AS June 2022 Paper 1 Q6
6. The surface of a horizontal tennis court is modelled as part of a horizontal plane, with the origin on the ground at the centre of the court, and
- \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors directed across the width and length of the court respectively
- \(\mathbf{k}\) is a unit vector directed vertically upwards
- units are metres
After being hit, a tennis ball, modelled as a particle, moves along the path with equation
\[\mathbf{r} = \left(-4.1 + 9\lambda - 2.3\lambda^2\right)\mathbf{i} + (-10.25 + 15\lambda)\mathbf{j} + \left(0.84 + 0.8\lambda - \lambda^2\right)\mathbf{k}\]where \(\lambda\) is a scalar parameter with \(\lambda \geqslant 0\)
Assuming that the tennis ball continues on this path until it hits the ground,
The direction in which the tennis ball is moving at a general point on its path is given by
\[(9 - 4.6\lambda)\mathbf{i} + 15\mathbf{j} + (0.8 - 2\lambda)\mathbf{k}\]The net of the tennis court lies in the plane \(\mathbf{r}.\mathbf{j} = 0\)
The maximum height above the court of the top of the net is 0.9 m.
Modelling the top of the net as a horizontal straight line,
With reference to the model,
| Scheme | Marks | AO |
|---|---|---|
| Need \(\mathbf{k}\) component to be zero at ground, so \(0.84 + 0.8\lambda - \lambda^2 = 0 \Rightarrow \lambda = \ldots\) | M1 | 1.1b |
| \(\lambda = -\dfrac{3}{5}, \dfrac{7}{5}\), but \(\lambda \geqslant 0\) so \(\lambda = \dfrac{7}{5}\) | A1 | 1.1b |
| (2) |
Notes
Accept any alternative vector notations throughout.
M1: Attempts to solve the quadratic from equating the \(\mathbf{k}\) component to zero.
A1: Correct value, must select positive root, so accept 1.4 oe.
Correct answer only M1 A1
| Scheme | Marks | AO |
|---|---|---|
| Direction is \((9 - 4.6 \times 1.4)\mathbf{i} + 15\mathbf{j} + (0.8 - 2 \times 1.4)\) \(= 2.56\mathbf{i} + 15\mathbf{j} - 2\mathbf{k}\) or \(\dfrac{64}{25}\mathbf{i} + 15\mathbf{j} - 2\mathbf{k}\) | B1ft | 2.2a |
| (1) |
Notes
B1ft: For \((2.56, 15, -2)\) o.e or follow through \((9 - 4.6 \times \text{‘}\lambda\text{’}, 15, 0.8 - 2 \times \text{‘}\lambda\text{’})\) for their \(\lambda\).
| Scheme | Marks | AO |
|---|---|---|
| Direction perpendicular to ground is \(a\mathbf{k}\), so angle to perpendicular is given by\[(\cos\theta) = \frac{a\mathbf{k}.(2.56\mathbf{i} + 15\mathbf{j} - 2\mathbf{k})}{a \times |2.56\mathbf{i} + 15\mathbf{j} - 2\mathbf{k}|}\ \text{ or }\ \frac{\begin{pmatrix}2.56\\ 15\\ -2\end{pmatrix}\bullet\begin{pmatrix}0\\ 0\\ a\end{pmatrix}}{\left|\begin{pmatrix}2.56\\ 15\\ -2\end{pmatrix}\right|\left|\begin{pmatrix}0\\ 0\\ a\end{pmatrix}\right|}\]or angle between \(\begin{pmatrix}2.56\\ 15\\ -2\end{pmatrix}\) and \(\begin{pmatrix}2.56\\ 15\\ 0\end{pmatrix}\) is given by\[(\cos\theta) = \frac{\begin{pmatrix}2.56\\ 15\\ -2\end{pmatrix}\bullet\begin{pmatrix}2.56\\ 15\\ 0\end{pmatrix}}{\left|\begin{pmatrix}2.56\\ 15\\ -2\end{pmatrix}\right|\left|\begin{pmatrix}2.56\\ 15\\ 0\end{pmatrix}\right|}\] | M1 | 1.1b |
| \[= \frac{-2}{\sqrt{2.56^2 + 15^2 + (-2)^2}}\ (= -0.130\ldots)\]Or\[= \frac{231.5536}{\sqrt{2.56^2 + 15^2 + (-2)^2}\sqrt{2.56^2 + 15^2 + (0)^2}} = 0.991\ldots\] | M1 | 1.1b |
| \(90^\circ - \arccos(\text{‘}-0.130\ldots\text{’}) = -7.48\ldots\) or \(\arccos(0.991\ldots)\) | ddM1 | 3.1b |
| So the tennis ball hits ground at angle of 7.5° (1d.p.) cao | A1 | 3.2a |
| Alternative Finds the length of the vector in the ij plane \(= \sqrt{2.56^2 + 15^2}\) | M1 | 1.1b |
| \(\tan\theta = \dfrac{2}{\sqrt{2.56^2 + 15^2}}\) | M1 | 1.1b |
| \(\theta = \arctan\left(\dfrac{2}{\sqrt{2.56^2 + 15^2}}\right)\) or \(\theta = 90 - \arctan\left(\dfrac{\sqrt{2.56^2 + 15^2}}{2}\right)\) | ddM1 | 3.1b |
| So the tennis ball hits ground at angle of 7.5° (1d.p.) | A1 | 3.2a |
| (4) |
Notes
M1: Recognises the angle between the perpendicular and direction vector is needed, and identifies the perpendicular as \(a\mathbf{k}\) for any non-zero \(a\) (including 1), and attempts dot product
Alternatively recognises the dot product of \((2.56, 15, -2)\) and \((2.56, 15, 0)\)
M1: Applies the dot product formula \(\dfrac{a \bullet b}{|a||b|}\) correctly between any two vectors, but must have dot product and modulus evaluated.
