M2 June 2007 Q6
6.
A golf ball \(P\) is projected with speed 35 m s\(^{-1}\) from a point \(A\) on a cliff above horizontal ground. The angle of projection is \(\alpha\) to the horizontal, where \(\tan\alpha = \dfrac{4}{3}\). The ball moves freely under gravity and hits the ground at the point \(B\), as shown in Figure 4.
The horizontal distance from \(A\) to \(B\) is 168 m.
By considering energy, or otherwise,
| Scheme | Marks |
|---|---|
| \(0 = (35\sin\alpha)^2 - 2gh\) | M1 A1 |
| \(h = 40\) m | A1 |
| (3) |
Notes
M1 Use of \(v^2 = u^2 + 2as\), or possibly a 2 stage method using \(v = u + at\) and \(s = ut + \dfrac{1}{2}at^2\)
A1 Correct expression. Alternatives need a complete method leading to an equation in h only.
A1 40(m) No more than 2sf due to use of \(g\).
| Scheme | Marks |
|---|---|
| \(x = 168 \Rightarrow 168 = 35\cos\alpha \cdot t\) \((\Rightarrow t = 8\) s\()\) | M1 A1 |
| At \(t = 8\), \(\ y = 35\sin\alpha \times t - \dfrac{1}{2}gt^2\ \ (= 28 \times 8 - \tfrac{1}{2} \cdot g \cdot 8^2 = -89.6\) m\()\) | M1 A1 |
| Hence height of \(A = 89.6\) m or 90 m | DM1 A1 |
| (6) |
Notes
M1 Use of \(x = u\cos\alpha \cdot t\) to find \(t\).
A1 \(168 = 35 \times \textit{their}\cos\alpha \times t\)
M1 Use of \(s = ut + \dfrac{1}{2}at^2\) to find vertical distance for their \(t\). (AB or top to B)
A1 \(y = 35\sin\alpha \times t - \dfrac{1}{2}gt^2\) (\(u, t\) consistent)
DM1 This mark dependent of the previous 2 M marks. Complete method for AB. Eliminate t and solve for s.
A1 cso.
(NB some candidates will make heavy weather of this, working from A to max height (40m) and then down again to B (129.6m))
OR : Using \(y = x\tan\alpha - \dfrac{gx^2\sec^2\alpha}{2u^2}\)
M1 formula used (condone sign error)
A1 \(x, u\) substituted correctly
M1 \(\alpha\) terms substituted correctly.
A1 fully correct formula
M1, A1 as above
| Scheme | Marks |
|---|---|
| \(\tfrac{1}{2}mv^2 = \tfrac{1}{2} \cdot m \cdot 35^2 + mg \cdot 89.6\) | M1 A1 |
| \(\Rightarrow v = 54.6\) or 55 m s\(^{-1}\) | A1 |
| (3) | |
| (12 marks) |
Notes
M1 Conservation of energy: change in KE = change in GPE. All terms present. One side correct (follow their h). (will probably work A to B, but could work top to B).
A1 Correct expression (follow their h)
A1 54.6 or 55 (m/s)
OR: M1 horizontal and vertical components found and combined using Pythagoras
\(v_x = 21\)
\(v_y = 28 - 9.8 \times 8\ (-50.4)\)
A1 \(v_x\) and \(v_y\) expressions correct (as above). Follow their \(h, t\).
A1 54.6 or 55
NB Penalty for inappropriate rounding after use of g only applies once per question.