# Coefficient of restitution question - four particles colliding [closed]

Four particles $A,B,C,D$ are equally spaced on a smooth horizontal table. Their masses are $\lambda m, m, m, m$ respectively. $A$ and $D$ are simultaneously projected, both with speed $u$, and collide with $B$ and $C$ respectively. In the following collision between $B$ and $C$, $B$ is brought to rest. The coefficient of restitution in each collision is $e$. Show that $e=\dfrac{\lambda-1}{3\lambda +1}$.

My attempt at a solution:

Collision between $A$ and $B$:

Let $v_1$ and $v_2$ be the speeds of $A$ and $B$ respectively in the same direction (right).

Conservation of momentum: $\lambda mu=\lambda mv_1+mv_2 \iff \lambda u=\lambda v_1+v_2$.

Coefficient of restitution: $e=\dfrac{v_2-v_1}{u}=\dfrac{\lambda(u-v_1)-v_1}{u}$.

Collision between $C$ and $D$:

Let $v_3$ and $v_4$ be the speeds of $D$ and $C$ respectively afterwards in the same direction (left).

Conservation of momentum: $mu=mv_3+mv_4 \iff u=v_3+v_4$.

Coefficient of restitution: $e=\dfrac{v_4-v_3}{u}=\dfrac{2v_4-u}{u}$.

Collision between $B$ and $C$:

Let $v_5$ be the speed of $C$ after the collision.

Conservation of momentum: $mv_2-mv_4=mv_5 \iff v_2-v_4=v_5$.

Coefficient of restitution: $e=\dfrac{v_5}{v_2-v_4}$.

At least one of these last two equations must be wrong, since they give $e=1$, but I can't see why. Where have I gone wrong? Are some of the other equations wrong?

I don't need a full solution - I'm sure I can get the answer once I've got the right equations.

## closed as off-topic by John Rennie, AccidentalFourierTransform, Kyle Kanos, Jon Custer, sammy gerbilJan 8 '18 at 1:10

This question appears to be off-topic. The users who voted to close gave this specific reason:

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• Please note that this site is not a place to obtain solutions to worked problems. Please see this Meta post on asking homework-like questions and this Meta post for "check my work problems". – Kyle Kanos Jan 7 '18 at 0:48
• @KyleKanos Thanks for clarifying this. I assumed it was like MSE where questions of this type are very much encouraged. – A. Goodier Jan 7 '18 at 10:49
• How could I have reworded this question to make it about a specific physics concept? Ask for clarification that I am calculating the coefficient of restitution correctly? – A. Goodier Jan 8 '18 at 16:35

The very last equation is wrong. The coefficient of restitution should be $\dfrac{v_5}{v_2+v_4}$ since the speeds $v_2$ and $v_4$ are in opposite directions.

• General Case - Two masses $m_1$ and $m_2$ with initial velocities $v_1$ and $v_2$ respectively collide (with convention that mass  is to the left of  and that $v_1>v_2$). They exchange an impulse (momentum) of magnitude $J$ such that after the collision their speeds are $$\begin{cases} v_1 - \frac{J}{m_1} & & \mbox{body } \\ v_2 + \frac{J}{m_2} & & \mbox{body } \end{cases}$$ The impulse is found from the coefficient of restitution $e$, the reduced mass $\frac{m_1 m_2}{m_1+m_2}$ of the pair and the relative impact speed $v_1-v_2$ $$J = (1+e) \frac{m_1 m_2}{m_1+m_2} (v_2-v_1)$$ For the general case the final speeds are thus: $$\begin{cases} v_1 - \frac{m_2 (1+e) (v_1-v_2)}{m_1+m_2} & & \mbox{body } \\ v_2 + \frac{m_1 (1+e) (v_1-v_2)}{m_1+m_2} & & \mbox{body } \end{cases}$$

• Collision of A into B - $m_1=\lambda m$, $m_2=m$, $v_1=v$, $v_2=0$ $$\begin{cases} v - \frac{(1+e) v}{\lambda+1} & & \mbox{body A} \\ 0 + \frac{\lambda (1+e) v}{\lambda + 1} & & \mbox{body B} \end{cases}$$

• Collision of D into C - $m_1= m$, $m_2=m$, $v_1=0$, $v_2=-v$ $$\begin{cases} 0 - \frac{(1+e) v}{2} & & \mbox{body C} \\ -v + \frac{ (1+e) v}{2} & & \mbox{body D} \end{cases}$$

• Collision of B and C - $m_1= m$, $m_2=m$, $v_1=0 + \frac{\lambda (1+e) v}{\lambda + 1}$, $v_2=0 - \frac{(1+e) v}{2}$ $$\begin{cases} - \frac{(1+e) v (e (3\lambda+1)-\lambda+1)}{4 (\lambda+1)} & & \mbox{body B} \\ \frac{ (1+e) v (e (3 \lambda +1)+\lambda-1)}{4 (\lambda+1)} & & \mbox{body C} \end{cases}$$

Now to make the final velocity of [B] zero you need

$$e (3\lambda+1)-\lambda+1) = 0$$

or $$\boxed{ e = \frac{\lambda-1}{3 \lambda+1} }$$

So it seems that my definition of COR is the inverse of the one used by the OP.

Relevant post Newtons Cradle, Collision Theory