Chapter 6 Linear Momentum - Lecturer-1

8/4/2019 Chapter 6 Linear Momentum - Lecturer-1

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1

LINEAR MOMENTUM ANDCOLLISONS

i. Able to explain and state the relation between linear

momentum and the Newton’s Second Law of motion.ii. Able to state the Principle of Conservation of Linear

Momentum.iii. Able to describe and analyze elastic and inelastic

collision in one dimension problems.iv. Able to describe and analyze impulse in a collision.

v. Able to apply the conservation of energy and momentumin solving collision problems.



2

Linear momentum, the product between mass and velocity. vector quantity.

Equation :

The S.I. unit : kg m s-1 The direction of the momentum is the same as the direction

of the velocity.

p

vm p



3

For closed system, the law of conservation of linearmomentum states that

“if the net external force acting on the system is zero,the total linear momentum of a system is constant”

0

0

0

- 0

constant

net

f i

f i

f i

F p

t

p

p p

p p

p p

The total of initial momentum = the total of final momentum



4

Example 1:

Figure above shows an object A of mass 200 g collides head-on withobject 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 :

1sm6

Au

AB

1sm3

Bu

f ip p

B B A A B B A A vmvmumum

20.1000.20030.10060.200 Av1sm3.5

A

v

1sm6;kg0.100;kg0.200 A B A umm

11 sm2;sm3 B B vu

to the left



5

Collision

Elastic InelasticCollision in which kinetic energy (aswell as momentum) is conserved

Collision in which kineticenergy is not conserved

1 2 2 2

m v1 1

m v

Collision Elastic Inelastic

Momentum

Kineticenergy

Completely

collision

f i

A B A B A B A B

p p

m u m u m v m v

f i

A B A B A B

p p

m u m u m m v

2 2 2 2

1 1

2 2

f i

A A B B A A B B

KE KE

m u m u m v m v

2 2 2 2

1 1

2 2

f i

f loss i

A A B B loss A A B B

KE KE

KE KE KE

m u m u KE m v m v

- - B A B Au u v v



Example 2:

Two bodies of masses 8 kg and 4 kg move along the x-axis inopposite directions with velocities of 11 ms-1 in the positive x-directionand 7 ms-1 in the negative x-direction respectively and collide headon. Determine the speed and direction of each after impact if

a) they stick together

b) the collision is perfectly elastic.

Ans: 5 m s-1 (to the right), V1 = - 1 m s-1 (to the left), V2 = 17 m s-1 (to the right)



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Impulse,

is defined as the product of a force, F and the time, t ORthe change of momentum.

is a vector quantity whose direction is the same as the

constant force on the object. SI unit: N s or kg m s -1

Impulse – momentum theorem:

Relation between kinetic energy and the magnitude ofmomentum:

J

-n e t

J F t p m v m u

The impulse exerted on a body is equal

to the change in the body’s momentum

2

2

2 1 1 1

2 2 2

m v pKE m v

m m



8

A 0.20 kg tennis ball strikes the wall horizontally with a speed of 100 m

s1 and it bounces off with a speed of 70 m s1 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 :

Example 3:

Wall (2)1

1sm100 1u

11sm70

1v

0 22 uv

kg0.201 m



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Solution :

a. From the equation of impulse that the force is constant,

Therefore the magnitude of the impulse is 34 N s.

b. Given the contact time,

12 p pdp J

111 uvm J

100700.20 J

sN34 J

s1010

3

dt dt F J av

3101034 avF

N3400

avF



10

An estimated force-time curve for a tennis ball of mass 60.0 g struck by

a racket is shown in figure above. Determinea. 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.

Example 4:

0.2 1.8 mst 0

kNF

1.0

18



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Solution :

a. From the force-time graph,

b. Given the ball’s initial speed,

graphunder thearea t F J

33 1018100.21.8

2

1

J

sN14.4 J

0u

uvmdp J

01060.014.4 3

v

1sm240 v

kg1060.0 3m

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Chapter 6 Linear Momentum - Lecturer-1