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January 29th, 2017, 02:20 AM   #1
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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:$$ v_A = \frac{u}{3}(7-8e)$$ 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:

$$I= (2m.5u) - 2m(\frac{u}{3}(7-8e))= \frac{16}{3}mu(1+e)$$

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:

$$ 3u - \frac{u}{3}(7-8e) = e(5u - (-3u))$$
$$\frac{9u}{3} -\frac{7u}{3} + \frac{8u}{3} = 8eu$$
$$\frac{9u}{3} -\frac{7u}{3} + \frac{8u}{3} = \frac{24}{3}eu$$
$$2u = 16eu$$
$$e=\frac{1}{8}$$

(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 02:25 AM.
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January 29th, 2017, 04:02 AM   #2
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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 ...

$\Delta p_A = 4mu-10mu = -6mu \implies J_{BA}=6mu$

$\dfrac{16}{3}mu \left(1+\dfrac{1}{8}\right)=6mu$
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$b$, $vb3u$, coefficient, collision, mechanics, restitution, speed



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