A2 June 2023 Q6
6. A particle \(P\) of mass \(m\) is falling vertically when it strikes a fixed smooth inclined plane. The plane is inclined to the horizontal at an angle \(\alpha\), where \(0 < \alpha \leqslant 45^\circ\)
At the instant immediately before the impact, the speed of \(P\) is \(u\).
At the instant immediately after the impact, \(P\) is moving horizontally with speed \(v\).
The coefficient of restitution between \(P\) and the plane is \(e\), where \(e > 0\)
| Scheme | Marks | AO |
|---|---|---|
 | ||
| CLM along the plane: | M1 | 3.1a |
| \((m)u\sin\alpha = (m)v\cos\alpha\) | A1 | 1.1b |
| Impulse-momentum perp to the plane: | M1 | 3.1a |
| \(I = m(v\sin\alpha - (-u\cos\alpha))\) | A1 | 1.1b |
| \(I = m\left(\dfrac{u\sin^2\alpha}{\cos\alpha} + u\cos\alpha\right) = \dfrac{mu}{\cos\alpha}(\sin^2\alpha + \cos^2\alpha) = mu\sec\alpha\) * | A1* | 2.2a |
| (5) |
Notes
M1: Correct no. of terms, dimensionally correct, mass × velocity, condone sin/cos confusion.
A1: Correct equation
M1: Dimensionally correct. Must be subtracting, but condone subtracting in the wrong order and sin/cos confusion
A1: Correct unsimplified equation
A1*: Given answer correctly obtained. Must be EXACT factorisation.
6(a) alt1
| Scheme | Marks | AO |
|---|---|---|
 | M1 M1 | 3.1a 3.1a |
| \(I\cos\alpha = m(0 - -u)\) | A1 A1 | 1.1b 1.1b |
| \(I = mu\sec\alpha\) * | A1* | 2.2a |
| (5) |
6(a) alt 2
| Scheme | Marks | AO |
|---|---|---|
Introduce and use an expression for \(e\) | ||
| CLM along the plane: | M1 | 3.1a |
| \(u\sin\alpha\) unchanged | A1 | 1.1b |
| Finds an expression for \(e\) together with Impulse-momentum perpendicular to the plane \(\quad \tan\alpha = \dfrac{eu\cos\alpha}{u\sin\alpha} \Rightarrow e = \tan^2\alpha\) and \(I = m(eu\cos\alpha - (-u\cos\alpha))\) | M1 | 3.1a |
| \(I = m(u\cos\alpha\tan^2\alpha - (-u\cos\alpha))\) | A1 | 1.1b |
| \(I = m\left(\dfrac{u\sin^2\alpha}{\cos\alpha} + u\cos\alpha\right) = \dfrac{mu}{\cos\alpha}(\sin^2\alpha + \cos^2\alpha) = mu\sec\alpha\) * | A1* | 2.2a |
| (5) |
6(a) alt 3
| Scheme | Marks | AO |
|---|---|---|
Use a vector approach and magnitude of impulse | ||
| CLM along the plane: | M1 | 3.1a |
| \((m)u\sin\alpha = (m)v\cos\alpha\) (this leads to \(v = u\tan\alpha\)) | A1 | 1.1b |
| Impulse-momentum as a vector equation followed by Pythagoras to find the magnitude. \(I = m\begin{pmatrix} -v \\ u \end{pmatrix}\) and \(|I| = m\sqrt{v^2 + u^2}\) | M1 | 3.1a |
| \(|I| = m\sqrt{u^2\tan^2\alpha + u^2}\) | A1 | 1.1b |
| \(I = m\sqrt{u^2(1 + \tan^2\alpha)} = m\sqrt{u^2\sec^2\alpha} = mu\sec\alpha\) * | A1* | 2.2a |
| (5) |
| Scheme | Marks | AO |
|---|---|---|
| NEL: \(eu\cos\alpha = v\sin\alpha\) | M1 | 3.4 |
| Squaring and adding their expressions for \(v\sin\alpha\) and \(v\cos\alpha\). | M1 | 1.1b |
| \(v^2 = u^2(\sin^2\alpha + e^2\cos^2\alpha)\) * | A1* | 1.1b |
| (3) |
Notes
M1: Attempt at NEL
M1: Squaring and adding their expressions for \(v\sin\alpha\) and \(v\cos\alpha\) to obtain \(v^2\).
A1*: Given answer correctly obtained. Must be EXACT.
| Scheme | Marks | AO |
|---|---|---|
| KE loss = \(\dfrac{1}{2}mu^2 - \dfrac{1}{2}mu^2(\sin^2\alpha + e^2\cos^2\alpha)\). | M1 | 2.1 |
| Use \(\sin^2\alpha + \cos^2\alpha = 1\) to give \(\qquad\) KE loss = \(\dfrac{1}{2}mu^2(1 - e^2)\cos^2\alpha\) * | A1* | 1.1b |
| (2) |
Notes
M1: Expression for difference of KE in terms of \(m\), \(u\), \(\alpha\) and \(e\)
A1*: Given answer correctly obtained. Factorisation must be EXACT.
| Scheme | Marks | AO |
|---|---|---|
| Use \(\tan^2\alpha = e\) oe to eliminate \(\alpha\) in given expression from (c) | M1 | 3.1a |
| KE Loss = \(\dfrac{1}{2}mu^2(1 - e)\) or \(\dfrac{1}{2}mu^2\dfrac{1}{1 + e}(1 - e^2)\) | A1 | 1.1b |
| (2) | ||
| (12 marks) |
Notes
M1: Complete method to eliminate \(\alpha\) e.g. using \(\tan^2\alpha = e\) to eliminate \(\alpha\)
Any trig identity used must be correct eg \(\sec^2\alpha = 1 + e\) or \(\cos^2\alpha = \dfrac{1}{1 + e}\)
A1: Correct answer.