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The Question

A particle of mass m is moving in the +x direction with a speed v. It is approaching a massive wall which is also moving in the +x direction, but with a speed u < v (see figure)

The particle is reflected from the wall. Assuming that the mass of the wall is very much greater than m we can neglect the change in the wall’s velocity in this collision. In the frame of reference in which the wall is stationary the magnitudes of the particle’s initial and final velocities are equal. Assuming u << v and that all velocities are non-relativistic, show that in the original frame of reference the changes in particle momentum p and kinetic energy KE are given by $∆p \approx −2mv$ and $∆KE \approx −2muv$.

Particle with velocity v and the wall receding at velocity u

My answer

The momentum change is as follows,

$$\Delta p = -m(v-u) - mv \approx -2mv \quad \text{as} \quad v \gg u$$

To calculate the kinetic energy change, I did the following:

$$\Delta KE = \frac{1}{2}m(v-u)^2 - \frac{1}{2}mv^2$$ $$\Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mv^2 - mvu + \frac{1}{2}mu^2$$ $$\Delta KE \approx -mvu$$

However this omits the factor of 2 in the question, where does this come from?

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    $\begingroup$ Your rebound velocity, $-v+u$, is incorrect. $\endgroup$ – Farcher Apr 5 at 11:36
  • $\begingroup$ What is the correct rebound velocity? $\endgroup$ – eineuler Apr 5 at 13:12
  • $\begingroup$ For an elastic collision there is a relationship between the speed of approach and the speed of separation. $\endgroup$ – Farcher Apr 5 at 13:27
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Start from the reference system of the wall; the speed of the ball is $v_w = v-u$

When it rebounces, still in this system, the speed is $v_{w \, after} = -v_w = u-v$

Go to the external reference, the speed of the rebounced ball is $v_{ext \, after} = v_{w \, after} + u = -v_w +u = u-v +u = -v +2w $ the 2 should fix your issue.

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