A2 June 2024 Q7
7.
A smooth solid hemisphere has radius \(r\) and the centre of its plane face is \(O\).
The hemisphere is fixed with its plane face in contact with horizontal ground, as shown in Figure 6.
A small stone is at the point \(A\), the highest point on the surface of the hemisphere.
The stone is projected horizontally from \(A\) with speed \(U\).
The stone is still in contact with the hemisphere at the point \(B\), where \(OB\) makes an angle \(\theta\) with the upward vertical.
The speed of the stone at the instant it reaches \(B\) is \(v\).
The stone is modelled as a particle \(P\) and air resistance is modelled as being negligible.
When \(P\) leaves the surface of the hemisphere, the speed of \(P\) is \(W\).
Given that \(U = \sqrt{\dfrac{2rg}{3}}\)
After leaving the surface of the hemisphere, \(P\) moves freely under gravity until it hits the ground.
At the instant when \(P\) hits the ground it is travelling at \(\alpha^\circ\) to the horizontal.
| Scheme | Marks | AO |
|---|---|---|
 | ||
| Energy equation | M1 | 3.1b |
| \(\dfrac{1}{2}mU^2 + mgr(1 - \cos\theta) = \dfrac{1}{2}mv^2\) | A1 | 1.1b |
| \(v^2 = U^2 + 2gr(1 - \cos\theta)\) | A1 | 1.1b |
| (3) |
Notes
M1: Equation for conservation of mechanical energy. All terms required and no extras; dimensionally correct. Condone sine / cosine confusion and sign errors.
A1: Correct unsimplified equation
A1: Or equivalent with \(v^2\) as subject
| Scheme | Marks | AO |
|---|---|---|
| Equation for circular motion at \(B\) | M1 | 2.1 |
| \(mg\cos\theta - R = \dfrac{mW^2}{r}\) | A1 | 1.1b |
| Use \(R = 0\) : \(mg\cos\theta = \dfrac{m\left(U^2 + 2gr(1 - \cos\theta)\right)}{r}\) | M1 | 3.3 |
| \(rmg\cos\theta = m\left(\dfrac{2rg}{3} + 2gr(1 - \cos\theta)\right) \quad \left(\Rightarrow \cos\theta = \dfrac{8}{9}\right)\) | A1 | 1.1b |
| \(\Rightarrow W^2 = rg\cos\theta = \dfrac{8}{9}rg\) * | A1* | 2.2a |
| (5) |
Notes
M1: Equation for circular motion. Dimensionally correct. Condone sine / cosine confusion and sign errors. Condone if \(R = 0\) seen or implied at this stage.
A1: Correct unsimplified equation
M1: Use \(R = 0\) in a relevant equation to obtain equation in \(r\), \(g\) and \(\cos\theta\) or \(W\)
A1: Any correct equation in \(r\), \(g\) and \(\cos\theta\) or \(r\), \(g\) and \(W\)
E.g. \(W^2 = \dfrac{2gr}{3} + 2gr - 2W^2\)
A1*: Obtain given result from correct exact working
| Scheme | Marks | AO |
|---|---|---|
| Energy equation | M1 | 3.1b |
| From \(B\): \(\dfrac{1}{2}mV^2 = \dfrac{1}{2}mW^2 + mgr\cos\theta\) or from \(A\): \(\dfrac{1}{2}mV^2 = \dfrac{1}{2}mU^2 + mgr\) | A1 | 1.1b |
| \(V^2 = v^2 + 2gr\cos\theta = \dfrac{8rg}{9} + \dfrac{16rg}{9} = \dfrac{8rg}{3}, \quad V = \sqrt{\dfrac{8rg}{3}}\) | A1 | 1.1b |
| (3) |
Notes
M1: Complete method to find the speed, e.g. by using conservation of energy or projectile motion. All terms required and no extras; dimensionally correct.
A1: Correct unsimplified equation(s)
A1: Any equivalent form.
| Scheme | Marks | AO |
|---|---|---|
 | ||
| Form an equation in \(\alpha\) | M1 | 3.1b |
| \(\cos\alpha^\circ = \dfrac{W\cos\theta}{V}\ \left(= \dfrac{\frac{8}{9}\sqrt{\frac{8rg}{9}}}{\sqrt{\frac{8rg}{3}}} = \dfrac{8\sqrt{3}}{27}\right)\) | A1ft | 1.1b |
| \(\alpha = 59(.12276...)\) | A1 | 1.1b |
| (3) | ||
| (14 marks) |
Notes
M1: Complete method to form a trig ratio for \(\alpha\)
A1ft: Correct use of their values to obtain a ratio for \(\alpha\). Ft their \(V\).
\(\left(\sin\alpha = \dfrac{\sqrt{537}}{27} = 0.858, \quad \tan\alpha = \dfrac{\sqrt{179}}{8} = 1.67\right)\)
A1: 59 or better