Chapter 6 Linear Momentum

8/4/2019 Chapter 6 Linear Momentum

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CHAPTER 6

1

LINEAR MOMENTUM ANDLINEAR MOMENTUM ANDCOLLISONSCOLLISONS



2

LEARNING OUTCOMESLEARNING OUTCOMES

i. Able to explain and state the relation between linearmomentum and the Newton¶s Second Law of motion.

ii. Able to state the Principle of Conservation of LinearMomentum.

iii. Able to describe and analyze elastic and inelasticcollision 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.



3

the product between mass and velocitythe product between mass and velocity.

vector quantity.

Equation :

The S.I. unit : kg m skg m s--11

The dir ection of the momentumdir ection of the momentum is the samesame as the dir ection dir ection of the velocityof the velocity.

p

T

vmTT

!

Linear momentum,



4

For closed system, the law of conservation of linear

momentum states that

³if the net external force acting on the system is zero,

the total linear momentum of a system is constant´

0

0

0

- 0

cons tan t

n e t

f i

f i

f i

F

p

t

p

p p

p p

p p

!

(!

(

( !

!

!

! § §

v

v

v

v v

v v

v v

The total of initial momentum = the total of final momentumThe total of initial momentum = the total of final momentum



5

Collision

Elastic Inelastic

Collision in which kinetic energy (as

well as momentum) is conserved

Collision in which kineticenergy is not conserved

11 222 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

!

!

§ §uv uv

v v v v

f i

A B A B A B

p p

m u m u m m v

!

!

§ §uv uv

v v v

2 2 2 2

1 1

2 2

f i

¡ ¡

¡ ¡

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 Bm u m u m v m v

!

!

- - B A B Au u v v !v v v v



6

Example 1:

Figure above 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 :Solution :

1sm6

!

Au

AB

1

sm3

!

Bu

§§ !

TT

B B A A B B A Avmvmumum !

20.1000.20030.10060.200 ! A

v

1sm3.5

! A

v

1sm6;kg0.100;kg0.200

!!! A B Aumm

11sm2;sm3

!! B Bvu

to the lef tto the lef t



7

Figure below shows a body A of mass mA = 5 k gmoving towar ds the r ight on a level f r ictionless sur face

with a velocity uA = 2 ms-1. It collides with a second

body B of mass mB = 3 k g moving towar ds the left

with a velocity uB = 2 ms-1

. If the collision iscompletely elastic, deter mine the velocities of A and B

after the collision.

uA uBmA mB

Example 2:



8

Fr om the pr inci ple of conser vation of momentum,mAuA + mBuB = mAvA + mBvB

(5)(2) + (3)(-2) = 5vA + 3vB

5vA + 3vB = 4 (i)

SOLUTIONSOLUTION

Since the collision is completely elastic,

vB - vA = - ( uB - uA )

= - ( - 2 ± 2)

vB - vA = 4 (ii)

Solving equation (i) and (ii) simultaneously, we get

vA = -1 m/s , vB = 3 m/s



9

A ball A, 2.2 k g is moving with u¶ = 4i ms-1 collidewith another ball B moving with a velocity u´ = -7i ms-

1. After the collision the velocity of A and B is v¶ = -

3i ms-1 and v ³ = 4i ms-1 . If the total momentum is

conser ve in the collision, calculate

a) The mass of ball B.

b) The velocity of A and B if both balls stick together

after the collision.

(Answers : 1.4 kg, - 0.28 m s-1)

Exercise 1:



10

Two identical balls B1

and B2

of masses 2k g and 5 k g

moving at 12 ms-1 and 4 m s-1 respectively along +X axis.

After colliding elastically B2

moves at 6 m s-1. Compute

i)v

elocity of B1 after collision,

ii) the change in momentum of B1.

iii)If B2 mo

ves in the opposite direction and the collisionis per f ectly elastic, calculate the velocities of the balls after

collision.

[ 7 ms-1 , - 10 ms-1 ,5.1 ms-1,10.9 ms-1 ]

Exercise 2:



11

Impulse,

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

is a vector quantityvector quantity whose dir ectiondir ection is the samesame as the

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

Impulse ± momentum theorem:

Relation between kinetic energy and the magnitude of momentum:

J T

-net

J t p mv mu! ( ! ( !v v v v v

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 pm v

m m

! ! !



12

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 :Solution :

Example 3:

Wall (2)11

1sm100

!1u

111

sm70

!1v

0!!22

uv

kg0.201 !m



13


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

Therefore the magnitude of the impulse is 3434 NN ss.

b. Given the contact time,

12p pdp J

111uvm J !

? A100700. 0 ! J

s N34! J

s1010

3

v!dt dt F

av!

3101034

v!av

F

N3400!av

F



14

A 0.25 kg ball moving in the +x direction at 13 m/s is hit by a bat. Its

final velocity is 19 m/s in the ±x direction. The bat acts on the ball for 0.01 s. Find the average force and the impulse experienced by the ball.

Example 4:

SolutionSolution

By taking the +x direction as positive, we have vi = 13 m/s and vf = - 19

m/s.

N81.

)3)(.(F

)1319)(.()1.(F

!

!

!

From the impulse equation,

The impulse = F(t = (-8 )( . 1) = -8. Ns



15

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 5:

0.2 1.8 mst

0

kN F

1.0

18



16


a. From the force-time graph,

b. Given the ball¶s initial speed,

graphunder thearea t F J !

331018100.21.8

2

1vv!

J

s N14.4! J

0!u

uvmdp

01060.014.43 v

v

1s240

!v

kg1060.03

v!m

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Chapter 6 Linear Momentum