M2 June 2015 Q7
7.

At time \(t = 0\), a particle is projected from a fixed point \(O\) on horizontal ground with speed \(u\) m s\(^{-1}\) at an angle \(\theta^\circ\) to the horizontal. The particle moves freely under gravity and passes through the point \(A\) when \(t = 4\) s. As it passes through \(A\), the particle is moving upwards at 20\(^\circ\) to the horizontal with speed 15 m s\(^{-1}\), as shown in Figure 3.
At the point \(B\) on its path the particle is moving downwards at 20\(^\circ\) to the horizontal with speed 15 m s\(^{-1}\).
The particle reaches the ground at the point \(C\).
| Scheme | Marks |
|---|---|
| After 4 seconds from O, horizontal speed \(= u\cos\theta\) | B1 |
| Vertical component of speed at \(A = u + at\) | M1 |
| \(= u\sin\theta - 4g\) | A1 |
| At \(A\), components are \(15\cos 20\) (horizontal) and \(15\sin 20\) (vertical) | B1 |
| \(u\cos\theta = 15\cos 20\) \(u\sin\theta = 15\sin 20 + 4g\) | DM1 |
| \(\theta = 72.4\ \ (72)\) | A1 |
| \(u = 46.5\ \ \ (47)\) | A1 |
| (7) |
Notes
M1 Complete method using suvat to find \(v\).
DM1 Form simultaneous equations in \(u\) and \(\theta\) and attempt to solve for \(u\) or \(\theta\). Depends on the previous M1
A1 Remember - A0 for the first overspecified answer
Alt7a
| After 4 seconds from O, horizontal speed \(= u\cos\theta\) | B1 |
| At \(t = 4\), \(s = vt - \dfrac{1}{2}gt^2\) | M1 |
| \(= 98.9\ldots\) | A1 |
| At A, components are \(15\cos 20\) (horizontal) and \(15\sin 20\) (vertical) | B1 |
| \(\dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - 2gh\) | DM1 |
| \(u = 46.5\ \ \ (47)\) | A1 |
| \(\theta = 72.4\ \ (72)\) | A1 |
M1 Complete method to find the vertical height at \(A\)
DM1 Conservation of energy. The equation needs to include all three terms but condone sign error(s).
A1 Remember - A0 for the first overspecified answer
A1 Beware inappropriate use of suvat
| Scheme | Marks |
|---|---|
| \(-15\sin 20 = 15\sin 20 - gt\ \ \ \) or \(\ \ 0 = 15\sin 20t - \dfrac{1}{2}gt^2\) | M1 |
| \(t = 1.05\) (s) or 1.0 (s) | A1 |
| (2) |
Notes
M1 Complete method using suvat or otherwise to find the time to travel from \(A\) to \(B\)
| Scheme | Marks |
|---|---|
| Total time \(= 4 + (1.05) + 4\) | B1ft |
| Range \(= 46.5 \times \cos 72.4 \times (8 + 1.05)\ \ \ (\text{or } 15\cos 20 \times 9.05)\) | M1 |
| \(= 128\) (m) or 127 (m) (130) | A1 |
| (3) | |
| (12 marks) |
Notes
B1ft Follow their \(t\) or \(\dfrac{2u\sin\theta}{g}\) for their \(u, \theta\)
M1 Correct method to find \(OC\) for their \(t\), \(u\) and \(\theta\)