Conservation of energy in collision if A hits B but fails to move B

Question

If body A hits body B but fails to move B what happens to its kinetic energy? There is no increase in potential energy. So what is accounting for this loss?

Answer 1

In a perfectly elastic collision (which is not possible in a macroscopic setting), in an ideal situation (since in reality B would have to move at least somewhat), then A would rebound with the same kinetic energy it had before. In reality, A and B would both have some kinetic energy after collision, and some kinetic energy would be lost as other forms of energy such as heat, noise, potential (macroscopically).

Answer 2

In order to satisfy the law of conservation of momentum, B has to acquire a velocity as a result of the collision. For conservation of momentum we have

$$m_{A}v_{Ai}=m_{A}v_{Af}+m_{B}v_{Bf}$$

$$v_{Bf}=\frac{m_{A}(V_{Ai}-V_{Af})}{m_B}$$

Where

$V_{Ai}$ is the initial velocity of A

$V_{Af}$, $V_{Bf}$ are the velocities of A and B after the collision.

Note that the larger the mass of B compared to A, the lower velocity of B after the collision. In order for the velocity of B to be zero after the collision, its mass, $m_B$, would have to be infinitely greater than A. So B has to move, even if very little.

As far as kinetic energy is concerned, it is only conserved in a perfectly elastic collision which would mean that the sum of the kinetic energies of A and B following the collision would equal the kinetic energy of A before the collision. That would give you a second equation with the two unknown velocities allowing you to compute the final velocities. Although collision problems often specify a perfectly elastic collision, as @Adrian Howard points out a perfectly elastic collision cannot be achieved on a macroscopic level.

If the collision were perfectly inelastic, then A would collide with B and the two would stick together with the same final velocity. That would also give you an additional equation of $V_{Af}$=$V_{Bf}$ which, when combined with the conservation of momentum equation, would allow you could calculate the final velocity of both A and B.

For anything in between a perfectly elastic and perfectly inelastic collision, you would need information on how much kinetic energy is lost (converted to heat, sound, etc.)to determine the final velocities. The coefficient of restitution would provide that information, but that's another subject.

Bottom line: The first equation for conservation of momentum, and the second derived from the first, hold as long as the two objects form an isolated system to the extent there are no external forces that can influence their momenta, and as long as they don't contain and source of potential energy that could be release as a result of the collision. In order to determine the final velocities, information is needed on the nature of the collision to obtain an additional equation for the two unknowns. Whether the collision is completely elastic, completely inelastic, or something in between, in order to satisfy conservation of momentum when you solve for the velocity of B after the collision, it will be some non zero value, no matter how small.

Hope this helps.

Answer 3

As Adrian has pointed out, the scenario you describe is an ideal scenario, but we can still analyze it. And as BobD discusses, having object $B$ not move has problems. One of which is that the velocity of object $A$ also will not change, and so your scenario describes a "collisionless collision".

A better way to approach the immovable object is to take the limits $m_B\to\infty$ and $v_B\to0$ while keeping $m_Bv_B=m_A(v_{Af}-v_{Ai})$ constant so that momentum is conserved as we take the limits. Then the kinetic energy after the collision would be $$K_f=\frac12\left(m_Av_{Af}^2+m_Bv_B^2\right)=\frac12\left(m_Av_{Af}^2+m_A(v_{Af}-v_{Ai})v_B\right)\to\frac12m_Av_{Af}^2$$

Therefore, the change in kinetic energy would only be determined by what happens to body $A$ due to the collision in this specific limit. If body A rebounds at a lower speed than it comes in with, then energy was lost due to the collision somehow, but body B remaining stationary does not necessarily mean energy was lost, at least in this ideal limit.

If there are energy losses, then Adrian's answer discusses some places the energy could go. Just because there is not a (noticeable) potential energy interaction between the two bodies does not mean the energy has nowhere to go.