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2017 Paper 3 Q11
D: 1700.0 B: 1484.0
momentum conservation of momentum kinetic energy sequential firing simultaneous firing inequality telescoping sum railway truck recoil projectile

A railway truck, initially at rest, can move forwards without friction on a long straight horizontal track. On the truck, \(n\) guns are mounted parallel to the track and facing backwards, where \(n>1\). Each of the guns is loaded with a single projectile of mass \(m\). The mass of the truck and guns (but not including the projectiles) is \(M\). When a gun is fired, the projectile leaves its muzzle horizontally with a speed \(v-V\) relative to the ground, where \(V\) is the speed of the truck immediately before the gun is fired.

  1. All \(n\) guns are fired simultaneously. Find the speed, \(u\), with which the truck moves, and show that the kinetic energy, \(K\), which is gained by the system (truck, guns and projectiles) is given by \[ K= \tfrac{1}{2}nmv^2\left(1 +\frac{nm}{M} \right) . \]
  2. Instead, the guns are fired one at a time. Let \(u_r\) be the speed of the truck when \(r\) guns have been fired, so that \(u_0= 0\). Show that, for \(1\le r \le n\,\), \[ u_r - u_{r-1} = \frac{mv}{M+(n-r)m} \tag{\(*\)} \] and hence that \(u_n < u\,\).
  3. Let \(K_r\) be the total kinetic energy of the system when \(r\) guns have been fired (one at a time), so that \(K_0 = 0\). Using \((*)\), show that, for \(1\le r\le n\,\), \[ K_r -K_{r-1} = \tfrac 12 mv^2 + \tfrac12 mv (u_r-u_{r-1}) \] and hence show that \[ K_n = \tfrac{1}{2}nmv^2 +\tfrac{1}{2}mvu_n \,. \] Deduce that \(K_n < K\).

Solution:

  1. \(\,\) \begin{align*} \text{COM}: && 0 &= nm(v-0) - Mu \\ \Rightarrow && u &= \frac{nm}{M}v \\ \\ \Rightarrow && \text{K.E.} &= \underbrace{\tfrac12 nmv^2}_{\text{projectiles}} + \underbrace{\tfrac12Mu^2}_{\text{guns and truck}} \\ &&&= \tfrac12nmv^2 + \tfrac12M \frac{n^2m^2}{M^2}v^2 \\ &&&= \tfrac12 nmv^2 \left (1 + \frac{nm}{M} \right) \end{align*}
  2. \(\,\) \begin{align*} \text{COM}: && ((n-r+1)m+M)u_{r-1} &= -m(v-u_{r-1})+ ((n-r)m+M)u_r \\ \Rightarrow && mv &= -((n-r+1)m+M-m)u_{r-1}+((n-r)m+M)u_r \\ \Rightarrow && u_r-u_{r-1} &= \frac{mv}{M+(n-r)m} \\ \\ && u_n &= \frac{mv}{M+(n-1)m} + \frac{mv}{M+(n-2)m} + \cdots + \frac{mv}{M} \\ &&&< \frac{mv}M + \cdots + \frac{mv}{M} \\ &&&= \frac{mn}{M}v = u \end{align*}
  3. \(\,\) \begin{align*} && K_r &= \underbrace{K_{r-1}-\frac12(m(n-r+1)+M)u_{r-1}^2}_{\text{energy of already dispersed projectiles}} + \frac12m(v-u_{r-1})^2 + \frac12(m(n-r)+M)u_r^2 \\ \Rightarrow && K_r-K_{r-1} &= \tfrac12(u_r^2-u_{r-1}^2)(M+m(n-r))+\tfrac12mv^2-mvu_{r-1} \\ &&&=\tfrac12mv^2+ \tfrac12(u_r+u_{r-1})mv-mvu_{r-1} \\ &&&= \tfrac12mv^2 + \tfrac12mv(u_r-u_{r-1}) \\ \\ && K_n &= \frac12nmv^2 + \tfrac12mv(u_n - u_0) \\ &&&= \tfrac12nmv^2 + \tfrac12mvu_n \\ &&&< \tfrac12nmv^2 + \tfrac12mvu \\ &&&= \tfrac12nmv^2 + \tfrac12mv \frac{nm}{M}v \\ &&&= \tfrac12nmv^2 \left (1 +\frac{m}{M} \right) \\ &&&\leq K \end{align*}

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