LinearMomentum Theory

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General Physics 1

APHY 111

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Linear Momentum

4.1 Definition of Linear Momentum

• We consider a body moving along the x-axis

• The momentum of a body with mass m and

velocity vx is defined as

2

• The unit of momentum is Kg.m/s

• Similarly we can define momentum along any other

axis

x x mv p =


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Example 1: A ball of mass 3 kg is moving on a smooth horizontal table with a

speed of 5m/s when it collides normally with a fixed smooth vertical wall. It

rebounds from the wall with a speed of 2 m/s. Find the change in momentum

of the ball.

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The change in momentum is also known as Impulse (ώθηση)

4.2 Momentum and Force

• The total force acting on an object can be written

as the rate of change of momentum:

dp

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• For a finite time interval this is the average force

t

p

∆

∆=ForceAverage

dt =


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Example 2: In a particular crash test, a car of mass 1500 kg collides with a wall.

The initial and final velocities of the car are v i=-15m/s and vf =2.6m/s, respectively.

If the collision lasts for 0.150 s, find the impulse caused by the collision and the

average force exerted on the car.

Impulse

−−−

−=∆= p p p i f

5

kg.m/s102.641500(17.6)

.

4×==∆ p

i f

N1076.1150.01064.2ForceAverage 5

4×=×=

∆

∆=t p

Exercise 1: A tennis ball of mass 0.1kg is moving horizontally with a speed of 10m/s

when it collides normally with a vertical racquet. It rebounds from the racquet with aspeed of 25m/s.

(a) Find the change in momentum of the ball.

(b) If the time of contact between the ball and the racquet is 0.2s, find the average

force the racquet exerts on the ball during the collision.

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4.3 Conservation of Momentum

•In a mechanical system the total momentum is

conserved

momentum before = momentum after

This is the law of conservation of momentum and is very

important for understanding collisions

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B B A A vmvm += beforemomentum

placetakescollision

B B A A vmvm ′+′=after momentum

B B A A B B A A vmvmvmvm ′+′=+⇒


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Exercise 2: A bullet of mass mB

= 5g is fired from a rifle of mass mR

= 2Kg. If the

gun, just after the explosion, bounces with 2.5 m/s, calculate

(a)the speed of the bullet just after the explosion

The bullet then strikes a block of mass M = 495g horizontally and remains stuck in

it (plastic collision), calculate

(b) The final speed of the group

(c) Calculate the loss in the kinetic energy during the last impact. Explain.

Solution

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Exercise 3: A body A of mass 5,5 kg moving at 6 m/s collides with a second body

B of mass 4,5 kg moving at 4 m/s in the same line but in the opposite direction

with body A. After collision the two bodies move together with a common velocity.

Find

(a) The velocity (magnitude and direction) of the composite body.

(b) The loss in mechanical energy of the system.

(c) The average force body B experiences during the collision.

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Exercise 4: A rocket of mass 100kg is moving horizontally with a speed of 400m/s.

Suddenly an explosion occurs and the rocket brakes into two parts of equal mass,

moving in the same direction. If one part has a speed of 200m/s,

(a) Calculate the speed of the second part.

(b) Find the energy released during this explosion.

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4.3 Elastic Collisions

•In an elastic collision both momentum and kinetic

energy are conserved

momentum before = momentum after

Kinetic energy before = kinetic energy after

In terms of e uations we have:

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Combining the above two equations we have that

which is the condition for elastic collision

ii f f uuuu 1221 −=−


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Exercise 5: Two bodies of masses 1kg and 2kg move along a straight line with

velocities u1=4m/s and u2=1m/s, respectively, and in the same direction. The bodies

collide elastically. Taking into account that for an elastic collision their velocities u’1and u’2 after collision satisfy the condition ,

Find their velocities u’1 and u’2 after the collision.1221 uuuu −=′−′

Solution

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a e ave rom conserva on o momen um

Using the condition

we have to solve a system of equations ([1] and [2])

[1] 6242 21

22112211

=+=′+′

+=′+′

uu

umumumum

[2] 321

1221

−=′−′

−=′−′

uu

uuuu

smu

smu

/3

/0

2

1

=′

=′

The ballistic pendulumThe figure shows a ballistic pendulum, a

system for measuring the speed of a bullet.

The bullet, with mass mB , is fired into a

block of wood with mass mw, suspended

like a pendulum, and makes a completely

inelastic collision with it. After the impact of

the bullet, the block swings up to a

maximum height y. Given the values of y,

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mB , and mw, what is the initial speed VI of

the bullet?


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Momentum conservation gives

The wooden block will then start moving until all of its kinetic energy has been

converted to potential energy. Energy conservation then gives

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,

the speed v1of the bullet

Exercise 6

A bullet, with mass mB=0.1 kg , is fired into a block of wood with mass mw=10 kg,

suspended like a pendulum, and makes a completely inelastic collision with it.

(a) If the initial velocity of the bullet is v B=300 m/s, calculate the common velocity v c of

the wooden block immediately after collision.

(b) After the impact of the bullet, the block swings up to a maximum height y. Calculate

the value of the maximum height y.


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Exercise 7:Consider the frictionless track ABC of the figure. A block of mass m1 =

6 kg is released from A. It makes a head-on elastic collision at B with the block ofmass m2 = 12 kg that it is initially at rest.

(a) Calculate the velocity of m1 just before the collision,

(b) Calculate the velocity of m1 just after the collision.

Taking into account that for an elastic collision their velocities u’1 and u’2 after

collision satisfy the condition1221 uuuu −=′−′


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LinearMomentum Theory

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