Note Chap 3 SP015 (Momentum and impulse) Norul Huda

Chapter 03 Physics

CHAPTER 3:
Momentum and impulse

(F2F:3.5 Hours)

1

Chapter 03 Physics

Overview: Momentum
and impulse

Linear momentum

Impulse

Conservation of
momentum

Collision

Elastic Inelastic

2

Chapter 03 Physics
Learning Outcome:

3.1 Momentum and impulse (1 hour)

At the end of this chapter, students should be able
to:

 Define momentum and impulse, J = F t.

 Solve problem related to impulse and impulse-
momentum theorem,

J = p = mv f − mvi

 Use F-t graph to determine impulse.

3

Chapter 03 Physics

3.1 Momentum and im pulse
3.1.1 Linear momentum, p

At the end of this lesson, students should be able to:

✓ Define momentum.

 is defined as the product between mass and velocity.

 is a vector quantity.

 Equation : 
p = mv

 S.I. unit : kg m s-1.

 The direction of the momentum is the same as the direction
of the velocity.

py  px = p cosθ = mv cosθ
p
p y = p sin θ = mv sin θ
 p4

x

 Chapter 03 Physics
3.1.2 Impulse, J

At the end of this lesson, students should be able to:

✓ Define impulse, J = F t.

 defined as the product of a force, F and the time, t

OR J = Ft

where J : impulse
t : time
F : impulsive force

 a vector quantity whose direction is the same as the
constant force on the object.

 S.I. unit of impulse : N s or kg m s−1.

 force = non constant t2
t1
J = Fdt = Favt

where Fav : average impulsive force 5

Chapter 03 Physics

3.1.3 Impulse-momentum theorem

At the end of this lesson, students should be able to:

✓ Solve problem related to impulse and impulse-momentum
theorem.

 Single constant force, F acts on an object in a short time

interval (collision),

 F = F = dp = constant
dt

J = Fdt = dp = p2 − p1

where p2 : final momentum
p1 : initial momentum
F : impulsive force

 state as the impulse of a body equals to the change of the

body momentum. 6

Chapter 03 Physics

 Component form :

Jx = ( Fav ) t = p2x − p1x = m(vx − ux )
x
consider 2-D

( ) ( )collision only J y = Fav y t = p2 y − p1y = m v y − u y

Jz = (Fav ) t = p2z − p1z = m(vz − uz )
z

7

Chapter 03 Physics

At the end of this lesson, students should be able to:

✓ Use F-t graph to determine impulse.

 Impulsive force, F against time, t graph

F

0Figure 3.1 t1 t2 t

Shaded area under the F−t graph = impulse

Picture 3.1 Picture 3.3
Picture 3.2
8

Chapter 03 Physics

Example 3.1 :
A 0.20 kg tennis ball strikes the wall horizontally with a speed of
100 m s−1 and it bounces off with a speed of 70 m s−1 in the
opposite direction.

a. Calculate the magnitude of impulse delivered to the ball by the

wall,

b. If the ball is in contact with the wall for 10 ms, determine the

magnitude of average force exerted by the wall on the ball.

Solution : m1 = 0.20 kg s −1
u1 =
100 m

1

Wall (2)

1 v2 = u2 = 0
v1 = 70 m s−1

9

Chapter 03 Physics

Solution :

a. Apply: J = dp = p2 − p1

J = m1(v1 − u1 )
J = (0.20)(− 70) − 100

J = −34 N s

Therefore the magnitude of the impulse is 34 N s.

b. Given the contact time, dt = 10  10 −3 s

( )J = Favdt

34 = Fav 10 10−3

Fav = 3400 N

10

Chapter 03 Physics

Example 3.2 :

F (kN )

18

0 0.2 1.0 1.8 t (ms )

Figure 3.2

An estimated force-time curve for a tennis ball of mass 60.0 g
struck by a racket is shown in Figure 3.2. Determine
a. the impulse delivered to the ball,
b. the speed of the ball after being struck, assuming the ball is

being served so it is nearly at rest initially.

