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Conservation of Momentum (Oxford AQA IGCSE Physics)

Revision Note

Leander Oates

Conservation of Momentum

  • The principle of conservation of momentum states that:

In a closed system, the total momentum before an interaction is equal to the total momentum after an interaction

  • In this context, an interaction can be either:

    • A collision i.e. where two objects collide with each other

    • An explosion i.e. where a stationary object explodes into two (or more) parts

Collisions

  • For a collision between two objects:

The total momentum before a collision = the total momentum after a collision

Momentum before and after a collision

Before the collision mass m has a velocity u to the right, and mass M is stationary, therefore the momentum before the collision is m times u. After the collision, mass m has a velocity v to the left, and mass M has a velocity V to the right. Therefore, the momentum after the collision is M times V minus m times v
The momentum of a system before and after a collision where the momentum before equals the momentum after

  • Before the collision:

    • In the example above, the total momentum is m × u

      • The direction to the right is taken as positive

      • Mass M is stationary, therefore it has no momentum, M × 0 = 0

  • After the collision:

    • In the example above, the total momentum is (M  × V) + (m  × -v)

      • Mass M also now has momentum

      • The velocity of m is now -v  (since it is now travelling to the left) and the velocity of M  is V

      • The total moment can written more simply as (M  × V) – (m × v)

Explosions

  • Examples of explosions include:

    • A person jumping off a skateboard

    • A cannon or bullet being fired

    • A firework exploding

    • A person being thrown from a vehicle

    • The separation of space vehicles

  • As with collisions, the conservation of momentum equations can be applied:

    • The mass of the separate objects = the mass of the original object

    • The initial velocity (and therefore, momentum) of the stationary object is 0

    • Total momentum before explosion = total momentum after explosion

  • Remember that velocity is a vector quantity with both magnitude and direction

Momentum of a dynamite explosion

A stationary object before an explosion has a momentum of zero. After the explosion, fragments of the object are scattered in all directions. The momentum after the explosion is also zero.
The sum of the masses of the exploded pieces equals the mass of the original dynamite and the sum of the velocities of the separate pieces equals zero.

  • In the example above, some scattered fragments will travel in a positive direction and others in a negative direction

  • When the velocities of all the scattered pieces are added together the resultant velocity = 0

Worked Example

A car and a van collide. The diagram shows the car and van, just before and just after the collision. Before the collision, the van was at rest.

Before the collision, a car of mass 990 kg travels towards the van at a velocity of 10 metres per second. The van of mass 4200 kg is stationary. After the collision, the car has a velocity of 2 metres per second towards the van. The velocity of the van is unknown.

The car initially moves at a speed of 10 m/s, but this reduces to 2 m/s after the collision.

The mass of the car is 990 kg and the mass of the van is 4200 kg.

Calculate the velocity of the van when it is pushed forward by the collision.

Answer:

Step 1: State the principle of the conservation of momentum

Total momentum before a collision = total momentum after a collision 

Step 2: Calculate the total momentum of the car and van before the collision

  • Recall the equation for momentum:

p space equals space m v

  • Initial momentum of the car:

p subscript c a r end subscript space equals space 990 space cross times space 10

p subscript c a r end subscript space equals space 9900 space kg space straight m divided by straight s

  • Initial momentum of the van:

    • The van is at rest so v = 0 m/s

p subscript v a n end subscript space equals space 4200 space cross times space 0

p subscript v a n end subscript space equals space 0 space kg space straight m divided by straight s

  • Total momentum before collision:

 p subscript b e f o r e end subscript space equals space p subscript c a r end subscript space plus space p subscript v a n end subscript

p subscript b e f o r e end subscript space equals space 9900 space plus space 0

p subscript b e f o r e end subscript space equals space 9900 space kg space straight m divided by straight s

Step 3: Calculate the total momentum of the car and van after the collision

  • Final momentum of the car:

p subscript c a r end subscript space equals space 990 space cross times space 2

p subscript c a r end subscript space equals space 1980 space kg space straight m divided by straight s

  • Final momentum of the van:

p subscript v a n end subscript space equals space 4200 space cross times space v

  • Total momentum after collision:

p subscript a f t e r end subscript space equals space 1980 space plus space 4200 v

Step 4: Rearrange the conservation of momentum equation for the velocity of the van

p subscript b e f o r e end subscript space equals space p subscript a f t e r end subscript

9900 space equals space 1980 space plus space 4200 v

9900 space minus space 1980 space equals space 4200 v

v space equals space fraction numerator 9900 space minus space 1980 over denominator 4200 end fraction

v space equals space 1.9 space straight m divided by straight s

Worked Example

A gun of mass 3 kg fires a bullet of 30 g at a speed of 175 m/s. 

A stationary gun before firing has a momentum of zero. After firing, the bullet of mass 30 g leaves the gun with a velocity of 175 metres per second in the forward direction. The gun of mass 3 kg recoils with an unknown velocity in the backwards direction.

As the bullet is fired, the gun moves back with a recoil velocity v.

Calculate the recoil velocity of the gun v.

Answer:

Step 1: List the known quantities:

  • Velocity of gun-bullet system before firing = 0 m/s

  • Mass of gun, mgun = 3 kg

  • Mass of bullet, mbullet = 30 g = 0.03 kg

  • Velocity of bullet after firing , vbullet = 175 m/s

  • Velocity of gun after firing = v

Step 2: State the conservation of momentum equation

momentum before firing = momentum after firing

0 space equals space open parentheses m subscript g u n end subscript space cross times space v close parentheses space plus space open parentheses m subscript b u l l e t end subscript space cross times space v subscript b u l l e t end subscript close parentheses

Step 3: Substitute the known quantities into the equation

0 space equals space open parentheses 3 cross times space v close parentheses space plus space open parentheses 0.03 cross times space 175 close parentheses

Step 4: Rearrange the equation to find the recoil velocity of the gun

negative 3 v space equals space 5.25

v space equals space fraction numerator 5.25 over denominator negative 3 end fraction

v space equals space minus 1.75 space straight m divided by straight s

Examiner Tip

If it is not given in the question already, drawing a diagram of before and after helps keep track of all the masses and velocities (and directions) in the conservation of momentum questions.

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    Leander Oates

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