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11.11 Conservation of momentum

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Conservation of momentum

In the absence of an external force acting on a system, momentum is conserved.

Conservation of Linear Momentum

The total linear momentum of an isolated system is constant. An isolated system has no forces acting on it from the outside.

This means that in an isolated system the total momentum before a collision or explosion is equal to the total momentum after the collision or explosion.

Consider a simple collision of two billiard balls. The balls are rolling on a frictionless surface and the system is isolated. So, we can apply conservation of momentum. The first ball has a mass m 1 and an initial velocity v i 1 . The second ball has a mass m 2 and moves towards the first ball with an initial velocity v i 2 . This situation is shown in [link] .

Before the collision.

The total momentum of the system before the collision, p i is:

p i = m 1 v i 1 + m 2 v i 2

After the two balls collide and move away they each have a different momentum. If the first ball has a final velocity of v f 1 and the second ball has a final velocity of v f 2 then we have the situation shown in [link] .

After the collision.

The total momentum of the system after the collision, p f is:

p f = m 1 v f 1 + m 2 v f 2

This system of two balls is isolated since there are no external forces acting on the balls. Therefore, by the principle of conservation of linear momentum, the total momentum before the collision is equal to the total momentum after the collision. This gives the equation for the conservation of momentum in a collision of two objects,

p i = p f
m 1 v i 1 + m 2 v i 2 = m 1 v f 1 + m 2 v f 2
where: m 1 is the mass of object 1 (kg)
m 2 is the mass of object 2 (kg)
v i 1 is the initial velocity of object 1 (m · s - 1 + direction)
v i 2 is the initial velocity of object 2 (m · s - 1 - direction)
v f 1 is the final velocity of object 1 (m · s - 1 - direction)
v f 2 is the final velocity of object 2 (m · s - 1 + direction)

This equation is always true - momentum is always conserved in collisions.

A toy car of mass 1 kg moves westwards with a speed of 2 m · s - 1 . It collides head-on with a toy train. The train has a mass of 1,5 kg and is moving at a speed of 1,5 m · s - 1 eastwards. If the car rebounds at 2,05 m · s - 1 , calculate the velocity of the train.

  2. We will choose to the east as positive.

  3. p i = p f m 1 v i 1 + m 2 v i 2 = m 1 v f 1 + m 2 v f 2 ( 1 , 5 kg ) ( + 1 , 5 m · s - 1 ) + ( 2 kg ) ( - 2 m · s - 1 ) = ( 1 , 5 kg ) ( v f 1 ) + ( 2 kg ) ( 2 , 05 m · s - 1 ) 2 , 25 kg · m · s - 1 - 4 kg · m · s - 1 - 4 , 1 kg · m · s - 1 = ( 1 , 5 kg ) v f 1 5 , 85 kg · m · s - 1 = ( 1 , 5 kg ) v f 1 v f 1 = 3 , 9 m · s - 1 eastwards
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A helicopter flies at a speed of 275 m · s - 1 . The pilot fires a missile forward out of a gun barrel at a speed of 700 m · s - 1 . The respective masses of the helicopter and the missile are 5000 kg and 50 kg. Calculate the new speed of the helicopter immmediately after the missile had been fired.

  1. helicopter and missile

  2. m 1 = 5000 kg

    m 2 = 50 kg

    v i 1 = v i 2 = 275 m · s - 1

    v f 1 = ?

    v f 2 = 700 m · s - 1

  3. The helicopter and missile are connected initially and move at the same velocity. We will therefore combine their masses and change the momentum equation as follows:

    p i = p f ( m 1 + m 2 ) v i = m 1 v f 1 + m 2 v f 2 ( 5000 kg + 50 kg ) ( 275 m · s - 1 ) = ( 5000 kg ) ( v f 1 ) + ( 50 kg ) ( 700 m · s - 1 ) 1388750 kg · m · s - 1 - 35000 kg · m · s - 1 = ( 5000 kg ) ( v f 1 ) v f 1 = 270 , 75 m · s - 1

    Note that speed is asked and not velocity, therefore no direction is included in the answer.

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A bullet of mass 50 g travelling horizontally at 500 m · s - 1 strikes a stationary wooden block of mass 2 kg resting on a smooth horizontal surface. The bullet goes through the block and comes out on the other side at 200 m · s - 1 . Calculate the speed of the block after the bullet has come out the other side.

  2. We will choose to the right as positive.

  3. p i = p f m 1 v i 1 + m 2 v i 2 = m 1 v f 1 + m 2 v f 2 ( 0 , 05 kg ) ( + 500 m · s - 1 ) + ( 2 kg ) ( 0 m · s - 1 ) = ( 0 , 05 kg ) ( + 200 m · s - 1 ) + ( 2 kg ) ( v f 2 ) 25 + 0 - 10 = 2 v f 2 v f 2 = 7 , 5 m · s - 1 in the same direction as the bullet the same direction as the bullet
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Source:  OpenStax, Siyavula textbooks: grade 11 physical science. OpenStax CNX. Jul 29, 2011 Download for free at http://cnx.org/content/col11241/1.2
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