11

Chapter 03 Physics

Solution : m = 60.0  10 −3 kg

a. Apply: J = area under the F − t graph

( )J = 1 (1.8 − 0.2)10−3 18 103
2

J = 14.4 N s

b. Given, u = 0

J = dp = m(v − u )

( )14.4 = 60.0 10 −3 (v − 0)

v = 240 m s−1

12

Chapter 03 Physics

Exercise 3.1 :

1. A steel ball with mass 40.0 g is dropped from a height of
2.00 m onto a horizontal steel slab. The ball rebounds to a
height of 1.60 m.
a. Calculate the impulse delivered to the ball during impact.
b. If the ball is in contact with the slab for 2.00 ms, determine
the average force on the ball during impact.

ANS. : 0.47 N s; 237. 1 N
2. A golf ball (m = 46.0 g) is struck with a force that makes an

angle of 45 with the horizontal. The ball lands 200 m away
on a flat fairway. If the golf club and ball are in contact for
7.00 ms, calculate the average force of impact. (neglect the
air resistance.)
ANS. : 293 N

13

Chapter 03 Physics

3.

Figure 3.3

A tennis ball of mass, m = 0.060 kg and a speed,
v = 28 m s−1 strikes a wall at a 45 angle and rebounds with
the same speed at 45 as shown in Figure 3.3. Calculate the
impulse given by the wall.
ANS. : 2.4 N s to the left or −2.4 N s

14

Chapter 03 Physics
Learning Outcome:

3.2 Conservation of linear momentum (2.5 hours)

At the end of this chapter, students should be able
to:

 State the principle of conservation of linear
momentum.

 Differentiate elastic and inelastic collisions.

 Apply the principle of conservation of linear
momentum in the elastic and inelastic
collisions for 1D and 2D collisions.

15

Chapter 03 Physics

3.2.1 Principle of conservation of linear momentum

At the end of this lesson, students should be able to:

✓ State the principle of conservation of linear momentum.

 states “In an isolated (closed) system, the total momentum

of that system is constant.”dp
dt
 In a Closed system, F = 0

 F = = 0


dp = 0

p2 − p1 = 0
p1 = p2 = constant

 px = constant

 p y = constant 16

Chapter 03 Physics

 we obtain

Total of initial momentum = Total of final momentum

OR

  =   f
pi p

17

Chapter 03 Physics

Linear momentum in one dimension collision

Example 3.3 : uB = 3 m s−1 u A = 6 m s−1

BA

Figure 3.4

Figure 3.4 shows an object A of mass 200 g collides head-on with

object B of mass 100 g. After the collision, B moves at a speed

of 2 m s-1 to the left. Determine the velocity of A after Collision.

Solution : mA = 0.200 1k;gv;Bm=B = 0.100 kg; uA = −6 m s −1
uB = 3 m s− −2 m s−1


 pi = p f
mAuA + mBuB = mAvA + mBvB

(0.200 )(− 6) + (0.100 )(3) = (0.200 )v A + (0.100 )(− 2)
v A = −3.5 m s −1 to the left 18

Chapter 03 Physics

Linear momentum in two dimension collision

Example 3.4 :  m2
m1 u1
50

Before collision 
m1 v1
Figure 3.5
After collision

A tennis ball of mass m1 moving with initial velocity u1 collides

with a soccer ball of mass m2 initially at rest. After the collision,
the tennis ball is deflected 50 from its initial direction with a

velocity v1 as shown in figure 3.6. Suppose that m1 = 250 g,
m2 = 900 g, u1 = 20 m s−1 and v1 = 4 m s−1. Calculate the

magnitude and direction of soccer ball after the collision.

