M2 June 2009 Q7
7.
A particle \(P\) of mass 2 kg is projected up a rough plane with initial speed 14 m s\(^{-1}\), from a point \(X\) on the plane, as shown in Figure 4. The particle moves up the plane along the line of greatest slope through \(X\) and comes to instantaneous rest at the point \(Y\). The plane is inclined at an angle \(\alpha\) to the horizontal, where \(\tan\alpha = \dfrac{7}{24}\). The coefficient of friction between the particle and the plane is \(\dfrac{1}{8}\).
(a) Use the work-energy principle to show that \(XY = 25\) m. (7)
After reaching \(Y\), the particle \(P\) slides back down the plane.
(b) Find the speed of \(P\) as it passes through \(X\). (4)
| Scheme | Marks |
|---|---|
| KE at \(X = \dfrac{1}{2}mv^2 = \dfrac{1}{2} \times 2 \times 14^2\) | B1 |
| GPE at \(Y =\) \(mgd\sin\alpha\left(= 2 \times g \times d \times \dfrac{7}{25}\right)\) | B1 B1 |
| Normal reaction \(R = mg\cos\alpha\) | M1 |
| Friction \(= \mu \times R = \dfrac{1}{8} \times 2g \times \dfrac{24}{25}\) | M1A1 |
| Work Energy: \(\ \dfrac{1}{2}mv^2 - mgd\sin\alpha = \mu \times R \times d\) or equivalent | |
| \(196 = \dfrac{14gd}{25} + \dfrac{6gd}{25} = \dfrac{20gd}{25}\) | A1 |
| \(d = 25\) m | |
| (7) |
| Scheme | Marks |
|---|---|
| Work Energy | |
| First time at \(X\): \(\ \dfrac{1}{2}mv^2 = \dfrac{1}{2}m14^2\) | |
| Work done \(= \mu \times R \times 2d = \dfrac{1}{8} \times 2g \times \dfrac{24}{25} \times 2d\) | |
| Return to \(X\): \(\ \dfrac{1}{2}mv^2 = \dfrac{1}{2}m14^2 - \dfrac{1}{8} \times 2g \times \dfrac{24}{25} \times 50\) | M1A1 |
| \(v = 8.9\) m s\(^{-1}\) (accept 8.85 m s\(^{-1}\)) | DM1A1 |
| (4) | |
| (11 marks) |
OR: Resolve parallel to \(XY\) to find the acceleration and use of \(v^2 = u^2 + 2as\)
| \(2a = 2g\sin\alpha - F_{\max} = 2g \times \dfrac{7}{25} - \dfrac{6g}{25} = \dfrac{8g}{25}\) | M1A1 |
| \(v^2 = (0+)2 \times a \times s = 8g\); \(v = 8.9\) (accept 8.85 m s\(^{-1}\)) | DM1;A1 |