcome the friction. Thus, the efficiency of
machine decreases due to friction.

4.19 Methods of Reducing Frictions

The loss of energy due to friction is undesirable. Following methods can be used to reduce friction:

i. Polishing: The interlocking and the irregularities between the two surfaces are minimized by
polishing. This reduces the friction and makes their life long.

ii. Lubrication: A substance (a solid or a liquid) which forms a thin layer between the two surfaces
in contact is called a lubricant. The lubricant fills the depressions present in the surfaces of contact
and hence the friction is reduced.

iii. Streamlining: When a body moves through a fluid (liquid or gas), the particles of the fluid move
behind the body in regular lines of flow called streamlines. The resistance offered by the fluid to
the body is minimum when its shape resembles with that of streamlines. Hence, the shape of fast
moving automobiles, aeroplanes, submarines etc are so designed that it resembles with the
streamline pattern and the resistance offered by the body would be minimum.

iv. Use of alloys: Alloys have low coefficients of friction. So, friction can be reduced by lining the
moving parts with alloys.

v. Avoiding moisture: The friction increases due to presence of moisture. So, moisture between the
two surfaces should be removed to reduce friction.

vi. Use of ball bearings or roller bearings: The rolling friction is much less than the sliding friction.
We can convert the sliding friction into rolling friction by using ball bearings or roller bearings in
the rotating parts of machines.

vii. By Proper choice of materials: The frictional force depends on the nature of materials of the
surfaces in contact. The friction between iron and concrete is much more then that between rubber
and concrete. So, tyres are made up of rubber but not of iron to reduce the friction.

Laws of Motion | 97

Boost for Objectives

• First law of motion gives idea of inertia. So it s also called law of inertia.

• First law gives qualitative definition of force.
• Quantity of motion contained in a body is called momentum. →p = m→v

• Momentum can be changed by changing mass or velocity or both after applying force.
• 2nd law of motion gives quantitative definition (measurement) of force.

• Third law of motion gives property of force that exist in pair.

• The action and reaction act on two different bodies and hence do not cancel each other.

• Impulse is the product of force and time. →I = ∫ → .dt = ∫→ddpt .dt = ∫d→p = → – → 
pf pi
F

Impulse = Change in momentum

• Impulse is given by area enclosed by time-force graph.

• When n bullets each of mass 'm' are fired per second with velocity v, then force required to hold the gun is F

= mnv. [IOM]

• dm
When sand is dropped on a horizontal conveyor belt at rate dt , then force required to maintain its constant

speed v is (F = ddmt v)

• When a jet of liquid (ρ = density) is escaping from a pipe (cross-sectional area = A) at velocity v, then force

needed to hold the pipe firmly and force on vertical wall due to jet is. F = dm v = Av2ρ = V2ρ
dt A

where V is the volume of liquid ejecting per second.

• When water rebounds with equal speed after impact, then F = Avρ[v – (-v)]
∴ F = 2Av2ρ

• When water jet is directed 'θ' with the horizontal on a vertical wall.
F = Av2ρcosθ (if final velocity becomes zero)
F = 2Av2ρcosθ (if collision is elastic)

• We can derive Newton's first and third laws of motion from second law. So, 2nd law is the real law of motion.

• Rocket works on the principle of conservation of linear momentum.

• When the force acting on a particle is always perpendicular to its velocity, then the path followed by the

particle is circular.

• The weight measured by the spring balance in a lift is equal to apparent weight.

• If one is marooned on a frictionless horizontal plane and cannot exert any horizontal force by pushing against

the surface. He/she can get of it by spitting or sneezing or throwing any object in opposite direction.

• If a man standing on a platform of a balance takes a quick step, then his weight first decreases and then

increases.

• Frictional force arises due to interaction force between molecules of two bodies and hence is of electrical

origin.

• Friction depends on nature of materials and nature of surfaces in contact.

• Friction is a non-conservative force.

• µr < µk < µ
i.e., Limiting friction > Kinetic friction > Rolling friction

• A ball is rolling on a horizontal surface towards east then frictional force on the ball acts towards east.

• If a ball sliding on a horizontal surface towards east then frictional force on the ball acts towards west.

• Limiting friction is directly proportional to normal reaction i.e., F ∝ R. Hence, if normal reaction is tripled,

limiting friction becomes 3 times.

• No work is done against static friction but work is done against kinetic friction.

• When applied force equal limiting friction, the body begins to slide.

• When one walks on the ground, the force of friction exerted by ground on him is in forward direction.

• It is difficult to move a cycle with brakes on because sliding friction is more than rolling friction.

