Question

A proton flies horizontally towards a positively charged wall. It approaches the wall at a perfect zero degree angle and rebounds at the exact same speed in the opposite direction. Naturally the proton will never touch the wall. Apparently the kinetic energy in is conserved and the momentum is not, but how is that possible?

can you a) show me how this works algebraically and b) explain what is responsible for this?

Answer 1

if mass is assumed to be constant(positively charged wall), the velocity of the centre of mass of the system has to be different after the collision for the kinetic energy to be different.

However, if the momentum of the system is conserved, the velocity of the centre of mass of the system should remain the same.

1) mass is not constant and velocity is different: in a completely inelastic collision the two objects (A: m =1, B m =

Suppose that $v_a = 6m/s$ and $v_b = E_k = 0.5 * 6^2 = 18, p_a = 1 * 6 = 6, v_{cm} = p/M = 2$

Momentum is conserved: $ p_{ab} = 6$ , from this datum you can calculate its velocity: $$v_{ab} = v_{cm}= \frac{6}{3} = 2$$ and $E_k = 0.5 * 2^2 *3 = 6 E_a = 2 + E_b = 4$.

Velocity of center of mass is the same, although KE has changed.

Please note that momentum is conserved because we are assuming that there is no friction.

...how can there be a change in kinetic energy of the system if there is no change in momentum?

A change of KE without a change of momentum is not only possible but very frequent, because as you noted p = mv momentum varies linearly and KE quadratically. You can get the same product by a wide range of factors: 6 = 6*1, = 3*2, = 2*3, = 1*6, = 0.5*12, etc., different factors give same momentum

All these factors give same values for m*v, but as the figure for v must be squared, you get all different values between momentum and energy, therefore the same factors give momentum = 6, but KE =3, =6, =9, =18, =72, etc, same momentum corresponds to many different values of KE

I hope this clarified all your doubts