Magnitude of Total Momentum

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  • #1
justagirl
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Say there was 2 collisions. There are 2 carts of identical mass. In the first case cart A starts at height h above the ground and crashes into cart B which is stationary on the flat ground. They stick together.

In the second case, cart A starts at height h/2 above the ground, cart B starts at height h/2 above on the ground on the other side. They fall in opposite directions and collide and stick together.

The problem asks that when the carts reach the flat ground, in which case is the magnitude of the total momentum greater.

When they ask about the magnitude, in case 2, should the momentum be
M(a)V(a) + M(b)V(b) = 2M(a)V(a) (which would be greater than case 1), or should I factor in the fact that they are traveling in opposite directions, in which case the total momentum would be zero?

The second question asks whether the total energy lost in greater in case 1 or case 2. The energy lost would be greater in case 2 if the momentum was 2M(a)V(a)... but then again this goes to question 1.

Thanks!
 

Answers and Replies

  • #2
Steve4Physics
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When they ask about the magnitude, in case 2, should the momentum be
M(a)V(a) + M(b)V(b) = 2M(a)V(a) (which would be greater than case 1), or should I factor in the fact that they are traveling in opposite directions, in which case the total momentum would be zero?

At the time of writing, this is a 17+ year old question. So I'm giving a fairly complete answer/explanation.

You must account for the opposite directions. The total momentum for case-2 is therefore zero. That's because momenta are vectors; they must be added as vectors. (If the magnitude of the resultant is needed, this is found after the vector-addition.)

That means forcase-2, the carts (now stuck together) will end-up stationary. You can reach the same conclusion by thinking about symmetry. There is no preferred direction of motion for the combined carts in case-2.

The second question asks whether the total energy lost in greater in case 1 or case 2. The energy lost would be greater in case 2 if the momentum was 2M(a)V(a)... but then again this goes to question 1.
This is badly worded (or maybe a trick) question. Energy is always conserved. So the total energy loss in both cases is zero.

However, the question was probably intended to be about loss of mechanical energy (ME). Mechanical energy is potential energy + kinetic energy.

In both cases, the starting ME is the same:
- for case-1 it is mgh+ 0 = mgh
- for case-2 it is mg(h/2) + mg(h/2) = mgh

In case-2 all the ME is lost, as the carts end-up stationary. (The ME gets entirely converted to heat (and maybe a little sound energy) as a result of the collision.)

In case 1, the stuck-together cars are still moving. So not all of the ME has been converted to heat.

So the loss of ME is greater in case 2.
 

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