Velocity not affecting heat produced by impact

Question

A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 $\text{m s}^{–1}$. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?

The answer given in the book I am working from is that heat produced due to impact will be same in both the cases (when the elevator is moving with uniform velocity and when it is stationery) but I am getting different answers.

Why is the amount of heat produced the same irregardless of velocity? I don't follow the reasoning.

Answer 1

Since all frames of reference are equally valid, in a uniformly descending, non accelerating elevator, all physical occurrences would be indistinguishable from an elevator considered at rest. So in both cases the bolt would release the same energy on impact with the floor. If you were in the moving elevator, you would see the bolt start it's fall at 0 velocity, fall 3 m at about 9.8 meters per second squared, and hit the floor. You would see the same in an elevator considered at rest. An external observer on the ground would see the bolt fall for the same amount of time in both cases, since no relativistic speeds are involved.

Answer 2

I do not have the full answer, but here is what I found out so far:

If we take the elevator as a frame of reference, it is clear that everything would be identical and thus the heat produced from the impact would be the same for a stationary elevator. However, if we consider the ground as a reference frame, it would be different as you have calculated.

Although energy (kinetic in particular) is not the same when seen from different references ( https://www.physicsforums.com/threads/change-in-kinetic-energy-between-reference-frames.728195/) the energy conservation should still hold (and it should yield the same heat energy produced) (Kinetic energy with respect to different reference frames) but this is only under the condition that momentum is conserved.

So I think the problem here is that the law of conservation of momentum is not followed, since the elevator speed remains constant after the impact. Perhaps if you take your system to be the elevator and the bolt. Applying the law of conservation of momentum at the impact , the final speed would not be equal to the initial speed but rather a bit higher. I think if it is modelled like that, and both the changes in energies of the elevator and the bolt are considered, you would end up with the same value of heat generated whether the elevator is stationary or moving at const. speed.

Answer 3

In the frame of the lift, the bolt is at rest initially, then accelerates at 10ms^-2 over 3m until it hits the floor.

From the relation v²=u² + 2as the velocity of the bolt just before impact is sqrt(2*g*3m) = sqrt(60) = 7.74596669, and its KE is 9J. All of the KE is converted into heat energy.

The result has been calculated in the frame of the lift, so it will be the same regardless of how the lift appears to be moving to observers in other reference frames (assuming the speeds involved are sufficiently low that relativistic effects can be ignored).

You might have been confused by the fact that from some other reference frame in which the lift is moving, the gain in KE appears to be different. To take the example given in the question, if the lift is moving at 7m/s, then to a stationary observer the bolt appears to accelerate from 7m/s to 14.74596669m/s. Its KE as measured in that reference frame changes from 7.35J at the start to c32.6J just before impact, an increase of 25.26J, which is greater than the KE it gained when measured from the rest frame of the lift, and you might wonder why all of that increase KE is not converted into heat during the impact. The answer is that in the rest frame of the lift the lift does not recoil from the collision, so the lift gains no KE from the collision. However, in any other frame the recoil of the lift needs to be taken into account. When that is done, you find that some of the KE lost by the bolt is transferred into KE of the lift- the amount available to convert into heat remains as it was in the lift's rest frame.

You will see some answers which say that the bolt starts at one velocity and ends at the same velocity, so it gains no KE, which means that all of the PE lost is converted into heat. That answer is misleading- the bolt gains KE continually as it falls, and it is the KE gained by the bolt which is converted into heat at the moment of impact.