M3 June 2014 (R) Q4
4.
A smooth sphere of radius \(a\) is fixed with a point \(A\) of its surface in contact with a fixed vertical wall. A particle is placed on the highest point of the sphere and is projected towards the wall and perpendicular to the wall with horizontal speed \(\sqrt{\dfrac{2ag}{5}}\), as shown in Figure 2.
The particle leaves the surface of the sphere with speed \(V\).
The particle strikes the wall at the point \(X\).
| Scheme | Marks |
|---|---|
| \(\tfrac{1}{2}mV^2 - \tfrac{1}{2}m\tfrac{2ag}{5} = mga(1 - \cos\theta)\) | M1 A1 A1 |
| \(mg\cos\theta = m\tfrac{V^2}{a}\) | M1 A1 |
| \(V = \sqrt{\dfrac{4ag}{5}}\) | DM1 A1 |
| (7) |
Notes
M1 energy equation
A1 correct difference of KEs
A1 fully correct equation
M1 NL2 towards the centre. May include \(R\)
A1 correct equation May include \(R\)
DM1 set \(R = 0\) and solve for \(V\) or \(V^2\)
A1 correct final answer with no errors in working
| Scheme | Marks |
|---|---|
| \(\cos\theta = \dfrac{4}{5}\) | B1 |
| \(t = \dfrac{a - a\sin\theta}{V\cos\theta}\ \ \left(= \sqrt{\dfrac{5a}{16g}}\right)\) | M1 A1 |
| \(s = Vt\sin\theta + \tfrac{1}{2}gt^2\) | M1 |
| \(= \sqrt{\dfrac{4ag}{5}}\sqrt{\dfrac{5a}{16g}}\,\dfrac{3}{5} + \tfrac{1}{2}g\left(\dfrac{5a}{16g}\right)\) | M1 A1 |
| \(= \dfrac{73a}{160}\) | A1 |
| \(AX = a\cos\theta - \dfrac{73a}{160}\) | M1 |
| \(= \dfrac{11a}{32}\) | A1 |
| (9) | |
| (16 marks) |
Notes
B1 for correct trig function for \(\theta\)
M1 using the horizontal distance and speed to obtain an expression for the time
A1 correct expression
M1 using \(s = ut + \dfrac{1}{2}at^2\) to get the vertical distance
M1 attempt at initial vertical velocity
A1 correct initial vertical velocity
A1 correct vertical distance
M1 attempt distance \(AX\)
A1 correct final answer