M1 January 2010 Q5
5. A particle of mass 0.8 kg is held at rest on a rough plane. The plane is inclined at 30\(^\circ\) to the horizontal. The particle is released from rest and slides down a line of greatest slope of the plane. The particle moves 2.7 m during the first 3 seconds of its motion. Find
(a) the acceleration of the particle, (3)
(b) the coefficient of friction between the particle and the plane. (5)
The particle is now held on the same rough plane by a horizontal force of magnitude \(X\) newtons, acting in a plane containing a line of greatest slope of the plane, as shown in Figure 3. The particle is in equilibrium and on the point of moving up the plane.
(c) Find the value of \(X\). (7)
| Scheme | Marks |
|---|---|
| \(s = ut + \tfrac{1}{2}at^2 \ \Rightarrow\ 2.7 = \tfrac{1}{2}a \times 9\) | M1 A1 |
| \(a = 0.6\ \ (\text{m s}^{-2})\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\nwarrow\) \(R = 0.8g\cos 30^\circ\ \ (\approx 6.79)\) | B1 |
| Use of \(F = \mu R\) | B1 |
| \(\swarrow\) \(0.8g\sin 30^\circ - \mu R = 0.8 \times a\) | M1 A1 |
| \(\left(0.8g\sin 30^\circ - \mu 0.8g\cos 30^\circ = 0.8 \times 0.6\right)\) | |
| \(\mu \approx 0.51\) accept 0.507 | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(\uparrow\) \(R\cos 30^\circ = \mu R\cos 60^\circ + 0.8g\) | M1 A2 (1,0) |
| \((R \approx 12.8)\) | |
| \(\rightarrow\) \(X = R\sin 30^\circ + \mu R\sin 60^\circ\) | M1 A1 |
| Solving for \(X\), \(X \approx 12\) accept 12.0 | DM1 A1 |
| (7) | |
| (15 marks) |
Alternative to (c)
| \(\nwarrow\) \(R = X\sin 30^\circ + 0.8 \times 9.8\sin 60^\circ\) | M1 A2 (1,0) |
| \(\swarrow\) \(\mu R + 0.8g\cos 60^\circ = X\cos 30^\circ\) | M1 A1 |
| \(X = \dfrac{\mu 0.8g\sin 60^\circ + 0.8g\cos 60^\circ}{\cos 30^\circ - \mu\sin 30^\circ}\) | |
| Solving for \(X\), \(X \approx 12\) accept 12.0 | DM1 A1 |