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2013 Paper 2 Q11
Three identical particles lie, not touching one another, in a straight line on a smooth horizontal surface. One particle is projected with speed \(u\) directly towards the other two which are at rest. The coefficient of restitution in all collisions is \(e\), where \(0 < e < 1\,\).
- Show that, after the second collision, the speeds of the particles are \(\frac12u(1-e)\), \(\frac14u (1-e^2)\) and \(\frac14u(1+e)^2\). Deduce that there will be a third collision whatever the value of \(e\).
- Show that there will be a fourth collision if and only if \(e\) is less than a particular value which you should determine.
Solution:
- First Collision:
By NEL, \(v_2 = v_1 + eu\), so \begin{align*} \text{COM}: && mu &= mv_1 + m(v_1 + eu) \\ \Rightarrow && 2mv_1 &= mu(1-e) \\ \Rightarrow && v_1 &= \frac12 u(1-e) \\ && v_2 &= \frac12 u(1-e) + eu \\ &&&= \frac12 u(1+e) \end{align*} The second collision is identical to the first except replacing \(u\) with \(\frac12u(1+e)\), therefore after that collision: \begin{align*} && \text{first particle} &= \frac12 u(1-e) \\ && \text{second particle} &= \frac12 \left (\frac12 u(1+e) \right)(1-e) \\ &&&= \frac14 u(1-e^2) \\ && \text{third particle} &= \frac12 \left (\frac12 u(1+e) \right)(1+e) \\ &&&= \frac14 u(1+e)^2 \end{align*} After all these collisions, all particles are moving in the same direction (since they all have positive velocity), but the first particle is now travelling faster than the second particle (as \(\frac12(1-e) < 1\)). Therefore they will collide again.
- The third collision:
The speed of approach will be \(\frac12u(1-e) - \frac14u(1-e^2) = \frac14u(1-e)(2 - (1+e)) = \frac14 u(1-e)^2\), therefore by NEL, \(w_2 = w_1 + \frac14ue(1-e)^2\) \begin{align*} \text{COM}: && m\frac12u(1-e) + m \frac14u(1-e^2) &= mw_1 + m\left (w_1 + \frac14ue(1-e)^2 \right) \\ \Rightarrow && \frac14u(1-e)(2+(1+e)) &= 2w_1 + \frac14ue(1-e)^2 \\ \Rightarrow && 2w_1 &= \frac14u(1-e)(3+e)-\frac14ue(1-e)^2 \\ &&&= \frac14u(1-e)(3+e-e(1-e)) \\ &&&= \frac14u(1-e)(3+e^2) \\ \Rightarrow && w_1 &= \frac18 u(1-e)(3+e^2) \\ && w_2 &= \frac18 u(1-e)(3+e^2) + \frac14ue(1-e)^2 \\ &&&= \frac18u(1-e)(3+e^2+2e(1-e)) \\ &&&= \frac18u(1-e)(3+2e-e^2) \\ &&&= \frac18u(1-e)(1+e)(3-e) \\ \end{align*} A fourth collision is possible, iff \begin{align*} && \frac18u(1-e)(1+e)(3-e)&> \frac14 u(1+e)^2 \\ \Leftrightarrow && (1-e)(3-e)&> 2 (1+e) \\ \Leftrightarrow &&3-4e-e^2&> 2+2e \\ \Leftrightarrow &&1-5e-e^2&>0 \\ \Leftrightarrow && e &< 3-\sqrt{2} \end{align*}
2006 Paper 2 Q10
Three particles, \(A\), \(B\) and \(C\), of masses \(m\), \(km\) and \(3m\) respectively, are initially at rest lying in a straight line on a smooth horizontal surface. Then \(A\) is projected towards \(B\) at speed \(u\). After the collision, \(B\) collides with \(C\). The coefficient of restitution between \(A\) and \(B\) is \(\frac12\) and the coefficient of restitution between \(B\) and \(C\) is \(\frac14\).
- Find the range of values of \(k\) for which \(A\) and \(B\) collide for a second time.
- Given that \(k=1\) and that \(B\) and \(C\) are initially a distance \(d\) apart, show that the time that elapses between the two collisions of \(A\) and \(B\) is \(\dfrac{60d}{13u}\,\).
Solution:
- After the first collision, it takes \(B\), \(\frac{d}{v_B} = \frac{d}{u} \frac{2(k+1)}{3} = \frac{4d}{3u}\) to collide with \(C\). During which time \(B\) and \(A\) have been moving apart with speed \(\frac12u\) and so are a distance \(\frac{2d}{3}\) apart. After the second collision, \(w_B = \frac{3(4\cdot 1 - 3)}{8(1+1)(1+3)}u = \frac{3}{64}u\) and \(v_A = \frac{1}{4}u\) so they are moving together at speed \(\frac{16-3}{64}u = \frac{13}{64}u\). It will take them \(\frac{2d}{3} \div \frac{13}{64}u = \frac{128d}{3 \times 13u}\) to do this for a total time of \(\frac{128d}{3 \times 13u} + \frac{4d}{3u} = \frac{(128+52)d}{3 \times 13 u} = \frac{60d}{13u}\)
2000 Paper 1 Q10
Three particles \(P_1\), \(P_2\) and \(P_3\) of masses \(m_{1}\), \(m_{2}\) and \(m_{3}\) respectively lie at rest in a straight line on a smooth horizontal table. \(P_1\) is projected with speed \(v\) towards \(P_2\) and brought to rest by the collision. After \(P_2\) collides with \(P_3\), the latter moves forward with speed \(v\). The coefficients of restitution in the first and second collisions are \(e\) and \(e'\), respectively. Show that \[ e'= \frac{m_{2}+m_{3}-m_{1}}{m_{1}}. \] Show that \(2m_1\ge m_2 +m_3\ge m_1\) for such collisions to be possible. If \(m_1\), \(m_3\) and \(v\) are fixed, find, in terms of \(m_1\), \(m_3\) and \(v\), the largest and smallest possible values for the final energy of the system.
1995 Paper 2 Q10
Three small spheres of masses \(m_{1},m_{2}\) and \(m_{3},\) move in a straight line on a smooth horizontal table. (Their order on the straight line is the order given.) The coefficient of restitution between any two spheres is \(e\). The first moves with velocity \(u\) towards the second whilst the second and third are at rest. After the first collision the second sphere hits the third after which the velocity of the second sphere is \(u.\) Find \(m_{1}\) in terms of \(m_{2},m_{3}\) and \(e\). deduce that \[ m_{2}e>m_{3}(1+e+e^{2}). \] Suppose that the relation between \(m_{1},m_{2}\) and \(m_{3}\) is that in the formula you found above, but that now the first sphere initially moves with velocity \(u\) and the other two spheres with velocity \(v\), all in the same direction along the line. If \(u>v>0\) use the first part to find the velocity of the second sphere after two collisions have taken place. (You should not need to make any substantial computations but you should state your argument clearly.)