19

Chapter 03 Physics

Solution : m1 = 0.25 0 kg m; ms2−1=; 0.900 kg; u1 = 20 m s −1 ;
u2 = 0; v1 =4 θ1 = 50

  
pix = p fx

m1u1x + m2u2x = m1v1x + m2v2x

(0.250 )(20 ) + 0 = (0.250 )(v1 cos θ1 ) + (0.900 )v2 x

( )5 = (0.250 ) 4 cos 50 + (0.900 )v2 x

v2x = 4.84 m s−1

20

Chapter 03 Physics
Solution :  
p fy
 piy =
( )0 = m1v1 y + m2v2 y

0 = (0.250 ) − 4 sin 50 + (0.900 )v2 y
v2 y = 0.851 m s−1

Magnitude of the soccer ball,

( ) ( )v2 =
v2x 2 + v2 y 2

v2 = (4.84 )2 + (0.851)2 = 4.91 m s −1

Direction of the soccer ball, y

θ2 = t9a.n97−1 v2 y  = ta n −1  0.851   9.97
v2 x  4.84  v
θ2 = x

21

Chapter 03 Physics

Exercise 3.2 :

1. An object P of mass 4 kg moving with a velocity 4 m s−1 collides
elastically with another object Q of mass 2 kg moving with a
velocity 3 m s−1 towards it.

a. Determine the total momentum before collision.

b. If P immediately stop after the collision, calculate the final

velocity of Q.

c. If the two objects stick together after the collision, calculate

the final velocity of both objects.

ANS. : 10 kg m s−1; 5 m s−1 to the right; 1.7 m s−1 to the right

2. A marksman holds a rifle of mass mr = 3.00 kg loosely in his
hands, so as to let it recoil freely when fired. He fires a

bullet of mass mb = 5.00 g horizontally with a velocity 300 m s-1.
Determine

a. the recoil velocity of the rifle,

b. the final momentum of the system.

ANS. : −0.5 m s−1; U think. 22

Chapter 03 Physics

3.

1.20 kg 1.80 kg

Before 1.40 m s-1
0.630 m s-1

After 23
Figure 3.6

In Figure 3.6 show a 3.50 g bullet is fired horizontally at two
blocks at rest on a frictionless tabletop. The bullet passes
through the first block, with mass 1.20 kg, and embeds
itself in the second block, with mass 1.80 kg. Speeds of
0.630 m s−1 and 1.40 m s-1, respectively, are thereby given
to the blocks. Neglecting the mass removed from the first
block by the bullet, determine

a. the speed of the bullet immediately after it emerges
the first block and from

b. the initial speed of the bullet.

ANS. : 721 m s−1; 937.4 m s−1

Chapter 03 Physics

4. A ball moving with a speed of 17 m s−1 strikes an identical ball
that is initially at rest. After the collision, the incoming ball has
been deviated by 45 from its original direction, and the struck
ball moves off at 30 from the original direction as shown in
Figure 3.7. Calculate the speed of each ball after the collision.

Figure 3.7

ANS. : 8.80 m s− 1; 12.4 m s−1

24

Chapter 03 Physics

3.2.2 Collision

At the end of this lesson, students should be able to:

✓ Differentiate elastic and inelastic collisions.
 is defined as an isolated event in which two or more bodies

(the colliding bodies) exert relatively strong forces on each
other for a relatively short time.

Collision

Elastic Inelastic

25

Chapter 03 Physics

Elastic collision

 is defined as one in which the total kinetic energy (as well
as total momentum) of the system is the same before and
after the collision.

 Head-on collision of two billiard balls.

Before collision 1 m1u1 m2u2 2

At collision 12

After collision m1v1 1 2 m2v2

Figure 3.8 26

Chapter 03 Physics

 The properties are

a. The total momentum is conserved.

  =   f
pi p

b. The total kinetic energy is conserved.

