M3 January 2010 Q7
7. A light elastic string has natural length \(a\) and modulus of elasticity \(\dfrac{3}{2}mg\). A particle \(P\) of mass \(m\) is attached to one end of the string. The other end of the string is attached to a fixed point \(A\). The particle is released from rest at \(A\) and falls vertically. When \(P\) has fallen a distance \(a + x\), where \(x > 0\), the speed of \(P\) is \(v\).
(a) Show that \(v^2 = 2g(a + x) - \dfrac{3gx^2}{2a}\). (4)
(b) Find the greatest speed attained by \(P\) as it falls. (4)
After release, \(P\) next comes to instantaneous rest at a point \(D\).
(c) Find the magnitude of the acceleration of \(P\) at \(D\). (6)
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{2}mv^2 + \dfrac{3mgx^2}{4a} = mg(a + x)\) | M1 A2 (1, 0) |
| leading to \(\quad v^2 = 2g(a + x) - \dfrac{3gx^2}{2a}\) * cso | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| Greatest speed is when the acceleration is zero \(T = \dfrac{\lambda x}{a} = \dfrac{3mgx}{2a} = mg \quad\Rightarrow\quad x = \dfrac{2a}{3}\) | M1 A1 |
| \(v^2 = 2g\left(a + \dfrac{2a}{3}\right) - \dfrac{3g}{2a} \times \left(\dfrac{2a}{3}\right)^2 \ \left(= \dfrac{8ag}{3}\right)\) | M1 |
| \(v = \dfrac{2}{3}\sqrt{(6ag)}\) accept exact equivalents | A1 |
| (4) |
Alternative to (b)
| \(v^2 = 2g(a + x) - \dfrac{3gx^2}{2a}\) | |
| Differentiating with respect to \(x\) \(2v\dfrac{\mathrm{d}v}{\mathrm{d}x} = 2g - \dfrac{3gx}{a}\) | |
| \(\dfrac{\mathrm{d}v}{\mathrm{d}x} = 0 \ \Rightarrow x = \dfrac{2a}{3}\) | M1 A1 |
| \(v^2 = 2g\left(a + \dfrac{2a}{3}\right) - \dfrac{3g}{2a} \times \left(\dfrac{2a}{3}\right)^2 \ \left(= \dfrac{8ag}{3}\right)\) | M1 |
| \(v = \dfrac{2}{3}\sqrt{(6ag)}\) accept exact equivalents | A1 |
| (4) |
Alternative approach using SHM for (b) and (c)
If SHM is used mark (b) and (c) together placing the marks in the gird as shown.
| Establishment of equilibrium position \(T = \dfrac{\lambda x}{a} = \dfrac{3mge}{2a} = mg \quad\Rightarrow\quad e = \dfrac{2a}{3}\) | bM1 bA1 |
| N2L, using \(y\) for displacement from equilibrium position \(m\ddot{y} = mg - \dfrac{\frac{3}{2}mg(y + e)}{a} = -\dfrac{3g}{2a}y\) | bM1 bA1 |
| \(\omega^2 = \dfrac{3g}{2a}\) | |
| Speed at end of free fall \(\quad u^2 = 2ga\) | cM1 |
| Using \(A\) for amplitude and \(v^2 = \omega^2\left(a^2 - x^2\right)\) \(u^2 = 2ga\) when \(y = -\dfrac{2}{3}a \quad\Rightarrow\quad 2ga = \dfrac{3g}{2a}\left(A^2 - \dfrac{4a^2}{9}\right)\) | cM1 |
| \(A = \dfrac{4a}{3}\) | cA1 |
| Maximum speed \(\quad A\omega = \dfrac{4a}{3} \times \sqrt{\left(\dfrac{3g}{2a}\right)} = \dfrac{2}{3}\sqrt{(6ag)}\) | cM1 cA1 |
| Maximum acceleration \(\quad A\omega^2 = \dfrac{4a}{3} \times \dfrac{3g}{2a} = 2g\) | cA1 |
| Scheme | Marks |
|---|---|
| \(v = 0 \quad\Rightarrow\quad 2g(a + x) - \dfrac{3gx^2}{2a} = 0\) | M1 |
| \(3x^2 - 4ax - 4a^2 = (x - 2a)(3x + 2a) = 0\) \(x = 2a\) | M1 A1 |
| At \(D\), \(\quad m\ddot{x} = mg - \dfrac{\lambda \times 2a}{a}\) ft their \(2a\) | M1 A1ft |
| \(\left|\ddot{x}\right| = 2g\) | A1 |
| (6) | |
| (14 marks) |
Alternative approach using SHM for (b) and (c)
If SHM is used mark (b) and (c) together placing the marks in the gird as shown.
| Establishment of equilibrium position \(T = \dfrac{\lambda x}{a} = \dfrac{3mge}{2a} = mg \quad\Rightarrow\quad e = \dfrac{2a}{3}\) | bM1 bA1 |
| N2L, using \(y\) for displacement from equilibrium position \(m\ddot{y} = mg - \dfrac{\frac{3}{2}mg(y + e)}{a} = -\dfrac{3g}{2a}y\) | bM1 bA1 |
| \(\omega^2 = \dfrac{3g}{2a}\) | |
| Speed at end of free fall \(\quad u^2 = 2ga\) | cM1 |
| Using \(A\) for amplitude and \(v^2 = \omega^2\left(a^2 - x^2\right)\) \(u^2 = 2ga\) when \(y = -\dfrac{2}{3}a \quad\Rightarrow\quad 2ga = \dfrac{3g}{2a}\left(A^2 - \dfrac{4a^2}{9}\right)\) | cM1 |
| \(A = \dfrac{4a}{3}\) | cA1 |
| Maximum speed \(\quad A\omega = \dfrac{4a}{3} \times \sqrt{\left(\dfrac{3g}{2a}\right)} = \dfrac{2}{3}\sqrt{(6ag)}\) | cM1 cA1 |
| Maximum acceleration \(\quad A\omega^2 = \dfrac{4a}{3} \times \dfrac{3g}{2a} = 2g\) | cA1 |