M3 January 2009 Q5
5.
One end \(A\) of a light elastic string, of natural length \(a\) and modulus of elasticity \(6mg\), is fixed at a point on a smooth plane inclined at 30\(^\circ\) to the horizontal. A small ball \(B\) of mass \(m\) is attached to the other end of the string. Initially \(B\) is held at rest with the string lying along a line of greatest slope of the plane, with \(B\) below \(A\) and \(AB = a\). The ball is released and comes to instantaneous rest at a point \(C\) on the plane, as shown in Figure 2.
Find
(a) the length \(AC\), (5)
(b) the greatest speed attained by \(B\) as it moves from its initial position to \(C\). (7)
| Scheme | Marks |
|---|---|
| Let \(x\) be the distance from the initial position of \(B\) to \(C\) GPE lost = EPE gained | |
| \(mgx\sin 30^\circ = \dfrac{6mgx^2}{2a}\) | M1 A1=A1 |
| Leading to \(x = \dfrac{a}{6}\) | M1 |
| \(AC = \dfrac{7a}{6}\) | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| The greatest speed is attained when the acceleration of \(B\) is zero, that is where the forces on \(B\) are equal. | |
| \((\nwarrow)\) \(T = mg\sin 30^\circ = \dfrac{6mge}{a}\) | M1 |
| \(e = \dfrac{a}{12}\) | A1 |
| CE \(\dfrac{1}{2}mv^2 + \dfrac{6mg}{2a}\left(\dfrac{a}{12}\right)^2 = mg\dfrac{a}{12}\sin 30^\circ\) | M1 A1=A1 |
| Leading to \(v = \sqrt{\left(\dfrac{ga}{24}\right)} = \dfrac{\sqrt{6ga}}{12}\) | M1 A1 |
| (7) | |
| (12 marks) |
Alternative approach to (b) using calculus with energy.
| Let distance moved by \(B\) be \(x\) | |
| CE \(\dfrac{1}{2}mv^2 + \dfrac{6mg}{2a}x^2 = mgx\sin 30^\circ\) | M1 A1=A1 |
| \(v^2 = gx - \dfrac{6g}{a}x^2\) | |
| For maximum \(v\) \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(v^2\right) = 2v\dfrac{\mathrm{d}v}{\mathrm{d}x} = g - \dfrac{12g}{a}x = 0\) | M1 A1 |
| \(x = \dfrac{a}{12}\) | |
| \(v^2 = g\left(\dfrac{a}{12}\right) - \dfrac{6g}{a}\left(\dfrac{a}{12}\right)^2 = \dfrac{ga}{24}\) | M1 |
| \(v = \sqrt{\left(\dfrac{ga}{24}\right)}\) | A1 (7) |
Alternative approach to (b) using calculus with Newton’s second law.
| As before, the centre of the oscillation is when extension is \(\dfrac{a}{12}\) | M1 A1 |
| N2L \(mg\sin 30^\circ - T = m\ddot{x}\) \(\dfrac{1}{2}mg - \dfrac{6mg\left(\frac{a}{12} + x\right)}{a} = m\ddot{x}\) | M1 A1 |
| \(\ddot{x} = -\dfrac{6g}{a}x \Rightarrow \omega^2 = \dfrac{6g}{a}\) | A1 |
| \(v_{\max} = \omega a = \sqrt{\left(\dfrac{6g}{a}\right)} \times \dfrac{a}{12} = \sqrt{\left(\dfrac{ga}{24}\right)}\) | M1 A1 (7) |