A2 October 2021 Q5
5.
Figure 1 represents the plan view of part of a horizontal floor, where \(AB\) and \(BC\) represent fixed vertical walls, with \(AB\) perpendicular to \(BC\).
A small ball is projected along the floor towards the wall \(AB\). Immediately before hitting the wall \(AB\) the ball is moving with speed \(v\ \text{m s}^{-1}\) at an angle \(\theta\) to \(AB\).
The ball hits the wall \(AB\) and then hits the wall \(BC\).
The coefficient of restitution between the ball and the wall \(AB\) is \(\dfrac{1}{3}\)
The coefficient of restitution between the ball and the wall \(BC\) is \(e\).
The floor and the walls are modelled as being smooth.
The ball is modelled as a particle.
The ball loses half of its kinetic energy in the impact with the wall \(AB\).
The ball loses half of its remaining kinetic energy in the impact with the wall \(BC\).
| Scheme | Marks | AO |
|---|---|---|
Use the model to find components of velocity after first impact: | B1 B1 | 1.1b 3.4 |
| Kinetic energy: \(\dfrac{1}{2} \times \dfrac{1}{2}mv^2 = \dfrac{1}{2}m\left(v^2\cos^2\theta + \dfrac{1}{9}v^2\sin^2\theta\right)\) | M1 | 3.1b |
| \(\dfrac{1}{2} = \dfrac{1}{9} + \dfrac{8}{9}\cos^2\theta\) | M1 | 1.1b |
| \(\dfrac{7}{16} = \cos^2\theta,\quad \cos\theta = \dfrac{\sqrt{7}}{4}\) | A1 | 1.1b |
| (5) |
Notes
B1 B1: Parallel component correct
Perpendicular component correct Check the diagram
M1: Equation for KE in \(v, \theta\). Dimensionally correct. Includes all components. Condone \(\tfrac{1}{2}\) used on wrong side
M1: Form and solve equation in \(\cos\theta\)
A1: Or exact equivalent
Alternative (a)
| Scheme | Marks | AO |
|---|---|---|
| Working with initial velocity \(\mathbf{v} = x\mathbf{i} - y\mathbf{j}\), after impact \(\mathbf{v} = x\mathbf{i} + \dfrac{1}{3}y\mathbf{j}\) | B1 B1 | 1.1b 3.4 |
| KE: \(\dfrac{1}{2} \times \dfrac{1}{2}m(x^2 + y^2) = \dfrac{1}{2}m\left(x^2 + \dfrac{1}{9}y^2\right)\) | M1 | 3.1b |
| \(y^2 = \dfrac{9}{7}x^2,\quad \dfrac{y}{x} = \tan\theta = \dfrac{3}{\sqrt{7}}\) | M1 | 1.1b |
| \(\cos\theta = \dfrac{\sqrt{7}}{4}\) | A1 | 1.1b |
| (5) |
| Scheme | Marks | AO |
|---|---|---|
Use the model to find components of velocity after second impact: | B1 B1 | 1.1b 3.4 |
| Kinetic energy: \(\dfrac{1}{4} \times \dfrac{1}{2}mv^2 = \dfrac{1}{2}m\left(e^2v^2\cos^2\theta + \dfrac{1}{9}v^2\sin^2\theta\right)\) or \(\dfrac{1}{2} \times \dfrac{1}{2}m\left(v^2\cos^2\theta + \dfrac{1}{9}v^2\sin^2\theta\right) = \dfrac{1}{2}m\left(e^2v^2\cos^2\theta + \dfrac{1}{9}v^2\sin^2\theta\right)\) | M1 | 3.1b |
| \(\dfrac{1}{4} = \dfrac{7}{16}e^2 + \dfrac{1}{9} \times \dfrac{9}{16}\) | M1 | 1.1b |
| \(\Rightarrow e^2 = \dfrac{3}{7},\quad e = \sqrt{\dfrac{3}{7}}\) | A1 | 1.1b |
| (5) | ||
| (10 marks) |
Notes
B1 B1: Parallel component correct
Perpendicular component correct
M1: Equation for KE in \(x, y\). Dimensionally correct. Includes all components. Condone \(\tfrac{1}{2}\) used on wrong side
M1: Use their \(\cos\theta\) to form and solve equation in \(e\)
A1: Or exact equivalent
Alternative (b)
| Scheme | Marks | AO |
|---|---|---|
| After second impact \(\mathbf{v} = -ex\mathbf{i} + \dfrac{1}{3}y\mathbf{j}\) | B1 B1 | 1.1b 3.4 |
| KE: \(\dfrac{1}{4} \times \dfrac{1}{2}m(x^2 + y^2) = \dfrac{1}{2}m\left(e^2x^2 + \dfrac{1}{9}y^2\right)\) | M1 | 3.1b |
| \(4e^2x^2 + \dfrac{4}{9}y^2 = x^2 + y^2,\quad 4e^2 = 1 + \dfrac{5}{9}\left(\dfrac{y}{x}\right)^2\) | M1 | 1.1b |
| \(\Rightarrow e^2 = \dfrac{3}{7},\quad e = \sqrt{\dfrac{3}{7}}\) | A1 | 1.1b |
| (5) |