M3 January 2009 Q7
7.
A particle is projected from the highest point \(A\) on the outer surface of a fixed smooth sphere of radius \(a\) and centre \(O\). The lowest point \(B\) of the sphere is fixed to a horizontal plane. The particle is projected horizontally from \(A\) with speed \(\tfrac{1}{2}\sqrt{(ga)}\). The particle leaves the surface of the sphere at the point \(C\), where \(\angle AOC = \theta\), and strikes the plane at the point \(P\), as shown in Figure 5.
(a) Show that \(\cos\theta = \tfrac{3}{4}\). (7)
(b) Find the angle that the velocity of the particle makes with the horizontal as it reaches \(P\). (8)
| Scheme | Marks |
|---|---|
| Let speed at \(C\) be \(u\) | |
| CE \(\dfrac{1}{2}mu^2 - \dfrac{1}{2}m\left(\dfrac{ag}{4}\right) = mga(1 - \cos\theta)\) | M1 A1 |
| \(u^2 = \dfrac{9ga}{4} - 2ga\cos\theta\) | |
| \(mg\cos\theta\ \ (+R) = \dfrac{mu^2}{a}\) | M1 A1 |
| \(mg\cos\theta = \dfrac{9mg}{4} - 2mg\cos\theta\) eliminating \(u\) | M1 |
| Leading to \(\cos\theta = \dfrac{3}{4}\) * | M1 A1 |
| (7) |
| Scheme | Marks |
|---|---|
| At \(C\) \(u^2 = \dfrac{9ga}{4} - 2ga \times \dfrac{3}{4} = \dfrac{3}{4}ga\) | B1 |
| \((\rightarrow)\) \(u_x = u\cos\theta = \sqrt{\left(\dfrac{3ga}{4}\right)} \times \dfrac{3}{4} = \sqrt{\left(\dfrac{27ga}{64}\right)} = 2.033\sqrt{a}\) | M1 A1ft |
| \((\downarrow)\) \(u_y = u\sin\theta = \sqrt{\left(\dfrac{3ga}{4}\right)} \times \dfrac{\sqrt{7}}{4} = \sqrt{\left(\dfrac{21ga}{64}\right)} = 1.792\sqrt{a}\) | M1 |
| \(v_y^2 = u_y^2 + 2gh \Rightarrow v_y^2 = \dfrac{21}{64}ga + 2g \times \dfrac{7}{4}a = \dfrac{245}{64}ga\) | M1 A1 |
| \(\tan\psi = \dfrac{v_y}{u_x} = \sqrt{\left(\dfrac{245}{27}\right)} \approx 3.012\ \ldots\) | M1 |
| \(\psi \approx 72^\circ\) awrt 72\(^\circ\) Or \(1.3^{c}\ \ \left(1.2502^{c}\right)\) awrt \(1.3^{c}\) | A1 |
| (8) | |
| (15 marks) |
Alternative for the last five marks
| Let speed at \(P\) be \(v\). | |
| CE \(\dfrac{1}{2}mv^2 - \dfrac{1}{2}m\left(\dfrac{ag}{4}\right) = mg \times 2a\) or equivalent | M1 |
| \(v^2 = \dfrac{17mga}{4}\) | M1 A1 |
| \(\cos\psi = \dfrac{u_x}{v} = \sqrt{\left(\dfrac{27}{64} \times \dfrac{4}{17}\right)} = \sqrt{\left(\dfrac{27}{272}\right)} \approx 0.315\) | M1 |
| \(\psi \approx 72^\circ\) awrt 72\(^\circ\) | A1 |
Note: The time of flight from \(C\) to \(P\) is \(\dfrac{\sqrt{235} - \sqrt{21}}{8}\sqrt{\left(\dfrac{a}{g}\right)} \approx 1.38373\sqrt{\left(\dfrac{a}{g}\right)}\)