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A particle P of mass 3m is moving in a straight line with speed u on a smooth horizontal table , a second particle Q of mass 2m is moving with speed 2u in the opposite direction of P on the same straight line .The particle P collides directly with the particle Q , the coefficient of restitution between P and Q is e , Show that the direction of motion of P is reversed as a result of the collision . My turn: let $u_P=u , u_Q = -2u $ $$e = \frac{v_Q - v_P}{u_Q - u_P}= \frac{v_Q -v_P}{-2u-u}= \frac{v_Q -v_P}{-3u}$$then $$v_Q -v_P = -3u e$$ we have $$M_P u_P + M_Q u_Q = M_P v_P + M_Q v_Q$$ $$3m\times u + 2m\times (-2u) = 3m v_P + 2m v_Q$$ $$-u = 3 v_P + 2 v_Q $$ By solving the two equations we get $$v_P = u e $$ which is positive quantity, so the direction of motion of $P$ is not reversed! Is there any mistake with my solution?

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The error that you have made is in your first set of equations and the problem probably arises because you are using relative velocities.

Look of a very simple case of two equal masses and an elastic collision between them so that the coefficient of restitution $e=1$.

Assume that mass $1$ is travelling to the right with velocity $u\, \hat r$ and is going to hit stationary mass $2$.
The velocity of mass 1 relative to mass $2$ is $u\, \hat r$.

After collision mass 1 stops and mass 2 moves off to the right with velocity $u\, \hat r$.
The velocity of mass 1 relative to mass $2$ is now $-u\, \hat r$ and you will note the change of sign.

Thus the ratio of the velocity of mass 1 relative to mass $2$ after collision to the velocity of mass 1 relative to mass $2$ before the collision is $-1$ not the expected, using your formula, $+1$.

So in fact if you are dealing with velocities your first equation should be $$e = -\left (\frac{v_Q - v_P}{u_Q - u_P}\right )= -\left(\frac{v_Q -v_P}{-2u-u}\right )= \frac{v_Q -v_P}{3u}$$

Another way of using the coefficient of restitution is to use the ratio of the relative speed of separation to the relative speed of approach.
In your case the speed of approach would be $3u$ (both particles are approaching each other) and the speed of separation would be $v_{\rm Q}-v_{\rm P}$ (taking the initial direction of particle $P$ as positive with $v_{\rm P}$ and $v_{\rm Q}$ as components of the velocities in that direction, after the collision particle $Q$ is "trying to" move away from particle $P$ and particle $P$ is "trying to" move towards particle $Q$.

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  • $\begingroup$ Do you mean that $-\frac{v_Q - v_P}{u_Q - u_P}$ ? @Farcher $\endgroup$ – Hussien Mohamed Jul 13 at 12:43

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