Momentum and Collisions

Linear momentum and impulse. Conservation of momentum

Showing 1-2 of 2 problems
2024 Paper 3 Q9
D: 1500.0 B: 1500.0

The origin \(O\) of coordinates lies on a smooth horizontal table and the \(x\)- and \(y\)-axes lie in the plane of the table. A smooth sphere \(A\) of mass \(m\) and radius \(r\) is at rest on the table with its lowest point at the origin. A second smooth sphere \(B\) has the same mass and radius and also lies on the table. Its lowest point has \(y\)-coordinate \(2r\sin\alpha\), where \(\alpha\) is an acute angle, and large positive \(x\)-coordinate. Sphere \(B\) is now projected parallel to the \(x\)-axis, with speed \(u\), so that it strikes sphere \(A\). The coefficient of restitution in this collision is \(\frac{1}{3}\).

  1. Show that, after the collision, sphere \(B\) moves with velocity \[\begin{pmatrix} -\frac{1}{3}u\bigl(1 + 2\sin^2\alpha\bigr) \\ \frac{2}{3}u\sin\alpha\cos\alpha \end{pmatrix}.\]
  2. Show further that the lowest point of sphere \(B\) crosses the \(y\)-axis at the point \((0, Y)\), where \(Y = 2r(\cos\alpha\tan\beta + \sin\alpha)\) and \[\tan\beta = \frac{2\sin\alpha\cos\alpha}{1 + 2\sin^2\alpha}.\]
A third sphere \(C\) of radius \(r\) is at rest with its lowest point at \((0, h)\) on the table, where \(h > 0\).
  1. Show that, if \(h > Y + 2r\sec\beta\), sphere \(B\) will not strike sphere \(C\) in its motion after the collision with sphere \(A\).
  2. Show that \(Y < 2r\sec\beta\). Hence show that sphere \(B\) will not strike sphere \(C\) for any value of \(\alpha\), if \(h > \dfrac{8r}{\sqrt{3}}\).

2016 Paper 2 Q9
D: 1600.0 B: 1473.6

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 \(a\).

Show 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 \(R \cdot c = \frac12 M \left ( \frac{m}{m+M}u\right)^2 \Rightarrow c = \frac{Mm}{(M+m)^2}a\).