June 2024 Paper 1 Q14
14 A man runs at a constant speed of \(4\ \text{m s}^{-1}\) along a straight horizontal road. A woman is standing on a bridge that spans the road. At the instant that the man passes directly below the woman she throws a ball with initial speed \(u\ \text{m s}^{-1}\) at \(\alpha^\circ\) above the horizontal. The path of the ball is directly above the road. The man catches the ball 2.4 s after it is thrown. At the instant the man catches it, the ball is 3.6 m below the level of the point of projection.
- The value of \(u\)
- The value of \(\alpha\)
| Scheme | Marks | AO |
|---|---|---|
| The size and shape of the ball are neglected | B1 | 2.4 |
| [1] |
Notes
B1: Sensible comment equivalent to “the object is modelled as a point mass”. Do not allow for a statement that includes correct and incorrect ideas. (see appendix)
Appendix: exemplar responses for Q14(a)
| Response | Mark |
|---|---|
| The mass is concentrated at the centre | B0 |
| The object has no mass | B0 |
| There is no air resistance | B0 |
| The weight acts at the centre [of mass] | B0 |
| There is no spin | B1 |
| The ball’s size and shape do not matter | B1 |
| Only its mass is taken into account so it doesn’t spin | B1 |
| The size of the ball is negligible and there are no external forces acting on the ball | B0 |
| Point mass means there is no air resistance | B0 |
| There is no air resistance because the object has no size | B0 |
| Scheme | Marks | AO |
|---|---|---|
| Vertical motion \(s = -3.6, a = -9.8, t = 2.4\) \(s = u_y t + \tfrac{1}{2}at^2\) \(-3.6 = 2.4u_y - 4.9 \times 2.4^2\) | M1 | 3.1b |
| Giving \(u_y = 10.26\ \text{m s}^{-1}\) | A1 | 1.1 |
| [2] |
Notes
M1: Use of suvat equation(s) with \(s = \pm 3.6\) and \(v \neq 0\) leading to a value for \(u_y\) Allow sign errors
A1: cao
Allow SC1 if M0 awarded and \(u\sin\alpha\) seen.
Or award SC2 for \(u\sin\alpha\) in (b) if 10.26 seen in (c)
| Scheme | Marks | AO |
|---|---|---|
| The horizontal velocity of the ball must be the same as the velocity of the man | B1 | 3.1b |
| \(u = \sqrt{4^2 + 10.26^2} = 11.01\) | B1 | 1.1 |
| \(\alpha = \tan^{-1}\dfrac{10.26}{4} = 68.7^\circ\) | M1 A1 | 1.1 1.1 |
| [4] |
Notes
B1: Soi eg \(u\cos\alpha = 4\)
B1: FT their (b) and their horizontal velocity
M1: Also allow from solving \(u\cos\alpha = 4\) or \(u\sin\alpha = 10.26\)
A1: FT their (b) and their horizontal velocity
Alternative solution for last 3 marks
| Scheme | Marks | AO |
|---|---|---|
| \(-3.6 = 2.4\dfrac{4}{\cos\alpha}\sin\alpha - 4.9 \times 2.4^2\) | M1 | |
| \(\alpha = 68.7^\circ\) | A1 | |
| So \(u = \left[\dfrac{4}{\cos 68.7}\right] = 11.01\) | B1 |
M1: Substitute for \(u\) in terms of \(\cos\alpha\) leading to a value for \(\tan\alpha\)
A1: FT their horizontal velocity
B1: FT their value for \(\alpha\)