M2 June 2018 Q7
7. A particle, of mass 0.3 kg, is projected from a point \(O\) on horizontal ground with speed \(u\). The particle is projected at an angle \(\alpha\) above the horizontal, where \(\tan\alpha = 2\), and moves freely under gravity. When the particle has moved a horizontal distance \(x\) from \(O\), its height above the ground is \(y\).
The particle hits the ground at the point \(A\), where \(OA = 36\) m.
The point \(B\) lies on the path of the particle. The direction of motion of the particle at \(B\) is perpendicular to the initial direction of motion of the particle.
| Scheme | Marks |
|---|---|
| Horizontal distance: \(\ x = u\cos\alpha t\) | B1 |
| Vertical distance: \(\ y = u\sin\alpha t - \dfrac{1}{2}gt^2\) | M1A1 |
| \(y = u\sin\alpha \times \dfrac{x}{u\cos\alpha} - \dfrac{g}{2} \times \left(\dfrac{x}{u\cos\alpha}\right)^2\) \(= x\tan\alpha - \dfrac{gx^2}{2u^2} \times \dfrac{1}{\cos^2\alpha} = 2x - \dfrac{gx^2}{2u^2} \times \dfrac{1}{1/5}\) | DM1 |
| \(= 2x - \dfrac{5g}{2u^2}x^2\) | A1 |
| (5) |
Notes
B1 \(\dfrac{1}{\sqrt{5}}ut\)
M1A1 \(\dfrac{2}{\sqrt{5}}ut - \dfrac{1}{2}gt^2\). Condone sign errors and sin/cos confusion
DM1 Substitute for \(t\) and \(\alpha\). Dependent on previous M1
A1 Obtain given answer from exact working
| Scheme | Marks |
|---|---|
| \(x = 36, y = 0\) : \(\ 0 = 2 - \dfrac{5g}{2u^2} \times 36\), | M1 |
| \(u^2 = \dfrac{5g \times 36}{4},\ \ \ \ u = 21\) (m s\(^{-1}\)) | A1 |
| (2) |
Notes
M1 Use given equation or a complete method using suvat to find \(u\).
A1 Accept \(\sqrt{45g}\)
| Scheme | Marks |
|---|---|
| Min speed \(= u\cos\alpha\) | M1 |
| Minimum KE: \(\dfrac{1}{2} \times 0.3 \times (u\cos\alpha)^2 = \dfrac{0.3}{2}\left(\dfrac{21}{\sqrt{5}}\right)^2 = 13.2\ (13)\) (J) | DM1 A1 |
| (3) |
Notes
M1 \(\left(u\cos\alpha = 21 \times \dfrac{1}{\sqrt{5}} = 9.39\ (\text{m s}^{-1})\right)\) Consistent with their B1 in (a)
DM1 A1 Dependent on previous M1
7c alt
| Max ht when \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0,\ \ \ x = \dfrac{2u^2}{5g}\ (= 18)\) | M1 |
| Conservation of energy: \(\dfrac{1}{2}mu^2 - mgh = \dfrac{1}{2}mv^2\) | M1 |
| \(= \dfrac{1}{2} \times 0.3 \times 21^2 - 0.3 \times g \times \dfrac{2 \times 21^2}{5g}\) \(= 13.2\) (J) | A1 |
M1 Or from \(\dfrac{1}{2} \times 36\) (symmetry)
| Scheme | Marks |
|---|---|
| Gradient of trajectory at \(B = -\dfrac{1}{2}\) | B1 |
| Differentiate and equate : \(\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = 2 - \dfrac{5g}{u^2}x = -\dfrac{1}{2}\) | M1A1 |
| solve for \(x\): \(\ \ -\dfrac{1}{2} = 2 - \dfrac{5g}{u^2}x,\ \ \dfrac{5}{2} = \dfrac{5gx}{u^2}\) | DM1 |
| \(x = \dfrac{21^2}{2g} = 22.5\ (23)\) (m) | A1 |
| (5) | |
| (15 marks) |
Notes
B1 Accept \(\dfrac{-1}{\tan\alpha}\)
M1A1 (their \(u\))
DM1 Dependent on previous M1
7d alt1
| Gradient of trajectory at \(B = -\dfrac{1}{2}\) | B1 |
| Use components of velocity: \(-\dfrac{1}{2} = \dfrac{u\sin\alpha - gt}{u\cos\alpha}\) | M1 |
| \(t = \dfrac{u\sin\alpha + \frac{1}{2}u\cos\alpha}{g}\left(= \dfrac{105}{2g\sqrt{5}}\right)\) | A1 |
| Horizontal distance: \(\ u\cos\alpha t = 22.5\ (23)\) (m) | DM1 A1 |
B1 Accept \(\dfrac{-1}{\tan\alpha}\)
A1 \((t = 2.40)\)
DM1 A1 Dependent on previous M1
7d alt2
| Gradient of trajectory at \(B = -\dfrac{1}{2}\) | B1 |
| \(v_y = -\dfrac{1}{2} \times 21 \times \dfrac{1}{\sqrt{5}},\ \ -\dfrac{21}{2\sqrt{5}} = 21 \times \dfrac{2}{\sqrt{5}} - gt\) | M1 |
| \(t = \dfrac{\frac{5}{2} \times \frac{21}{\sqrt{5}}}{g}\ (= 2.39)\) | A1 |
| Horizontal distance: \(\ u\cos\alpha t = 22.5\ (23)\) (m) | DM1 A1 |
B1 Can be implied by downward velocity \(\dfrac{21\sqrt{5}}{2}\) or \(\dfrac{u\cos\alpha}{\tan\alpha}\)
M1 Use suvat to find \(t\)
DM1 A1 Dependent on previous M1
(Corrected from the printed mark scheme: in 7d alt1 and alt2 the horizontal distance is printed as \(u\cos\theta t\); the angle is \(\alpha\).)
7d alt3
| \(\begin{pmatrix}u\cos\alpha\\u\sin\alpha\end{pmatrix} \cdot \begin{pmatrix}u\cos\alpha\\u\sin\alpha - gt\end{pmatrix} = 0\) | B1 |
| \(u^2\left(\cos^2\alpha + \sin^2\alpha\right) - u\sin\alpha gt = 0\) \(\Rightarrow u = \sin\alpha.gt\) | M1A1 |
| Horizontal distance: \(u\cos\alpha.t = u\cos\alpha \times \dfrac{u}{g\sin\alpha} = \dfrac{21^2}{2g} = 22.5\) | M1A1 |
B1 Scalar product = 0
M1A1 Must have –gt in second vector. Solve for \(t\)