TF015 Ch 3 Momentum and Impulse

8/6/2019 TF015 Ch 3 Momentum and Impulse

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3.1 Conservation of Linear Momentum and Impulse

a) Define momentum.

b) State the principle of conservation of linear momentum.

c) State the conditions for elastic and inelastic collisions.

d) Apply the principle of conservation of momentum in

elastic and inelastic collisions.

e) Define and use the coefficient of restitution,

to determine the types of collisions.

f) Define impulse, and use F-tgraph to

determine impulse.

pJTT

(!

LEARNING OUTCOMES

!12

12

uu

vvek

CHAPTER 3

MOMENTUM & IMPULSE(2 Hours)


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Principle of Conservation of Linear Momentum

When the net external force acting on a system is

zero (isolated system), the total momentum of that

system is constant (or conserved).

constant00 system !!!7 pdt

dpFnet

T

Hence, for an isolated system:

! finalinitial ppTT


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We often use the principle of conservation of

momentum to solve problems where objectscollide or explode apart.


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Collision

an isolated event in which 2 or more bodies (thecolliding bodies) exert relatively strong forces on

each other for a relatively short time.

In a collision, the change in the momentum of the

system is zero because no external forces are acting

on the system.

Momentum is conserved for any collision

!finalinitial

pp

TT

There are 2 types of collision :

(1) Elastic

(2) Inelastic


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In an inelastic collision,

The momentum of a system is conserved.

But its kinetic energy is not conserved.

( part of the initial kinetic energy would have been

transformed to other forms of energy like sound & heat. )

! finalinitial ppTT

finalintial KK 7{7

In an elastic collision,

Momentum and kinetic energy are conserved.

! finalinitial ppTT

finalintial KK 7!7


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For completely / perfectly inelastic collision,

objects stick together after impact. The

objects have the same final velocity.

vmmumum )( 212211 !Apply Principle Conservation of Momentum,


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Elastic Collision in one dimension motion

Consider 2 particles that undergo an elastic head-on

collision.

In this case, both momentum and kinetic energy are

conserved.According to the Principle of Conservation of

Momentum:

22112211 vmvmumum !


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Knowing that kinetic energy is conserved for the

system:

222222112211

2

1

2

1

2

1

2

1vmvmumum !

Elastic Collision in two dimension motion

- can be analyzed by using the fact that momentum is

a vector quantity.- consider an elastic collision between 2 objects in

which the collision is oblique (not head-on)

collision.


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before after

The initial velocity for object m1 is u1 along the x axis

and the object with mass m2 is initially at rest.


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After the collision, two objects move off in different

direction.

According to Principle of Conservation ofMomentum:

Momentum along x-axis:

afterxbeforex pp 7!722211111 coscos UU vmvmum !

Momentum along y-axis:

afterybeforey pp 7!7222111 sinsin0 UU vmvm !


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ExampleA 200 g tennis ball moving with a velocity of 15 m s1

along x axis collides with a stationary ball of 800 g asshown in figure below. After the collision, tennis ball is

scattered at an angle of 45 from its original direction with

velocity of 5 m s1. Find the final velocity (magnitude and

direction) of the struck ball.


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Solution

Conservation of Momentum along x-axis:

)1(8661.2cos

2929.2cos8.0

cos8.07071.03

cos)8.0(45cos5)2.0()15()2.0(

22

22

22

22

.!

!

!

!

U

U

U

U

v

v

v

v

afterxbeforex pp 7!7

22211111 coscos UU vmvmum !

Conservation of Momentum along y-axis:

afterybeforey pp 7!7

222111 sisi vmvm !


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)2(8839.0sin

7071.0sin8.0

sin)8.0(45sin5)2.0(0

22

22

22

.!

!

!

U

U

U

v

v

v

8661.2

8839.0

cos

sin:

)1(

)2(

22

22

!U

U

v

v

r!

!

14.17

3084.0tan

2

2

U

U

Substitute 2 into equation (2):

1sm32947.0

8839.0

8839.014.17sin

2

2

!!

!r

v

v


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Example ( PYQ Session 2004/05 )

Object A of mass 8 kg moving at 4 m s1 collides with

another object B of mass 6 kg moving at 5 m s1 in theopposite direction. After the collision, object A moves

opposite to its initial direction at 0.1 m s1.

(i) What is the velocity of B after the collision?

(ii) Show that the collision is inelastic.