98 | Essential Physics

• When a bicycle is in motion and pedaled, the force of friction exerted by ground on the two wheels is such
that it acts in the backward direction on the front wheel and in the forward direction on the rear wheel.

• When the bicycle is not pedaled, frictional force on both wheels is in backward direction.
• For a smooth surface, frictional force and hence coefficient of sliding friction are both zero.
• Fast moving vehicles are streamlined shaped to reduce friction.
• It is easier to pull a lawn roller then to push it because pulling decreases normal reaction.
• To avoid slipping while walking on ice, one should take smaller steps because of smaller friction.
• When a block sides down the plane with a constant speed, then the inclination of the plane is equal to angle of

repose (α) and µ = tanα.
• Acceleration of a body sliding in an inclined plane of inclination θ is

i. For downward motion = (sinθ –µcosθ)g
ii. For upward motion a = (sinθ + µcosθ)g
• A body of mass m is placed on the floor of a lift. If the lift falls freely and the body is pulled horizontally,
then the frictional force offered by the body is zero.

Short Questions with Answers

1. Why is it easier to pull a body than to push it? [HSEB 2072 C, 2069 S]

Fcos6θ0° F F

R

R

θ60° 6θ0°

m Fsin6θ0° m Fsin6θ0°
Ff Ff

θθ
Fcosθ
mg F

mg

Fig. Pulling Fig. Pushing

During Pulling, During Pushing
R + F cos θ = mg R = mg + F cos θ
R = mg – F cos θ

The reaction (R) gives the apparent weight of a body. During pulling, the apparent weight decreases but

during pushing, the apparent weight increases. Also, force of friction is directly proportional to normal
reaction R (i.e. apparent weight). Thus, the force of friction during pulling [F1 = µR = µ(mg – Fcosθ)]
becomes less than that during pushing [F2 = µR = µ (mg + Fcosθ)]. So, it is easier to pull a body than to
push it.

2. Why does a cricketer lower his hand while catching cricket ball? [HSEB 2072 D]
Ans:
The impulse of a force is, Impulse = F × t = change in linear momentum; which remains constant.
3. While the cricketer lower his hand, then time of impact (t) to catch the ball increases. When t increases
Ans: then for same impulse, the reactional force F decreases. Hence, the hands are not hurt severely and it
also avoids from "catch miss".

A firecracker at rest explodes, sending fragments off in all directions. Initially, the firecracker has

zero momentum, but after the explosion the fragments flying off each other have quite a lot of

momentum. Hasn't momentum been created? If not, explain why not? [HSEB 2070 B]

From linear momentum conservation principle, the total linear momentum of a system remains constant

in the absence of external force. Before explosion, the firecracker is at rest and hence its linear
momentum is zero. After explosion, the different fragments of it more in different directions. The linear
momentum is a vector quantity. Hence, the vector sum of linear momenta of these fragments in all
directions becomes zero after explosion too. Hence, there is no creation of linear momentum.

Laws of Motion | 99

4. A man drops his briefcase in an elevator but it does not fall to the floor. What can be concluded
Ans:
about the situation? [HSEB 2069 A]
5.
Ans: If a man drops his briefcase in an elevator but it does not fall to the floor, then it can be concluded that

6. the elevator is in free fall situation. In this case, both the elevator and the briefcase are in free fall
Ans: situation and they have same acceleration, g = 9.8 ms–2. Here, both the briefcase and elevator are

7. dropped from different heights with equal acceleration, so, they never meet.
Ans:
In rain, a scooter may slip on the turning of a road, why? [HSEB 2069 A]
8.
Ans: The friction between road and tyres provides the necessary centripetal force for a vehicle to take a turn
9.
Ans: on the turning of a road. In rain, the frictional force between road and tyres, is very less. This less value

10. of frictional force can't provide the sufficient centripetal force for the scooter to take a turn on the
Ans:
turning. Hence, in rainy day, the scooter may slip on the turning of a road.
11.
Ans: When a large heavy truck collides with a passenger car, the occupants of the car are more likely

to be hurt than the truck driver, why? [HSEB 2069 B, 2066]

During collision between a large heavy truck and a passenger car, the car comes at rest suddenly and the

time of impact (t) for car is short. We know, impulse = Force × time of impact. Due to small time of

impact for car, a large amount of force acts on the car, called impulsive force. Due to this reason, the

occupants of the car are more likely to be hurt than the truck driver.

Why does a heavy riffle not kick as strongly as a light rifle using the same cartridges?

[HSEB 2067 S]

Before firing, both the rifle and the cartridges are at rest and hence total linear momentum is zero. From

linear momentum conservation principle, the total linear momentum after firing is also zero. That is,

m1v1 + m2v2 = 0

or, m1v1 = – m2v2
Hence, the rifle and the cartridges (bullet) move in opposite directions after firing with equal momenta.

