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2016 Paper 1 Q10
D: 1484.0
B: 1500.0
momentum
collisions
coefficient of restitution
Newton's law of restitution
simultaneous collisions
sequential collisions
inequality
conservation of momentum
Four particles \(A\), \(B\), \(C\) and \(D\) are initially at rest on a smooth horizontal table. They lie equally spaced a small distance apart, in the order \(ABCD\), in a straight line. Their masses are \(\lambda m\), \(m\), \(m\) and \(m\), respectively, where \(\lambda>1\). Particles \(A\) and \(D\) are simultaneously projected, both at speed \(u\), so that they collide with \(B\) and \(C\) (respectively). In the following collision between \(B\) and \(C\), particle \(B\) is brought to rest. The coefficient of restitution in each collision is \(e\).
- Show that \(e = \dfrac {\lambda-1}{3\lambda+1}\) and deduce that \(e < \frac 13\,\).
- Given also that \(C\) and \(D\) move towards each other with the same speed, find the value of \(\lambda\) and of \(e\).
Solution:
- \begin{align*} \text{NEL}: && w_C &= e(v_B - v_C) \\ \text{COM}: && mv_B+ mv_C &= m w_C \\ \Rightarrow && w_C &= v_B + v_C\\ \Rightarrow && e(v_B - v_C) &= (v_B + v_C) \\ \Rightarrow && (1-e)v_B &= -(1+e)v_C \\ \Rightarrow && (1-e) \frac{\lambda(1+ e)}{1+\lambda} &= (1+e) \frac{1+e}{2} \\ \Rightarrow && 2\lambda - 2\lambda e &= 1+\lambda + e + \lambda e \\ \Rightarrow && (3\lambda +1)e &= \lambda - 1 \\ \Rightarrow && e &= \frac{\lambda -1}{3\lambda + 1} \\ &&&< \frac{\lambda - 1 + \frac{4}{3}}{3\lambda + 1} \\ &&& = \frac13 \end{align*}
- Since they move towards each other at the same speed \(w_C = - v_D\) \begin{align*} && w_C &= - v_D \\ \Rightarrow && v_B + v_C &= -(v_C+eu) \\ \Rightarrow && -eu &= v_B +2v_C \\ &&&= \frac{\lambda(1+ e)}{1+\lambda} u -(1+e)u \\ \Rightarrow && 1 &= \frac{\lambda(1+e)}{1+\lambda} \\ \Rightarrow && 1+\lambda &= \lambda \left ( 1 + \frac{\lambda -1}{3\lambda+1} \right) \\ &&&= \lambda \frac{4\lambda}{3\lambda +1} \\ \Rightarrow && 1+4\lambda + 3\lambda^2 &= 4\lambda^2 \\ \Rightarrow && 0 &= \lambda^2 - 4\lambda - 1 \\ \Rightarrow && \lambda &= \frac{4 \pm \sqrt{20}}{2} \\ &&&= 2\pm \sqrt{5} \\ \Rightarrow && \lambda &= 2 + \sqrt{5} \\ && e &= \frac{1+\sqrt{5}}{7+3\sqrt{5}} \\ &&&=\sqrt{5}-2 \end{align*}
2009 Paper 2 Q10
D: 1600.0
B: 1500.0
momentum
collisions
coefficient of restitution
simultaneous collisions
conservation of momentum
two-dimensional
constraint equations
\(\,\)
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\rput[tl](-2.09,-0.3){\(P_1\)} \rput[tl](-1.11,-0.3){\(P_2\)} \rput[tl](-0.55,-0.6){\(P_3\)} \rput[tl](-0.55,-1.6){\(P_4\)} \rput[tl](0.07,0.3){\(O\)} \psline{->}(-2.1,0.4)(-1.6,0.4) \psline{->}(0.4,-1.9)(0.4,-1.4) \rput[tl](-1.55,0.45){\(u\)} \rput[tl](0.32,-1.2){\(u\)} \begin{scriptsize} \psdots[dotsize=18pt 0,dotstyle=*](-1,0) \psdots[dotsize=18pt 0,dotstyle=*](-2,0) \psdots[dotsize=18pt 0,dotstyle=*](0,-0.7) \psdots[dotsize=18pt 0,dotstyle=*](0,-1.7) \end{scriptsize} \end{pspicture*}
Four particles \(P_1\), \(P_2\), \(P_3\) and
\(P_4\), of masses \(m_1\), \(m_2\), \(m_3\) and \(m_4\), respectively, are arranged
on smooth horizontal axes as shown in the diagram.
Initially, \(P_2\) and \(P_3\) are stationary, and
both \(P_1\) and \(P_4\) are moving towards \(O\) with speed \(u\).
Then \(P_1\) and \(P_2\) collide, at the same moment as
\(P_4\) and \(P_3\) collide. Subsequently, \(P_2\) and
\(P_3\) collide at \(O\), as do \(P_1\) and \(P_4\) some time later.
The coefficient of
restitution between each pair of particles is~\(e\), and \(e>0\).
Show that initially \(P_2\) and \(P_3\) are equidistant from \(O\).