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2010 Paper 3 Q11
D: 1700.0 B: 1469.5

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 \(R< (M+m)\mu g\).

2004 Paper 1 Q10
D: 1516.0 B: 1516.0

A particle is travelling in a straight line. It accelerates from its initial velocity \(u\) to velocity \(v\), where \(v > \vert u \vert > 0\,\), travelling a distance \(d_1\) with uniform acceleration of magnitude \(3a\,\). It then comes to rest after travelling a further distance \(d_2\,\) with uniform deceleration of magnitude \(a\,\). Show that

  1. if \(u>0\) then \(3d_1 < d_2\,\);
  2. if \(u<0\) then \(d_2 < 3d_1 < 2d_2\,\).
Show also that the average speed of the particle (that is, the total distance travelled divided by the total time) is greater in the case \(u>0\) than in the case \(u<0\,\). \noindent {\bf Note:} In this question \(d_1\) and \(d_2\) are distances travelled by the particle which are not the same, in the second case, as displacements from the starting point.

1988 Paper 1 Q12
D: 1484.0 B: 1471.5

A skater of mass \(M\) is skating inattentively on a smooth frozen canal. She suddenly realises that she is heading perpendicularly towards the straight canal bank at speed \(V\). She is at a distance \(d\) from the bank and can choose one of two methods of trying to avoid it; either she can apply a force of constant magnitude \(F\), acting at right-angles to her velocity, so that she travels in a circle; or she can apply a force of magnitude \(\frac{1}{2}F(V^{2}+v^{2})/V^{2}\) directly backwards, where \(v\) is her instantaneous speed. Treating the skater as a particle, find the set of values of \(d\) for which she can avoid hitting the bank. Comment briefly on the assumption that the skater is a particle.


Solution: Suppose she applies a force of magnitude \(\frac{1}{2}F(V^{2}+v^{2})/V^{2}\) backwards, then \begin{align*} && M v \frac{dv}{dx} &= -\frac{1}{2}F(V^{2}+v^{2})/V^{2} \\ \Rightarrow && M\int_{V}^0 \frac{2v}{V^2+ v^2} \d v &= - \frac{F}{V^2} x \\ \Rightarrow && M \left [ -\log(V^2+v^2) \right]_0^V &= -\frac{Fx}{V^2} \\ \Rightarrow && -M \ln 2&= -\frac{Fx}{V^2} \end{align*} Therefore she will stop quickly enough if \(d > \frac{V^2M \ln 2}{F}\) If she attempts to use the right-angled method, then she will travel a distance at most \(r\) where \(r\) is the radius of her circle: \begin{align*} && F &= M \frac{V^2}{r} \\ \Rightarrow && r &= \frac{MV^2}{F} \end{align*} Therefore she can always avoid the wall if \(d > \frac{MV^2}{F}\). There are two potential issues with being a particle. Firstly we would need to account for any variation in the distance to the wall (which could be accounted for by changing \(d\)). Secondly when she enters circular motion she will rotate and therefore we might need to consider her inertia as well as just her velocity when modelling.