Solution

Given: Before collision :

mA= 8 kg , uA = 4 m s

1

mB = 6 kg , uB = 5 m s1 ( opposite direction

with object A )


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after collision :

vA = 0.1 m s1 ( opposite to its initial direction ) vB = ?

(i) Using Principle of Conservation of Momentum:

1sm467.0

8.26

6)1.0(8)5(6)4(8

!

!

!

!

7!7

B

B

B

BBAABBAA

afterbefore

v

v

v

vmvmumum

pp

(ii) For an inelastic collision, kinetic energy of the

system is conserved

22

2

1

2

1BBAAbefore uuKE !


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J139

7564

)5)(6(2

1)4)(8(

2

1 22

!

!

!

22

21

21 BBAAafter vvKE !

J6943.0

6543.004.0

)467.0)(6(2

1)1.0)(8(

2

1 22

!

!

!

KE before KE after inelastic collision


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Example

ub

A stationary 2.50 kg block of wood suspended by a wire

of negligible mass. A 0.01 kg bullet is fired into the

block. The block (with the bullet in it) swings to a

maximum height of 0.65 m above the initial position.

Find the initial speed the bullet is fired.


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Solution

Given: m1 = 0.01 kg, m2 = 2.50 kg

hf= 0.65 m

before collision, block is stationary u2 = 0

find u1 = ?

Completely inelastic collision

fvmmumum )( 212211 !

fvu )50.201.0(0)(01.0 1 !)1(2511 .fvu !


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Apply Conservation of Mechanical energy

ff ghmmvmm )()(2

121

2

21 !

)65.0)(81.9(51.2)51.2(2

1 2

!fv-1sm57.3!fv

Substitute into (1) :

)57.3(2511 !u-1sm07.8961 !u


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Coefficient of RestitutionIs the ratio of the difference in velocities before

and after the collision.

!

12

12

uu

vvek

where

v1 & v2 velocity of objects after collision

u1 & u2 velocity of objects before collision

The value ofek is always between 0 and 1.

A perfectly elastic collision has ek= 1

A perfectly inelastic collision has ek= 0


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Example

A ball with the mass of 1 kg moves to the right at 4 m s1

and collides with a stationary ball with the mass of 2 kg.

Calculate the final velocity of each ball if the collision is

elastic.

Solution

A B

uA= 4 m s1

before after

A B

vA= ? vB = ?


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Conservation of momentum

fi pp 7!7

BBAABBAA vmvmumum !

BA vv 2)0(2)4(1 !

)1(24 .BA vv !

The coefficient of restitution

! 12

12

uu

vv

ek

! 40AB vv

)2(4 .! BA vv


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Equating (1) & (2) :

424 ! BB vv-1sm67.2!Bv

( Ball B moves to the right )

4! BA vv

467.2 !Av

-1

sm33.1!Av

From:

( Ball A moves to the left )


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Impulse

When 2 objects such as baseball and a bat, a hammer

and a nail or even two cars collide, they can exert

large forces on one another in a short period of time,t.

Forces of this type, which exist only over a very shorttime during the impact are often called impulsive force


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Impulse of the force equals to the change inthe

momentum of the particles.

JT

FT

pJTT

(!

initialfinal ppTT

!before during after


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From Newtons 2nd law of motion

dtdp!

dtdp !

dtdp

t

t

p

p

f

i

!2

1

Integrating :

where pi momentum of body at time ti

pf momentum of body at time tf


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Impulse isequaltotheareaunderthe Ft graph.

Iftheforce is constant (does not vary withtime)

ortaking theaverageforcefortime interval t(

tJ (!T

dtFpp

t

t

if

!

2

1

fi ttJ tofromgrapht-Funderthearea!T


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Solution

Taking motion to the right as +

(a) Impulse = area under the graph F-t

(b)

20003.02

1vv!

sN3!JT

pJTT

(!

mumv !

))22((06.03 ! v-1

sm2

8!v( to the right )


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Follow Up exercise

A karate student tries to break a board.

How can the abrupt stop of the hand generate so much force

to break the board?

Assume that the hand has a mass of 0.35 kg and that thespeeds of the hand just before and just after hitting the

board are 10 m s1 and 0 m s1 respectively. What is the

average force exerted by the fist on the board if the contact

time is only 0.30 ms?

TF015 Ch 3 Momentum and Impulse

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Transcript of TF015 Ch 3 Momentum and Impulse