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January 29th, 2017, 03:20 AM  #1 
Member Joined: Jun 2015 From: Warwick Posts: 34 Thanks: 1  The coefficient of restitution for $B$, with speed $v_B=3u$, after a collision.
Two particles, $A$ of mass $2m$ and $B$ of mass $m$, moving on a smooth horizontal table in opposite directions with speeds $5u$ and $3u$ respectively, collide directly. Find their velocities after the collision in terms of $u$ and the coefficient of restitution $e$. Show that the magnitude of the impulse exerted by $B$ on $A$ is $$\frac{16}{3}mu(1+e)$$ Find the value of $e$ for which the speed of $B$ after the collision is $3u$. The main question here is that the value of $e$ should be $\frac{2}{7}$, but I get $\frac{1}{8}$. Workings so far: For the first part, using conservation of momentum and taking movement of the particles to the right as positive: $$(2m . 5u) + (m .(3u) = 2mv_A + mv_B$$ from which we get $$7mu = 2mv_A + mv_B$$ and Newton's Experimental Law: $$v_B  v_A = e(5u  (3u))$$or $$v_B  v_A = 8eu$$ From these equations we have:$$ v_A = \frac{u}{3}(78e)$$ and $$ v_B = \frac{u}{3}(7+16e)$$ and the magnitude of the impulse exerted by $B$ on $A$, using $I = mv_A  mu_A$ is: $$I= (2m.5u)  2m(\frac{u}{3}(78e))= \frac{16}{3}mu(1+e)$$ Now, for $e$, when the speed of $B$ after the collision is $v_B = 3u$: using $$v_B  v_A = e(u_A  u_B)$$ and substituting the values in and solving for $e$ we get: $$ 3u  \frac{u}{3}(78e) = e(5u  (3u))$$ $$\frac{9u}{3} \frac{7u}{3} + \frac{8u}{3} = 8eu$$ $$\frac{9u}{3} \frac{7u}{3} + \frac{8u}{3} = \frac{24}{3}eu$$ $$2u = 16eu$$ $$e=\frac{1}{8}$$ (the value of $e$ should be $\frac{2}{7}$, but I get $\frac{1}{8}$). Can anyone get $e=\frac{2}{7}$? Last edited by Statistics132; January 29th, 2017 at 03:25 AM. 
January 29th, 2017, 05:02 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,298 Thanks: 1129 
agree with your determination of $e$ ... $p_0 = 2m(5u) + m(3u) = 7mu$ $p_f = 7mu = 2m(v_f) + m(3u) \implies v_f = 2u$ $e=\dfrac{\text{opening speed}}{\text{closing speed}} = \dfrac{u}{8u} = \dfrac{1}{8}$ I also noted that $e=\dfrac{1}{8}$ works in the given impulse expression ... $\Delta p_A = 4mu10mu = 6mu \implies J_{BA}=6mu$ $\dfrac{16}{3}mu \left(1+\dfrac{1}{8}\right)=6mu$ 

Tags 
$b$, $vb3u$, coefficient, collision, mechanics, restitution, speed 
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