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2011 Paper 1 Q10
D: 1516.0 B: 1484.0
momentum elastic collision restitution free fall energy conservation vertical motion bouncing particle one-dimensional collision

A particle, \(A\), is dropped from a point \(P\) which is at a height \(h\) above a horizontal plane. A~second particle, \(B\), is dropped from \(P\) and first collides with \(A\) after \(A\) has bounced on the plane and before \(A\) reaches \(P\) again. The bounce and the collision are both perfectly elastic. Explain why the speeds of \(A\) and \(B\) immediately before the first collision are the same. The masses of \(A\) and \(B\) are \(M\) and \(m\), respectively, where \(M>3m\), and the speed of the particles immediately before the first collision is \(u\). Show that both particles move upwards after their first collision and that the maximum height of \(B\) above the plane after the first collision and before the second collision is \[ h+ \frac{4M(M-m)u^2}{(M+m)^2g}\,. \]

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2009 Paper 1 Q11
D: 1500.0 B: 1500.0
momentum collisions coefficient of restitution kinetic energy loss one-dimensional collision conservation of momentum Newton's law of restitution algebraic manipulation

Two particles move on a smooth horizontal table and collide. The masses of the particles are \(m\) and \(M\). Their velocities before the collision are \(u{\bf i}\) and \(v{\bf i}\,\), respectively, where \(\bf i\) is a unit vector and \(u>v\). Their velocities after the collision are \(p{\bf i}\) and \(q{\bf i}\,\), respectively. The coefficient of restitution between the two particles is \(e\), where \(e<1\).

  1. Show that the loss of kinetic energy due to the collision is \[ \tfrac12 m (u-p)(u-v)(1-e)\,, \] and deduce that \(u\ge p\).
  2. Given that each particle loses the same (non-zero) amount of kinetic energy in the collision, show that \[ u+v+p+q=0\,, \] and that, if \(m\ne M\), \[ e= \frac{(M+3m)u + (3M+m)v}{(M-m)(u-v)}\,. \]

Solution:

  1. \begin{align*} \text{COM}: && mu + Mv &= mp + Mq \\ \Rightarrow && m(u-p) &= M(q-v) \\ \text{NEL}: && q-p &= e(u-v) \\ && q +ev &= p+eu \\ && \Delta \text{ k.e.} &= \frac12 m u^2 + \frac12 M v^2 -\frac12 m p^2 - \frac12 M q^2 \\ &&&= \frac12m (u^2 - p^2)+\frac12M(v^2-q^2) \\ &&&= \frac12m (u^2 - p^2)+\frac12M(v-q)(v+q) \\ &&&= \frac12m(u^2-p^2) - \frac12 m(u-p)(v+q) \\ &&&= \frac12 m(u-p) \left ( u+p-v-q\right) \\ &&&= \frac12 m(u-p) \left (u-v+(p-q)\right) \\ &&&= \frac12 m(u-p) \left (u-v-e(u-v)\right) \\ &&&= \frac12m(u-p)(u-v)(1-e) \end{align*} Since the loss in energy is positive, and \(m\), \(u-v\) and \(1-e\) are all positive, so is \(u-p\), ie \(u \geq p\)
  2. \begin{align*} && \frac12 m u^2 - \frac12mp^2 &= \frac12Mv^2 - \frac12Mq^2 \\ && \frac12 m(u-p)(u+p) &= \frac12 M (v-q)(v+q) \\ && \frac12 m (u-p)(u+p) &= -\frac12 m(u-p)(v+q) \\ \Rightarrow && u+p+v+q &= 0 \end{align*} \begin{align*} && p+q &= -(u+v)\\ &&mp+Mq &= mu+Mv \\ \Rightarrow && (M-m)q &= mu+Mv+mu+mv\\ \Rightarrow && q &= \frac{(M+m)v+2mu}{M-m} \\ \Rightarrow && (m-M)p &= mu+Mv+Mu+Mv \\ \Rightarrow && p &= -\frac{(M+m)u+2Mv}{M-m} \\ \\ && e &= \frac{q-p}{u-v} \\ &&&= \frac{(M+m)v+2mu+(M+m)u+2Mv}{(u-v)(M-m)} \\ &&&= \frac{(3M+m)v+(3m+M)u}{(u-v)(M-m)} \end{align*}

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