Momentum
Forces and Motion

Explosion of two trolleys

Practical Activity for 14-16 PRACTICAL PHYISCS

Class practical

Another example of an inelastic interaction.

Apparatus and Materials

  • Dynamics trolleys, 2
  • Blocks of wood (to act as stops), 2

Health & Safety and Technical Notes

Read our standard health & safety guidance

Runways are not required for this experiment. Use the normal bench-top. A block of wood at each end of the bench acts as a stop. Position the blocks so that each trolley moves the same distance after the explosion before hitting its stop.

The buffer rods in the trolleys are pushed in against the built-in compression springs.

Procedure

  1. Put two dynamics trolleys together in the middle of the bench.
  2. Release one of the buffer rods by a smart tap on the vertical release rod. The trolleys will fly apart.

Teaching Notes

  • As the trolleys will be moving in opposite directions, it is not possible to compensate for friction. Therefore, it is not worthwhile to use ticker-tape for this experiment. Instead, blocks of wood are used to act as stops at measured distances along the bench. Since this experiment deals with speeds (for momenta) rather than accelerations, this method of comparing distances travelled will suffice. With trolleys of equal masses, the trolleys hit their stops simultaneously.
  • The experiment can be repeated by doubling the mass of one trolley (put an extra trolley on top). Students now decide for themselves how to rearrange the blocks (or the starting point) so that the trolleys again hit the blocks simultaneously. They can find, from the masses and distances, whether the trolleys acquired equal and opposite amounts of momentum.
  • This is an interesting experiment because the total momentum before the explosion is zero. Because of the vector nature of momentum, the total momentum after the explosion is also zero.
  • Give some other examples which use the idea of recoil, e.g.:
    • Firing a cannon ball
    • Firing a bullet from a rifle
    • Pushing a boat away from a bank.
Momentum
appears in the relation p=mv F=dp/dt λ=h/p ΔxΔp>ℏ/2
has the special case Angular Momentum
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