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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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2010 Paper 3 Q11
D: 1700.0 B: 1469.5
momentum collisions friction Newton's second law relative motion bullet and block kinematics deceleration energy conservation embedding

A bullet of mass \(m\) is fired horizontally with speed \(u\) into a wooden block of mass \(M\) at rest on a horizontal surface. The coefficient of friction between the block and the surface is \(\mu\). While the bullet is moving through the block, it experiences a constant force of resistance to its motion of magnitude \(R\), where \(R>(M+m)\mu g\). The bullet moves horizontally in the block and does not emerge from the other side of the block.

  1. Show that the magnitude, \(a\), of the deceleration of the bullet relative to the block while the bullet is moving through the block is given by \[ a= \frac R m + \frac {R-(M+m)\mu g}{M}\, . \]
  2. Show that the common speed, \(v\), of the block and bullet when the bullet stops moving through the block satisfies \[ av = \frac{Ru-(M+m)\mu gu}M\,. \]
  3. Obtain an expression, in terms of \(u\), \(v\) and \(a\), for the distance moved by the block while the bullet is moving through the block.
  4. Show that the total distance moved by the block is \[ \frac{muv}{2(M+m)\mu g}\,. \]
Describe briefly what happens if \(R< (M+m)\mu g\).

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