M2 January 2008 Q3
3. A car of mass 1000 kg is moving at a constant speed of 16 m s\(^{-1}\) up a straight road inclined at an angle \(\theta\) to the horizontal. The rate of working of the engine of the car is 20 kW and the resistance to motion from non-gravitational forces is modelled as a constant force of magnitude 550 N.
(a) Show that \(\sin\theta = \dfrac{1}{14}\). (5)
When the car is travelling up the road at 16 m s\(^{-1}\), the engine is switched off. The car comes to rest, without braking, having moved a distance \(y\) metres from the point where the engine was switched off. The resistance to motion from non-gravitational forces is again modelled as a constant force of magnitude 550 N.
(b) Find the value of \(y\). (4)
| Scheme | Marks |
|---|---|
| \(20\,000 = 16F\ \ (F = 1250)\) | M1 A1 |
| \(\nearrow\) \(F = 550 + 1000 \times 9.8\sin\theta\) ft their \(F\) | M1 A1ft |
| Leading to \(\sin\theta = \tfrac{1}{14}\) * cso | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| N2L \(\nearrow\) \(550 + 1000 \times 9.8 \times \sin\theta = 1000a\) \(\left(550 + 1000 \times 9.8 \times \tfrac{1}{14} = 1000a\right)\) or \(1250 = 1000a\) \((a = (-)1.25)\) | M1 A1 |
| \(v^2 = u^2 + 2as \ \Rightarrow\ 16^2 = 2 \times 1.25 \times y\) | M1 |
| \(y \approx 102\) accept 102.4, 100 | A1 |
| (4) | |
| (9 marks) |
Alternative to (b)
| Work-Energy \(\tfrac{1}{2} \times 1000 \times 16^2 - 1000 \times 9.8 \times \tfrac{1}{14}y = 550y\) | M1 M1 A1 |
| \(y \approx 102\) accept 102.4, 100 | A1 |
(4)