M3 June 2013 (R) Q5
5.
Part of a hollow spherical shell, centre \(O\) and radius \(r\), forms a bowl with a plane circular rim. The bowl is fixed to a horizontal surface at \(A\) with the rim uppermost and horizontal.
The point \(A\) is the lowest point of the bowl. The point \(B\), where \(\angle AOB = \alpha\) and \(\tan\alpha = \dfrac{3}{4}\), is on the rim of the bowl, as shown in Figure 2. A small smooth marble \(M\) is placed inside the bowl at \(A\), and given an initial horizontal speed \(\sqrt{(gr)}\). The motion of \(M\) takes place in the vertical plane \(OAB\).
(a) Show that the speed of \(M\) as it reaches \(B\) is \(\sqrt{\left(\dfrac{3}{5}gr\right)}\). (4)
After leaving the surface of the bowl at \(B\), \(M\) moves freely under gravity and first strikes the horizontal surface at the point \(C\). Given that \(r = 0.4\) m,
(b) find the distance \(AC\). (8)
| Scheme | Marks |
|---|---|
| Use of Energy at A = energy at B | |
| \(\dfrac{1}{2}mu^2 = \dfrac{1}{2}mv^2 + mgh,\ \ \dfrac{1}{2}mgr = \dfrac{1}{2}mv^2 + mg \times r(1 - \cos\alpha)\) | M1 A1A1 |
| \(= \dfrac{1}{2}mv^2 + mg \times r \times \dfrac{1}{5}\) | |
| \(v^2 = gr - \dfrac{2gr}{5} = \dfrac{3gr}{5}\) | |
| \(v = \sqrt{\dfrac{3gr}{5}}\) *AG* | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| v1 Horizontal component of speed at \(B\) and at \(C\) = their \(v\cos\theta\) | M1 |
| Vertical component of speed at B = their \(v\sin\theta\) | M1 |
| Conservation of energy gives speed at C \(= \sqrt{\dfrac{2g}{5}}\) | |
| Vertical component of speed at C \(= \sqrt{\dfrac{2g}{5} - \dfrac{16 \times 6g}{25^2}} \approx 1.5539..\) | M1A1 |
| \(v = u + at \Rightarrow\) \(t = \dfrac{1.5539\ldots + 0.92017\ldots}{g} \approx 0.252..\) seconds | M1A1 |
| Horizontal distance \(= \dfrac{3}{5} \times 0.4 + 1.22689.. \times 0.252\ldots = 0.55\) (m) | M1A1 |
| (8) | |
| (12 marks) |
(b) v2
| Horizontal component of speed at \(B\) and at \(C\) = their \(v\cos\theta\) | M1 |
| Vertical component of speed at B = their \(v\sin\theta\) | M1 |
| \(s = ut + \dfrac{1}{2}at^2:\ -\dfrac{1}{5} \times 0.4 = -\dfrac{2}{25} = \sqrt{\dfrac{6g}{25}} \times \dfrac{3}{5}t - \dfrac{1}{2}gt^2\) | M1A1 |
| \(4.9t^2 - .92017..t - 0.08 = 0\) | |
| \(t = \dfrac{0.920 + \sqrt{0.920^2 + 0.32 \times 4.9}}{9.8} = 0.252\ldots\ldots\) | M1A1 |
| Horizontal distance \(= \dfrac{3}{5} \times 0.4 + 1.22689.. \times 0.252\ldots = 0.55\) (m) | M1A1 |
(b) v3
| Horizontal component of speed at \(B\) and at \(C\) = their \(v\cos\theta\) | M1 |
| Vertical component of speed at B = their \(v\sin\theta\) | M1 |
| \(s = ut + \dfrac{1}{2}at^2:\ -\dfrac{1}{5} \times 0.4 = -\dfrac{2}{25} = \sqrt{\dfrac{6g}{25}} \times \dfrac{3}{5}t - \dfrac{1}{2}gt^2\) | M1A1 |
| \(4.9t^2 - .92017t - 0.08 = 0\) | |
| Horizontal distance from B \(= 1.22689\ldots \times t = x\) | |
| Form quadratic in x by substituting for \(t\) above | M1 |
| \(3.255x^2 - 0.75x - 0.08 = 0\) | |
| \(x = \dfrac{0.75 + \sqrt{0.75^2 + 4 \times 3.255 \times 0.08}}{2 \times 3.255} = 0.3097\ldots\) | M1A1 |
| Horizontal distance \(= \dfrac{3}{5} \times 0.4 + 0.3097\ldots = 0.55\) (m) | A1 |