M4 June 2017 Q4
4. [In this question, the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are in a vertical plane, \(\mathbf{i}\) being horizontal and \(\mathbf{j}\) being vertically upwards.]
A line of greatest slope of a fixed smooth plane is parallel to the vector \((-4\mathbf{i} - 3\mathbf{j})\). A particle \(P\) falls vertically and strikes the plane. Immediately before the impact, \(P\) has velocity \(-7\mathbf{j}\) m s\(^{-1}\). Immediately after the impact, \(P\) has velocity \((-a\mathbf{i} + \mathbf{j})\) m s\(^{-1}\), where \(a\) is a positive constant.
| Scheme | Marks |
|---|---|
| Components parallel to the plane unchanged: \(\left(\begin{pmatrix}0\\-7\end{pmatrix}\cdot\dfrac{1}{5}\begin{pmatrix}-4\\-3\end{pmatrix} = \begin{pmatrix}-a\\1\end{pmatrix}\cdot\dfrac{1}{5}\begin{pmatrix}-4\\-3\end{pmatrix}\right)\) \(\Rightarrow 21 = 4a - 3\) | M1 |
| \(a = 6\) | A1 |
| (2) |
Notes
M1 Use of scalar product. Do not need to see \(\dfrac{1}{5}\)
A1 *Given Answer*
4a alt
| Components parallel to the plane: \(7\sin\theta = v\cos(\theta + \alpha)\) | |
| \(7\sin\theta = v(\cos\theta\cos\alpha - \sin\theta\sin\alpha),\) \(\Rightarrow 7\tan\theta = a - \tan\theta\) | M1 |
| \(8\tan\theta = a = 6\) | A1 |
| (2) |
\(\theta + \alpha = 46.3\ldots^\circ\)
M1 Equate components and form an equation in \(a\) and \(\theta\)
| Scheme | Marks |
|---|---|
| Component of \(-7\mathbf{j}\) parallel to the plane \(= \dfrac{21}{|4\mathbf{i} + 3\mathbf{j}|}\) | M1 |
| \(= 4.2\) | A1 |
| \(\sqrt{49 - 4.2^2} = \sqrt{31.36}\) | M1 |
| \(\sqrt{36 + 1 - 4.2^2} = \sqrt{19.36}\) | A1 |
| \(\sqrt{19.36} = e\times\sqrt{31.36}\) | DM1 |
| \(e = 0.786\) | A1 |
| (6) | |
| (8 marks) |
Notes
M1 Scalar product of \(-7\mathbf{j}\) and unit vector parallel to plane
M1 Use Pythagoras to find components perpendicular to the plane
A1 Both correct
DM1 Use if impact law. Dependent on preceding M mark
Alt 4b
| Component of \(-7\mathbf{j}\) perpendicular to the plane \(= \dfrac{1}{5}\begin{pmatrix}0\\-7\end{pmatrix}\cdot\begin{pmatrix}-3\\4\end{pmatrix}\) | M1A1 |
| Component of \(-a\mathbf{i} + \mathbf{j}\) perpendicular to the plane \(= \dfrac{1}{5}\begin{pmatrix}-6\\1\end{pmatrix}\cdot\begin{pmatrix}-3\\4\end{pmatrix}\) | M1A1 |
| Impact law: \(e = \dfrac{\frac{1}{5}\times 22}{\frac{1}{5}\times 28} = \dfrac{22}{28} = \dfrac{11}{14}\ (= 0.786)\) | DM1 A1 |
| (6) |
Alt 4b
| Components perpendicular to the plane: \(e\times 7\cos\theta = v\sin(\theta + \alpha)\) | M1 |
| \(e\times 7\cos\theta = v(\sin\theta\cos\alpha + \cos\theta\sin\alpha)\) | A1 |
| Substitute for \(\alpha\): \(7e = 6\tan\theta + 1\) | M1A1 |
| Solve for \(e\): \(7e = 6\times\dfrac{3}{4} + 1 = \dfrac{11}{2},\quad e = \dfrac{11}{14}\) | DM1 A1 |
| (6) |
\(\theta + \alpha = 46.3\ldots^\circ,\ v = \sqrt{37}\)
Alt 4b
| Components perpendicular to the plane: \(e\times 7\cos\theta = v\sin(\theta + \alpha)\) | M1 |
| \(e\times 7\cos\theta = v(\sin\theta\cos\alpha + \cos\theta\sin\alpha)\) | A1 |
| Divide and substitute for \(\alpha\): \(e\cot\theta = \tan(\theta + \alpha) = \dfrac{\tan\theta + \tan\alpha}{1 - \tan\theta\tan\alpha}\) | M1 |
| \(= \dfrac{\frac{3}{4} + \frac{1}{6}}{1 - \frac{3}{4\times 6}} = \dfrac{3\times 6 + 4}{4\times 6 - 3}\) | A1 |
| Solve for \(e\): \(e = \dfrac{22}{21}\times\dfrac{3}{4} = \dfrac{11}{14}\) | DM1 A1 |
| (6) |
Alt 4b
| Parallel: \(7\sin\theta = \sqrt{37}\cos(\theta + \alpha)\) Perpendicular: \(e7\cos\theta = \sqrt{37}\sin(\theta + \alpha)\) | M1A1 |
| \(49\sin^2\theta = 37\cos^2(\theta + \alpha)\) \(\Rightarrow \sin^2(\theta + \alpha) = 1 - \dfrac{49}{37}\sin^2\theta\) | M1 |
| \(49e^2\cos^2\theta = 37\sin^2(\theta + \alpha) = 37 - 49\sin^2\theta\) | A1 |
| \(e^2 = \dfrac{37 - 49\times\frac{9}{25}}{49\times\frac{16}{25}} = \dfrac{121}{196},\quad e = \dfrac{11}{14}\) | M1A1 |
| (6) |
M1A1 Pair of equations
M1 Square and substitute to eliminate \(\theta + \alpha\)
M1A1 Substitute for \(\theta\) to obtain \(e\).