Newton's Law Of Restitution

I've done part (a) but cannot understand the method of how part (b) is done. I've written down the answer I obtained (which is correct) to part (a). Can someone help? Thanks in advance.

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** Q:**

A small smooth ball $\displaystyle A$ of mass $\displaystyle m$ is moving on a horizontal table with speed $\displaystyle u$ when it collides directly with another small smooth ball $\displaystyle B$ of mass $\displaystyle 3m$ which is at rest on the table. The balls have the same radius and the coefficient of restitution between the balls is $\displaystyle e$. The direction of motion of $\displaystyle A$ is reversed as a result of the collision.

(a) Find, in terms of $\displaystyle e$ and $\displaystyle u$. the speeds of $\displaystyle A$ and $\displaystyle B$ immediately after the collision.

In the subsequent motion $\displaystyle B$ strikes a vertical wall, which is perpendicular to the direction of motion of $\displaystyle B$, and rebounds. The coefficient of restitution between $\displaystyle B$ and the wall is $\displaystyle \frac{3}{4}$.

Given that there is a second collision between $\displaystyle A$ and $\displaystyle B$,

(b) find the range of values of $\displaystyle e$ for which the motion described is possible.

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A:

(a)

$\displaystyle v_1 = \frac{1}{4} u (3e - 1)$

$\displaystyle v_2 = \frac{1}{4}u(e + 1)$