1 problem found
A ball of mass \(m\) is thrown vertically upwards from the floor of a room of height \(h\) with speed \(\sqrt{2kgh},\) where \(k>1.\) The coefficient of restitution between the ball and the ceiling or floor is \(a\). Both the ceiling and floor are level. Show that the kinetic energy of the ball immediately before hitting the ceiling for the \(n\)th time is \[ mgh\left(a^{4n-4}(k-1)+\frac{a^{4n-4}-1}{a^{2}+1}\right). \] Hence show that the number of times the ball hits the ceiling is at most \[ 1-\frac{\ln[a^{2}(k-1)+k]}{4\ln a}. \]
Solution: \begin{align*} && \text{energy when projected} &= \frac12 m(2kgh) \\ &&&= kghm \\ && \text{energy when hitting ceiling the first time} &= mgh + \frac12 m v^2 \\ \text{COE}: && kghm &= mgh + \frac12 mv^2 \\ \Rightarrow && v^2 &= 2gh(k-1) \end{align*} It will rebound with speed \(\sqrt{2gh(k-1)}a\). \begin{align*} && \text{energy when rebounding from ceiling} &=gh(k-1)a^2 + mgh \\ && \text{energy before hitting the floor} &= \frac12 mv^2 \\ \text{COE}: && gh(k-1)a^2 + mgh &= \frac12 mv^2 \\ \Rightarrow && v^2 &= 2gh((k-1)a^2+1) \end{align*} The ball will rebound with kinetic energy \(m gh((k-1)a^2+1)a^2 = mgh((k-1)a^4+a^2)\) And will reach the ceiling with kinetic energy \(mgh((k-1)a^4+a^2-1)\). When \(n = 1\), the kinetic energy (before hitting the ceiling for the first time) is \(mgh(k-1)\). Suppose \(s_n\) is the expression for the kinetic energy divided by \(mgh\), ie \(s_1 = k-1\), then: Clearly \(s_1 = k-1 = a^{4\cdot1-4}(k-1) + \frac{a^{4\cdot-4}-1}{a^2+1}\), so our hypothesis holds for \(n=1\). Suppose it is true for \(n\), then the \(n+1\)th time it will be: \begin{align*} s_{n+1} &= s_n a^4+a^2-1 \\ &= \left ( a^{4n-4}(k-1)+\frac{a^{4n-4}-1}{a^{2}+1} \right) a^4 + a^2 - 1 \\ &= a^{4(n+1)-4}(k-1) + \frac{a^{4(n+1)-4}-a^4}{a^2+1} + \frac{a^4-1}{a^2+1} \\ &= a^{4(n+1)-4}(k-1) + \frac{a^{4(n+1)-4}-a^4+a^4-1}{a^2+1} \\ &= a^{4(n+1)-4}(k-1) + \frac{a^{4(n+1)-4}-1}{a^2+1} \end{align*} Which is our desired expression, therefore it is true by induction. We wont reach the ceiling if this energy is not positive, ie: \begin{align*} && 0 &\leq a^{4n-4}(k-1)+\frac{a^{4n-4}-1}{a^{2}+1} \\ \Rightarrow && \frac{1}{a^2+1}&\geq a^{4n-4}\left (k - 1 + \frac{1}{a^2+1} \right) \\ \Rightarrow && a^{4n-4} &\geq \frac{1}{a^2+1} \cdot \frac{1}{k - 1 + \frac{1}{a^2+1}} \\ \Rightarrow && a^{4n-4} &\geq \frac{1}{(k-1)(a^2+1)+1} \\ \Rightarrow && 4(n-1) \ln a &\geq - \ln[(k-1)(a^2+1)+1] \\ \underbrace{\Rightarrow}_{\ln a < 0} && (n-1) &\leq \frac{ - \ln[(k-1)(a^2+1)+1]}{4\ln a} \\ \Rightarrow && n & \leq 1 -\frac{ \ln[(k-1)(a^2+1)+1]}{4\ln a} \\ &&&= 1 -\frac{ \ln[(k-1)a^2+k]}{4\ln a} \end{align*}