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When two objects collide, each one changes its velocity. One may slow down, the other may speed up, and one may even bounce backwards. But one thing does not change: the total momentum of the two objects added together.
total momentum before = total momentum after
This rule is called the conservation of momentum. "Conserved" means the total stays the same. Momentum can pass from one object to another, but none is gained and none is lost.
The rule only works when no external forces act. The forces the objects put on each other during the collision do not count. In exam questions the collision is so quick that you can ignore external forces, unless the question tells you otherwise.
Remember, the momentum of each object is its mass multiplied by its velocity, in kg m/s. To get the total, work out the momentum of each object and add them together.
Key terms:
There are three kinds of question. The method is the same for all of them.
The objects stick and move off as one object. Add their masses to get the mass afterwards.
A railway wagon of mass 3000 kg rolls along a level track at 4 m/s. It hits a stationary wagon of mass 1000 kg. The wagons couple together. Calculate their velocity after the collision.
Take the direction the first wagon is moving as positive.
Total momentum before = (3000 × 4) + (1000 × 0) = 12 000 kg m/s
Mass after = 3000 + 1000 = 4000 kg
Total momentum after = 4000 × v
4000 × v = 12 000
v = 12 000 / 4000
v = 3 m/s in the direction the first wagon was moving
The same momentum is now shared by a bigger mass, so the velocity is smaller. The stationary wagon had zero momentum before, but you should still write it in your working.
When the objects do not stick, there are two velocities afterwards. The question gives you one of them and asks for the other.
A basketball of mass 0.5 kg rolls to the right at 4 m/s. It hits a stationary medicine ball of mass 2 kg. The basketball bounces straight back at 2 m/s. Calculate the velocity of the medicine ball after the collision.
Take right as positive. After the collision the basketball moves to the left, so its velocity is −2 m/s.
Total momentum before = (0.5 × 4) + (2 × 0) = 2 kg m/s
Total momentum after = (0.5 × (−2)) + (2 × v) = −1 + (2 × v)
−1 + (2 × v) = 2
2 × v = 3
v = +1.5 m/s
The answer is positive, so the medicine ball moves to the right.
This is where the signs matter. If you forget the minus sign for the basketball, you get the wrong answer of 0.5 m/s. You can check the right answer: afterwards the medicine ball has 2 × 1.5 = +3 kg m/s and the basketball has −1 kg m/s. Add them and you get 2 kg m/s, the same as before.
In the explosions you will meet in this course, everything starts at rest. Nothing is moving, so the total momentum before is zero. Momentum is conserved, so the total momentum after must also be zero.
How can moving objects have a total momentum of zero? They move in opposite directions. One part has positive momentum and the other has negative momentum of exactly the same size. Add them together and they cancel out.
This explains recoil. When a cannon fires a ball forwards, the cannon rolls backwards. The cannon has a much bigger mass than the ball, so it moves back much more slowly.
A stationary cannon of mass 500 kg fires a ball of mass 5 kg at 100 m/s. Calculate the recoil velocity of the cannon.
Take the direction of the ball as positive.
Total momentum before = 0 kg m/s
(5 × 100) + (500 × v) = 0
500 × v = −500
v = −1 m/s
The minus sign shows that the cannon moves backwards.
A child of mass 30 kg jumps off a stationary raft at 2 m/s. The raft moves the opposite way at 0.5 m/s. Calculate the mass of the raft.
Take the direction of the child as positive, so the raft has a velocity of −0.5 m/s.
Total momentum before = 0 kg m/s
(30 × 2) + (m × (−0.5)) = 0
0.5 × m = 60
m = 120 kg
Some questions ask for a momentum, not a velocity. In worked example 4 the child has a momentum of +60 kg m/s, so the momentum of the raft must be −60 kg m/s. The two must add up to zero.
The specification suggests an investigation of momentum during collisions, using trolleys and light gates. It is an optional extra, not a required practical, but it helps to know how it works.
In a real experiment the momentum after is often a little smaller. Friction is an external force, and it slows the trolleys between the light gates.
(a) State what is meant by the conservation of momentum. [2 marks]
(b) A van of mass 1500 kg travels along a straight road at 12 m/s. It collides with a stationary car of mass 900 kg. The two vehicles lock together.
(i) Calculate the momentum of the van before the collision. [2 marks]
(ii) Calculate the velocity of the van and the car just after the collision. [3 marks]
(c) A firework is at rest for a moment at the top of its flight. It explodes into two pieces. Piece A has a mass of 0.3 kg and moves to the right at 20 m/s. Piece B has a mass of 0.2 kg.
(i) State the total momentum of the two pieces just after the explosion. [1 mark]
(ii) Calculate the velocity of piece B just after the explosion. [3 marks]
(a) The total momentum before a collision or explosion is equal to the total momentum after it (1), as long as no external forces act (1).
(b)(i) p = m × v = 1500 × 12 (1)
p = 18 000 kg m/s (1)
(b)(ii) Mass after = 1500 + 900 = 2400 kg (1)
2400 × v = 18 000 (1)
v = 18 000 / 2400 = 7.5 m/s (1)
(c)(i) Zero, 0 kg m/s (1)
(c)(ii) Taking right as positive: momentum of A = 0.3 × 20 = 6 kg m/s (1)
6 + (0.2 × v) = 0 (1)
v = −6 / 0.2 = −30 m/s, so piece B moves to the left at 30 m/s (1)
In part (b)(ii) the vehicles lock together, so divide by the total mass of 2400 kg, not 1500 or 900. In part (c)(ii) start by writing "total momentum before = 0". The minus sign then appears by itself.