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2000 Paper 3 Q9
D: 1700.0 B: 1500.0
momentum elastic collision kinetic energy oblique collision two-dimensional collision vector dot product inequality conservation of energy conservation of momentum

Two small discs of masses \(m\) and \(\mu m\) lie on a smooth horizontal surface. The disc of mass \(\mu m\) is at rest, and the disc of mass \(m\) is projected towards it with velocity \(\mathbf{u}\). After the collision, the disc of mass \(\mu m\) moves in the direction given by unit vector \(\mathbf{n}\). The collision is perfectly elastic.

  1. Show that the speed of the disc of mass \(\mu m\) after the collision is \ \ $ \dfrac {2\mathbf{u} \cdot \mathbf{n}}{1+\mu}. $
  2. Given that the two discs have equal kinetic energy after the collision, find an expression for the cosine of the angle between \(\bf n\) and \(\bf u\) and show that \(3-\sqrt8\le \mu \le 3+\sqrt8\).

Solution:

  1. In the direction of \(\mathbf{n}\), Conservation of momentum gives: \(m \mathbf{u} \cdot \mathbf{n} = m v_m + \mu m v_{\mu m}\) Newton's experimental law gives: \(\frac{\mathbf{u} \cdot \mathbf{n}}{v_{\mu m} - v_m} = 1\) Therefore \begin{align*} && \mathbf{u} \cdot \mathbf{n} &= v_m + \mu v_{\mu m} \\ && \mathbf{u} \cdot \mathbf{n} &= v_{\mu m} - v_m \\ \Rightarrow && 2 \mathbf{u} \cdot \mathbf{n} &= (1 + \mu)v_{\mu m} \\ \Rightarrow && v_{\mu m} &= \frac{2 \mathbf{u} \cdot \mathbf{n}}{1 + \mu} \\ \end{align*}
  2. Kinetic energy after the collision for the second mass is: \(\frac12 m \mu \frac{4 (\mathbf{u} \cdot \mathbf{n})^2 }{(1 + \mu)^2}\) For the first mass the final speed (in the direction \(\mathbf{n}\) is: \(\mathbf{u} \cdot \mathbf{n}- \frac{2 \mathbf{u} \cdot \mathbf{n}}{1 + \mu} = \frac{(\mu - 1) \mathbf{u} \cdot \mathbf{n}}{1 + \mu}\) It's velocity perpendicular to \(\mathbf{n}\) is unchanged, which is \(\mathbf{u} - (\mathbf{u} \cdot \mathbf{n}) \mathbf{n}\), so it's speed squared is \(\mathbf{u}\cdot \mathbf{u} - (\mathbf{u} \cdot \mathbf{n})^2\) Therefore the total kinetic energy is: \(\frac12 m \frac{(\mu - 1)^2 (\mathbf{u} \cdot \mathbf{n})^2}{(1 + \mu)} + \frac12 m (\mathbf{u}\cdot \mathbf{u} - (\mathbf{u} \cdot \mathbf{n})^2)\) Therefore since the kinetic energies are equal we have: \begin{align*} && \frac12 m \frac{(\mu - 1)^2 (\mathbf{u} \cdot \mathbf{n})^2}{(1 + \mu)} + \frac12 m (\mathbf{u}\cdot \mathbf{u} - (\mathbf{u} \cdot \mathbf{n})^2) &= \frac12 m \mu \frac{4 (\mathbf{u} \cdot \mathbf{n})^2 }{(1 + \mu)^2} \\ \Rightarrow && \frac{(\mu - 1)^2 (\mathbf{u} \cdot \mathbf{n})^2}{(1 + \mu)} + (\mathbf{u}\cdot \mathbf{u} - (\mathbf{u} \cdot \mathbf{n})^2) &= \mu \frac{4 (\mathbf{u} \cdot \mathbf{n})^2 }{(1 + \mu)^2} \\ \Rightarrow && (\mathbf{u} \cdot \mathbf{n})^2 \l 1 + \frac{4\mu}{(1+ \mu)^2} - \frac{(1-\mu)^2}{(1+ \mu)^2} \r &= (\mathbf{u} \cdot \mathbf{u}) \\ \Rightarrow && (\mathbf{u} \cdot \mathbf{n})^2 \l \frac{(1 + \mu)^2 + 4\mu - (1-\mu)^2}{(1+ \mu)^2} \r &= (\mathbf{u} \cdot \mathbf{u}) \\ \Rightarrow && (\mathbf{u} \cdot \mathbf{n})^2 \l \frac{8\mu}{(1+ \mu)^2} \r &= (\mathbf{u} \cdot \mathbf{u}) \\ \Rightarrow && \cos^2 \theta &=\frac{(1 + \mu)^2}{8\mu} \\ \end{align*} We need \begin{align*} && \frac{(1 + \mu)^2}{8\mu} & \leq 1 \\ \Rightarrow && 1 +2 \mu + \mu^2 \leq 8 \mu \\ \Rightarrow && 1 - 6 \mu + \mu^2 \leq 0 \end{align*} This quadratic has roots at \(3 \pm \sqrt{8}\) and therefore our quadratic inequality is satisfied if: \(\boxed{3 - \sqrt{8} \leq \mu \leq 3 + \sqrt{8}}\)

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