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2016 Paper 2 Q9
D: 1600.0 B: 1473.6
momentum conservation of momentum energy conservation bullet and block resistance force relative motion work-energy theorem collisions

A small bullet of mass \(m\) is fired into a block of wood of mass \(M\) which is at rest. The speed of the bullet on entering the block is \(u\). Its trajectory within the block is a horizontal straight line and the resistance to the bullet's motion is \(R\), which is constant.

  1. The block is fixed. The bullet travels a distance \(a\) inside the block before coming to rest. Find an expression for \(a\) in terms of \(m\), \(u\) and \(R\).
  2. Instead, the block is free to move on a smooth horizontal table. The bullet travels a distance \(b\) inside the block before coming to rest relative to the block, at which time the block has moved a distance \(c\) on the table. Find expressions for \(b\) and \(c\) in terms of \(M\), \(m\) and \(a\).

Solution:

  1. Since \(R\) is constant, \(F=ma \Rightarrow \text{acc} = \frac{R}{m}\) and \(v^2 = u^2 + 2as\) so \(0 = u^2 - 2 \frac{R}{m}a\), ie \(a = \frac{m u^2}{2R}\)
  2. By conservation of momentum, the bullet/block combination will eventually be travelling at \(v = \frac{m}{m+M}u\). The bullet will slow down to this speed in a time of \(\frac{m}{m+M}u = u - \frac{R}{m} T \Rightarrow T = \frac{Mm}{R(m+M)}u\) and will travel \(b+c = \frac{\left ( 1 - \left ( \frac{m}{m+M} \right)^2\right)u^2m}{2R}= \left ( 1 - \left ( \frac{m}{m+M} \right)^2\right)a\). The block will travell \(c = \frac12 \frac{R}{M} \frac{M^2m^2u^2}{R^2(m+M)^2} = \frac{Mm}{(m+M)^2}a\) Therefore the \(b = \left ( 1 - \left ( \frac{m}{m+M} \right)^2\right)a - \frac{Mm}{(m+M)^2}a = \frac{M}{M+m}a\)
Work done by friction is the relative gain in KE for the block. ie \(R \cdot c = \frac12 M \left ( \frac{m}{m+M}u\right)^2 \Rightarrow c = \frac{Mm}{(M+m)^2}a\).

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