M1 June 2007 Q6
6.
Two particles \(P\) and \(Q\) have mass 0.5 kg and \(m\) kg respectively, where \(m \lt 0.5\). The particles are connected by a light inextensible string which passes over a smooth, fixed pulley. Initially \(P\) is 3.15 m above horizontal ground. The particles are released from rest with the string taut and the hanging parts of the string vertical, as shown in Figure 4. After \(P\) has been descending for 1.5 s, it strikes the ground. Particle \(P\) reaches the ground before \(Q\) has reached the pulley.
(a) Show that the acceleration of \(P\) as it descends is 2.8 m s\(^{-2}\). (3)
(b) Find the tension in the string as \(P\) descends. (3)
(c) Show that \(m = \dfrac{5}{18}\). (4)
(d) State how you have used the information that the string is inextensible. (1)
When \(P\) strikes the ground, \(P\) does not rebound and the string becomes slack. Particle \(Q\) then moves freely under gravity, without reaching the pulley, until the string becomes taut again.
(e) Find the time between the instant when \(P\) strikes the ground and the instant when the string becomes taut again. (6)
| Scheme | Marks |
|---|---|
| \(s = ut + \tfrac{1}{2}at^2\ \ \Rightarrow\ \ 3.15 = \tfrac{1}{2}a \times \tfrac{9}{4}\) | M1 A1 |
| \(a = 2.8\ \ (\text{m s}^{-2})\) \(\ast\) cso | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| N2L for \(P\): \(0.5g - T = 0.5 \times 2.8\) | M1 A1 |
| \(T = 3.5\ \ (\text{N})\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| N2L for \(Q\): \(T - mg = 2.8m\) | M1 A1 |
| \(m = \dfrac{3.5}{12.6} = \dfrac{5}{18}\) \(\ast\) cso | DM1 A1 |
| (4) |
| Scheme | Marks |
|---|---|
| The acceleration of \(P\) is equal to the acceleration of \(Q\). | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(v = u + at\ \ \Rightarrow\ \ v = 2.8 \times 1.5\) (or \(\ v^2 = u^2 + 2as\ \ \Rightarrow\ \ v^2 = 2 \times 2.8 \times 3.15\)) \((v^2 = 17.64,\ v = 4.2)\) | M1 A1 |
| \(v = u + at\ \ \Rightarrow\ \ 4.2 = -4.2 + 9.8t\) | DM1 A1 |
| \(t = \dfrac{6}{7}\), 0.86, 0.857 (s) | DM1 A1 |
| (6) | |
| (17 marks) |