 Ki =  K f

OR

1 m1u12 + 1 m2u 2 = 1 m1v12 + 1 m 2 v 22
2 2 2 2 2

27

Chapter 03 Physics

Inelastic (non-elastic) collision

 is defined as one in which the total kinetic energy of the
system is not the same before and after the collision (even
though the total momentum of the system is conserved).

 Completely inelastic collision of two billiard balls.

Before collision 1 m1u1 u2 = 0

2

m2

At collision 12

After collision v
(stick together)
12
28
Figure 3.9

 Caution: Chapter 03 Physics

⚫ Not all the inelastic collision is stick together.

⚫ In fact, inelastic collisions include many situations in which
the bodies do not stick.

 The properties are

a. The total momentum is conserved.

  =  
pi pf

b. The total kinetic energy is not conserved.
c. The total energy is conserved.

   Ei = E f OR K i = K f + losses energy

29

Case study 1: Chapter 03 Physics
Before

After

Type of collision and why? 30

Case study 2: Chapter 03 Physics
Before

After

Type of collision and why? 31

Chapter 03 Physics

Example 3.5 :

Two titanium spheres approach each other head-on with the same
speed of 3 m s-1 and collide elastically. After the collision, one of
the spheres, whose mass is 500 g, remains at rest. Calculate the
mass of the other sphere.

Solution : m1= 0.500 kg; u1=3 m s-1; u2=-3 m s-1; v1=0
u2 2
Before collision 1 u1

v1 = 0 v2 = ?

After collision 12

By using :    m2 = ?
pi = p f
m) 31u+1 m+2m( −2u32) = m1v1 + m v
( 2 2
= m2v2
0.50 0 32

Chapter 03 Physics

Solution :

m2v2 + 3m2 = 1.5
v2 = 1.5 − 3m2
(1)
m2

collision is elastic  K = Kinitial final

1 m1u12 + 1 m2u22 = 012+m121v1m2 2+v2122 m2v22
2 1 2 =
1 (0.500)(3)2 + 2 m2 (3)2
2
m2v22 − 9m2 = 4.5
(2)

33

Chapter 03 Physics

Solution :
By substituting eq. (1) into eq. (2), therefore

m2  1.5 − 3m2 2 − 9m2 = 4.5
 m2 
  m2 = 0.167 kg

34

Chapter 03 Physics

Example 3.6 :

A ball of mass 0.3 kg is dropped from a height of 2.00 m above a
tile floor and rebounds to a height of 1.30 m.
a. Determine the ball’s speed just before and after strike the floor.

b. State the type of the collision between ball and floor. Give

reason. (Given g = 9.81 m s−2)

Solution : m1 = 0.3 kg; h0 = 2.00 m; h1 = 1.30 m
1 u=0
a. i. Before collision,

1 v=0 sy = −h0 = −2.00 m
2.00 m
v12 = u2 − 2gsy
v1 ' 1.30 m
v12 = 0 − 2(9.81)(−2.00)

11 v1 = 6.26 m s−1

v1 Floor (2) 35

Chapter 03 Physics

Solution :

a. ii. After collision, sy = h1 = 1.30 m

v2 = (v1 ')2 − 2 gsy
0 = (v1 ')2 − 2(9.81)(1.30 )

v1 ' = 5.05 m s −1

b. The initial total kinetic energy of the system just before collision

is K initial = 1 m1u12 + 1 m1u 2 2
2 2
1 1
= 2 m1v12 + 0= 2 ( 0.3) ( 6.26 )2

= 5.88 J

36

Chapter 03 Physics

Solution :

b. The final total kinetic energy of the system just after collision is

 K final = 1 m1v12 + 1 m1v2 2
2 2

= 1 m1 ( v1 ')2 + 0
2

= 1 (0.3)(5.05)2

2

= 3.83 J

 K initial  K final

the collision between ball and floor is inelastic.

37

Chapter 03 Physics

THE END.

Next Chapter…

CHAPTER 4 :
FORCES

38