M2 June 2016 Q6
6. [In this question, \(\mathbf{i}\) is a horizontal unit vector and \(\mathbf{j}\) is an upward vertical unit vector.]
A particle \(P\) is projected from a fixed origin \(O\) with velocity \((3\mathbf{i} + 4\mathbf{j})\) m s\(^{-1}\). The particle moves freely under gravity and passes through the point \(A\) with position vector \(\lambda(\mathbf{i} - \mathbf{j})\) m, where \(\lambda\) is a positive constant.
| Scheme | Marks |
|---|---|
| Horizontal motion: \(\ x = 3t\) | B1 |
| Vertical motion: \(\ \ y = 4t - \dfrac{g}{2}t^2\) | M1 A1 |
| \(\left(y = 4 \times \dfrac{x}{3} - \dfrac{g}{2} \times \dfrac{x^2}{9}\right),\ \ \lambda = -\left(\dfrac{4\lambda}{3} - \dfrac{g\lambda^2}{18}\right)\) | M1 |
| \(,\ \ \dfrac{7\lambda}{3} = \dfrac{g\lambda^2}{18}\) | M1 |
| \(\lambda = \dfrac{42}{g}\) or 4.3 (4.29) | A1 |
| (6) |
Notes
M1 Correct use of suvat. Condone sign error(s)
M1 Use \(y = -x\) and form an equation in one variable
M1 solve for \(\lambda\)
A1 Not \(\dfrac{30}{7}\)
alta
| Horizontal motion: \(\ x = 3t\) | B1 |
| Vertical motion: \(\ \ y = 4t - \dfrac{g}{2}t^2\) | M1 A1 |
| \(\Rightarrow -3t = 4t - \dfrac{1}{2}gt^2,\ \ \left(t = \dfrac{14}{g}\right)\) | M1 |
| \(\lambda = 3t\) | M1 |
| \(\lambda = 4.3\ \ \ \ (4.29)\) | A1 (6) |
M1 Correct use of suvat. Condone sign error(s)
M1 Use \(y = -x\) and form an equation in one variable
M1 Solve for \(\lambda\)
| Scheme | Marks |
|---|---|
| At A: \(\ \ v_{\rightarrow}\ 3\) (m s\(^{-1}\)) | B1 |
| \(v_{\uparrow}\ \ 4 - g \times \dfrac{14}{g}\) | M1 |
| \(= -10\) (m s\(^{-1}\)) | A1 |
| Speed \(= \sqrt{(\text{their } 10)^2 + (3)^2}\) | DM1 |
| \(= \sqrt{109}\) (m s\(^{-1}\)) | A1 |
| \(\tan^{-1}\left(\dfrac{\text{their } 10}{3}\right)\) or \(\tan^{-1}\left(\dfrac{3}{\text{their } 10}\right)\) | DM1 |
| Direction \(= 73.3^\circ\) below the horizontal | A1 |
| (7) | |
| (13 marks) |
Notes
M1 Complete method using suvat to find \(v_{\uparrow}\) with their \(t\) or \(\lambda\)
A1 Accept +10 with direction confirmed by diagram
DM1 Dependent on the first M1 in (b)
A1 (10.4) Allow for \(v_{\uparrow} = 10\)
DM1 Use trig to find a relevant angle. Dependent on the first M1 in (b)
A1 (1.28 radians) Accept direction \(3\mathbf{i} - 10\mathbf{j}\). Do not accept a bearing
Alt 6b
| Loss in GPE : \(\ mg\lambda = 42m\) | B1 |
| Gain in KE : \(\ \dfrac{1}{2}mv^2 - \dfrac{1}{2}m \times 25\) | M1 A1 |
| Solve for \(v\): \(\ 42 = \dfrac{1}{2}v^2 - \dfrac{25}{2}\) | M1 |
| \(v = \sqrt{109}\) | A1 |
| \(v\cos\theta = 3\) | M1 |
| \(\theta = 73.3^\circ\) below the horizontal | A1 (7) |
M1 Terms must be dimensionally correct. Condone sign error.
M1 Use trig. to find a relevant angle
A1 (7) Accept correct angle marked correctly on a diagram.