Two particles A and B are on a smooth horizontal floor with B between A and a vertical wall. The masses
of A and B are 4kg and 11 kg respectively. Initially, B is at rest and A is moving towards B with a speed of
$u$ ms–1. A collides directly with B. The coefficient of restitution between A and B is e.After the collision between A and B the direction of motion of A is reversed. B subsequently collides directly with the vertical wall. The coefficient of restitution between B and the wall is $-\frac{1\:}{2\:}e$
ii) Given that there is a second collision between A and B, find the range of possible values of e
So I've attempted the question and got that A keeps moving in the positive direction with a speed of $\frac{1}{15}u\left(4-11e\right)$
The speed of B after it collides with the wall is $-\frac{2}{15\:}eu\left(1+e\right)$. So shouldn't that mean there will be a collision as long as the velocity of A is positive, ie, if $4-11e>0$?
The mark scheme says that they only collide again if
$4-11e<0$. I don't understand how that makes sense. Wouldn't the velocity of A then be negative, as in, it's moving away from B? In that case the velocity of B would have to be greater than the velocity of A right? But then the mark scheme says they only collide is the velocity of A is greater than B.
$\frac{1}{15}u\left(4-11e\right)>-\frac{2}{15}eu\left(1+e\right)$
Is there perhaps something I'm missing here?
Best Answer
Per situation description, the direction of A is reversed after the A-B collision. Thus $4−11e<0$ so that the negative forward speed gives a backwards motion.
Then after the wall collision $B$ has (still) to be faster than $A$, requiring $$ 2e(1+e)>(11e-4) >0, $$ assuming that your computations of the collusions are correct.