Elastic collision of trolleys

Class practical

Students investigate what happens when two trolleys collide and no energy is lost in the collision.

Apparatus and materials

Per student pair

Dynamics trolleys, 3

Runway

Ticker-tape vibrator

Health & Safety and Technical notes

The runways are heavy and long. They need to be handled with care to avoid damage to students on nearby equipment. Runways are best lifted into position by 2 people. Place a barrier to ensure the trolleys do not roll off the end of the bench.

Procedure

1 Two vibrators can be used, one for each tape, if desired.

2 Multiflash photographs can be taken of the collision or use a camcorder and play back frame by frame. A dowel covered in aluminium foil is attached to the trolleys. The light should point down the track. Position the camera (with mechanical strobe if needed) opposite where the collision will take place.

a Set up the trolley board and compensate it for friction. It can only be compensated if the trolleys continue to move in the same direction.

b Put two trolleys on the board, each with a ticker-tape attached. Run both tapes under the same vibrator, but with a separate carbon paper for each. Place one trolley at rest about halfway along the runway. Hold the other trolley at the beginning of the runway.

c Give the trolley at the top of the runway an initial push, leaving it to run with constant velocity until it collides with the other trolley.

d Use the ticker-tapes to calculate the speed of the moving trolley before the collision, and of each of the speeds of the two trolleys after collision.

e Calculate the total forward momentum before and after the collision.

Teaching notes

1 You could have the following discussion before or after the experiment:

Here are two things which are going to collide. If during the collision A pushes B to the right and B pushes A to the left with equal and opposite forces, then B receives that force F from A and has a change in momentum (F)t and A has a change of momentum given by this force (-F)t.

The time during which A pushes B must be just the same as the time during which B pushes A. So we have changes in momentum of Ft and –Ft. The total change in momentum is Ft +(-Ft) = 0 = (mv)B + (mv)A. The total momentum in any kind of interaction, in a closed system, is therefore always the same. We say momentum is 'conserved'. This is an application of Newton's third law.

Of course all participants must be included. When a car comes to rest it loses a lot of momentum and that momentum seems to disappear but the Earth gains an equal amount of momentum carried off by friction. The accounting is harder to do and so in elementary physics this momentum loss is rarely calculated but is discussed as part of the complete system. Even momentum in electromagnetic fields needs to be included.

2 If the trolleys have equal masses and are arranged with spring buffers so that the collision is elastic, the moving trolley is brought practically to rest and the stationary trolley picks up the full motion. It is probably better to start with a collision of unequal masses – otherwise velocities are confused with momenta. The trolley that is used as a projectile should be loaded up with an equal trolley so that it has twice the mass of the stationary trolley.

3 The relative masses on the trolleys can be varied either by piling up more trolleys or loading them with weights and weighing the trolley and weights on a balance.

On collision the trolleys do not stick together, but the lighter one bounces forward with greater speed while the original one moves more slowly after the collision. Ticker-tape records are obtained of the motion both before and after the collision and these are analyzed to see whether momentum is conserved or not.

4 In these experiments, if the friction compensation is well done, the colliding bodies have constant speeds before collision and constant speeds (of different size) after collision. Before the experiment, make it clear to students that they are to measure speeds, not accelerations. The main concern is whether momentum is conserved in the collision.

5 Either 'hard' collisions or collisions using the spring-loaded rod on the trolley can be used. Both types of collision should be tried, to see whether the time of collision makes any difference to the conservation.

Results:
i A double trolley mass, 2 M collides with a single trolley mass M.

The same force F causes the first trolley to slow down and the second trolley, originally stationary, to speed up while the spring is in contact with both of them. The speed of both trolley sets is determined by their masses.

ii A single trolley collides with a stationary single trolley. In this case their masses are the same and the moving trolley slows down and stops and the originally stationary one speeds up until its velocity is the same as the one which was originally moving

iii Avoid using a single trolley colliding with a double trolley, because the single trolley bounces back and tangles the ticker-tape whilst the double trolley moves off at a slow velocity.

iv Hard collisions, without the spring extended, show what happens when the time of interaction is reduced.