AS June 2019 Q3
3. A particle, \(P\), of mass \(m\) kg is projected with speed \(5\ \text{m s}^{-1}\) down a line of greatest slope of a rough plane. The plane is inclined to the horizontal at an angle \(\alpha\), where \(\sin\alpha = \dfrac{3}{5}\)
The total resistance to the motion of \(P\) is a force of magnitude \(\dfrac{1}{5}mg\)
Use the work-energy principle to find the speed of \(P\) at the instant when it has moved a distance 8 m down the plane from the point of projection. (7)
| Scheme | Marks | AO |
|---|---|---|
| Work done \(= \dfrac{1}{5}mg \times 8 \quad (15.68m)\) | B1 | 3.4 |
| PE Loss \(= 8mg\sin\alpha \quad (47.04m)\) | B1 | 1.1b |
| KE Gain = Difference of two KE terms | M1 | 3.4 |
| \(= \dfrac{1}{2}mv^2 - \dfrac{1}{2}m5^2\) | A1 | 1.1b |
| Work done against friction = PE Loss – KE Gain | M1 | 2.1 |
| \(\dfrac{1}{5}mg \times 8 = 8mg\sin\alpha - \left(\dfrac{1}{2}mv^2 - \dfrac{1}{2}m5^2\right)\) | A1 | 1.1b |
| \(v = 9.4\) or \(9.37\ (\text{m s}^{-1})\) | A1 | 1.1b |
| (7) | ||
| (7 marks) |
Notes
The question instructs candidates to use the work-energy principle, so suvat methods will not score the second M1.
B1: Work done against friction seen or implied
B1: PE loss seen or implied
NB: B1B1 for \(\left(\dfrac{3}{5}mg - \dfrac{1}{5}mg\right) \times 8 \quad \left(= \dfrac{16}{5}mg\right)\)
M1: Difference in two KE terms seen or implied (allow KE loss)
A1: Correct unsimplified expression. Allow \(\pm\)
M1: Work-energy equation with all terms. Must be dimensionally correct but condone sign errors
A1: Correct unsimplified equation
A1: 2 sf or 3 sf (after use of \(g = 9.8\))