ddM1: Dependent on both previous marks. A correct method to proceed to the required angle, usually \(90^\circ - \arccos(\text{‘}-0.130\ldots\text{’})\) as shown in scheme but may e.g. use \(\sin\theta\) instead of \(\cos\theta\) in formula.
Alternatively is using dot product of \((2.56, 15, -2)\) and \((2.56, 15, 0)\) finds arccos(0.991…)
A1: For 7.5° cao
Alternative
M1: Finds the length of the vector in the ij plane.
M1: Finds the tan of any angle the
ddM1: Dependent on both previous marks. Finds the required angle
A1: For 7.5° cao
| Scheme | Marks | AO |
|---|---|---|
| In same plane as net when \(\mathbf{r}.\mathbf{j} = 0\),\[\begin{pmatrix}-4.1 + 9\lambda - 2.3\lambda^2\\ -10.25 + 15\lambda\\ 0.84 + 0.8\lambda - \lambda^2\end{pmatrix}\bullet\begin{pmatrix}0\\ 1\\ 0\end{pmatrix}\ \text{leading to } -10.25 + 15\lambda = 0 \Rightarrow \lambda = \ldots\]\[\left(= \frac{41}{60} = 0.683333\ldots\right)\] | M1 | 3.1b |
| So is at position\[\left(-4.1 + 9 \times \frac{41}{60} - 2.3\left(\frac{41}{60}\right)^2\right)\mathbf{i} + 0\mathbf{j} + \left(0.84 + 0.8 \times \frac{41}{60} - \left(\frac{41}{60}\right)^2\right)\mathbf{k}\] | M1 | 1.1b |
| \(=\) awrt \(0.976\mathbf{i} +\) awrt \(0.920\mathbf{k}\) or \(=\) awrt \(0.976\mathbf{i} + 0.92\mathbf{k}\) (to 3 s.f.) or \(=\) awrt \(0.976\mathbf{i} + \dfrac{3311}{3600}\mathbf{k}\) | A1 | 1.1b |
| (3) |
Notes
M1: Attempts to find value of \(\lambda\) that gives zero \(\mathbf{j}\) component.
M1: Uses their value of \(\lambda\) in the equation of the path to find position.
A1: Correct position.
| Scheme | Marks | AO |
|---|---|---|
| Modelling as a line, height of net is 0.9m along its length so as 0.92 > 0.9 the ball will pass over the net according to the model. | B1ft | 3.2a |
| (1) |
Notes
B1ft: States that 0.920 > 0.9 so according to the model the ball will pass over the net. Follow through on their \(\mathbf{k}\) component and draws an appropriate conclusion. May stay the value of k > 0.92
| Scheme | Marks | AO |
|---|---|---|
| Identifies a suitable feature of the model that affects the outcome | M1 | 3.2b |
| And uses it to draw a compatible conclusion. For example
| A1 | 2.2b |
| (2) | ||
| (13 marks) |
Notes
M1: There must be some reference to the model to score this mark. See scheme for examples. It is likely to be either the ball is not a particle, or the top of the net is not a straight line. Accept references to the ball crossing a long way from the middle.
Do not accept reasons such as “there may be wind/air resistance” as these are not referencing the given model.
A1: For a reasonable conclusion based on their reference to the model.
For example
The ball is not a particle; therefore, it will not go over the net is M1A0 as not explained why – needs reference to radius/diameter