Since, light rifle has small mass, so it will recoil with large velocity than a heavy rifle. Hence, a heavy

rifle doesn't kick as strongly as a light rifle.

Why a man getting out of a moving bus must run in the same direction for a certain distance?

[HSEB 2065]

Due to inertia of motion, the man getting out of a moving bus tends to move along the direction of

moving bus. The feet of the man comes to rest but his upper part is still in motion due to inertia of

motion. Hence, the man must run in the direction of bus for a certain distance to prevent from falling.

Explain why a coin placed on a cardboard covering a glass falls into the glass when the cardboard

is pulled suddenly to one side? [HSEB 2063]

A body tends to continue in its original state due to the property of inertia. Before pulling the cardboard,

both the cardboard and the coin possess the inertia of rest. When the cardboard is pulled suddenly to

one side, the coin placed over it, tries to remain in the state of rest and it falls into the glass due to

inertia of rest. Here, at the same instant, the cardboard comes in motion but the coin tends to remain at rest.

Explain how Newton's first law of motion follows from the second law? [HSEB 2061]

The second law of motion gives the measurement of force i.e.,

F = ma

In the absence of external force, F = 0

or, ma = 0

∴ a=0 [since, m ≠ 0]

This implies that a body at rest will remain at rest and a body in uniform motion in a straight line will

continue to do so in the absence of external force. This is the first law of motion. Hence, Newton's first

law of motion follows from second law of motion.

Explain how Newton's third law of motion follows from the second law?

Consider a collision (mutual interaction) between two bodies A and B. Let,

FBA = Force (action) exerted by A on B during interaction.

FAB = Force (reaction) exerted by B on A during interaction

dPA = Rate of change of momentum of A.
dt

100 | Essential Physics

dPB = Rate of change of momentum of B.
dt

From Newton's second law of motion.

FBA = dPB and FAB = dPA
dt dt

Adding, FBA + FAB = dPB + dPA
dt dt

d . . . (1)
or, FBA + FAB = dt (PB + PA)

According to Newton's second law of motion, the external force F acting on the bodies is,

d . . . (2)
F = dt (PB + PA)

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

FBA + FAB = F . . . (3)
In the absence of external force, F = 0

or, FBA + FAB = 0
∴ FAB = – FBA
This is Newton's third law of motion. Hence, the third law is contained in second law of motion.

12. Why is the kinetic friction less than the limiting friction? [HSEB 2061]
Ans:
The force of interlock between projections and depressions of two surfaces in contact causes friction. In
13.
Ans: the static (limiting) friction, the projections and depressions between two surfaces are strongly bounded
14.
Ans: (interlocked). In the case of kinetic friction (sliding friction), the projections and depressions between

15. two surfaces are weakly interlocked. So, a large force is required to start a motion (limiting friction)
Ans:
than to continue the motion (Kinetic friction). Hence, kinetic friction is always less than the limiting

friction.

Can a body be regarded in a state of rest as well as in motion at the same time? Give an example.

Yes. Rest and motion are not absolute terms, but they are relative terms. Rest and motion are

comparative states of bodies. For example, the passengers in a moving bus are at rest with respect to

each other but they are in motion with respect to surrounding objects like buildings, trees etc. So, a

body can be regarded in a state of rest as well as in motion at the same time.

Is friction a necessary evil? Explain. [HSEB 2059]

Friction is Necessity (Advantages)

i. It will not be possible to walk without friction between ground and our feet. When the ground

becomes slippery after rain, it is made rough by spreading sand over it to increase friction.

ii. Tyres of vehicles are made rough to increase friction.

iii. The various parts of a machine are capable to rotate due to the friction between the belt and

pulley.

iv. It would be impossible to write, to climb, to fix a nail, to drive etc. in the absence of friction.

Friction is an Evil (Disadvantages)

i. Wear and tear in different parts of a machine is due to friction.

ii. The friction between different parts of rotating machines produces heat and causes damage to

them.

iii. Extra power is necessary to apply to a machine to overcome the friction. Thus, the efficiency of

machine decreases due to friction.

If a moving bullet striking a block of wood on a frictionless table embeds inside it, what happens

to the K.E. of the bullet? [HSEB 2058]

A moving bullet possesses kinetic energy due to its motion. When the bullet embeds inside a wooden

block on a frictionless table, then K.E. of bullet is shared between the block and the bullet assuming no

energy loss during the collision. There is no opposing force (friction) to reduce the magnitude of

velocity on the frictionless table. Therefore, they continue to move on the frictionless table for ever with

constant velocity.