Problems

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Solution: We can `ignore' the fact that they are both accelerating, because the acceleration is the same for both object so it will "cancel" out. Therefore the time taken is the same as if the object has to travel distance \(d\) at speed \(u\), ie \(d/u\). \begin{align*} && u_A &= u - g \frac{d}{u} \\ && u_B &= -g\frac{d}{u} \end{align*}

The speed of approach is \(u\), therefore the speed of separation is \(eu\), in particular \(v_B = v_A + eu\) \begin{align*} \text{COM}: && m\left (u-g\frac{d}{u} \right)+m\left (-g\frac{d}{u} \right) &= mv_A + m(v_A + eu) \\ \Rightarrow && 2v_A &= u - 2g\frac{d}{u}-eu \\ \Rightarrow && v_A &= \frac12 (1-e)u - \frac{gd}{u} \\ \Rightarrow && v_B &= \frac12 (1+e)u - \frac{gd}{u} \\ \\ && \text{initial k.e.} &= \frac12 m \left (u-g\frac{d}{u} \right)^2 + \frac12 m \left (-g\frac{d}{u} \right)^2 \\ &&&= \frac12m \left (u^2 -2gd + \frac{2g^2d^2}{u^2} \right) \\ && \text{final k.e.} &= \frac12 m \left ( \frac12 (1-e)u - \frac{gd}{u}\right)^2 + \frac12 m \left ( \frac12 (1+e)u - \frac{gd}{u}\right)^2 \\ &&&= \frac12 m \left (\frac14 \left ( (1-e)^2+(1+e)^2\right)u^2 - gd \left ((1-e)+(1+e) \right) +\frac{2g^2d^2}{u^2}\right) \\ &&&= \frac12 m \left (\frac12(1+e^2)u^2-2gd+ \frac{2g^2d^2}{u^2}\right) \\ \Rightarrow && \text{loss k.e.} &= \frac12m \left ( u^2 - \frac12(1+e^2)u^2\right) \\ &&&= \frac14mu^2(1-e^2) \end{align*}