-2
$\begingroup$

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.

$\endgroup$

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

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

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, AccidentalFourierTransform, Kyle Kanos, Jon Custer, sammy gerbil
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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". $\endgroup$ – Kyle Kanos Jan 7 '18 at 0:48
  • $\begingroup$ @KyleKanos Thanks for clarifying this. I assumed it was like MSE where questions of this type are very much encouraged. $\endgroup$ – A. Goodier Jan 7 '18 at 10:49
  • $\begingroup$ 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? $\endgroup$ – A. Goodier Jan 8 '18 at 16:35
-1
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$
  • General Case - Two masses $m_1$ and $m_2$ with initial velocities $v_1$ and $v_2$ respectively collide (with convention that mass [1] is to the left of [2] and that $v_1>v_2$). They exchange an impulse (momentum) of magnitude $J$ such that after the collision their speeds are $$ \begin{cases} v_1 - \frac{J}{m_1} & & \mbox{body [1]} \\ v_2 + \frac{J}{m_2} & & \mbox{body [2]} \end{cases}$$ 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: $$ \begin{cases} v_1 - \frac{m_2 (1+e) (v_1-v_2)}{m_1+m_2} & & \mbox{body [1]} \\ v_2 + \frac{m_1 (1+e) (v_1-v_2)}{m_1+m_2} & & \mbox{body [2]} \end{cases}$$

  • 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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.