M2 January 2007 Q2
2. A car of mass 800 kg is moving at a constant speed of 15 m s\(^{-1}\) down a straight road inclined at an angle \(\alpha\) to the horizontal, where \(\sin\alpha = \tfrac{1}{24}\). The resistance to motion from non-gravitational forces is modelled as a constant force of magnitude 900 N.
When the car is travelling down the road at 15 m s\(^{-1}\), the engine is switched off. The car comes to rest in time \(T\) seconds after the engine is switched off. The resistance to motion from non-gravitational forces is again modelled as a constant force of magnitude 900 N.
| Scheme | Marks |
|---|---|
| \(F + 800g\sin\alpha = 900\) | M1 |
| \(F = 573\tfrac{1}{3}\) | A1 |
| \(W = 573\tfrac{1}{3} \times 15 = 8600\) | M1 |
| \(= 8.6\) kW | A1 |
| (4) |
Notes
NB. Going up hill is an error, not a Misread
| Scheme | Marks |
|---|---|
| N2L \(800 \times 9.8 \times \dfrac{1}{24} - 900 = 800a\ \ *\) | M1 |
| \(a = -\dfrac{43}{60}\) awrt −0.72 | A1 |
| \(0 = 15 - \dfrac{43}{60}T\) | M1 |
| \(T \approx 21\) accept 20.9 | A1cso |
| (4) | |
| (8 marks) |
Notes
* If they are using their F from (a) then they need to have scored the M1 in (a) in order to score the M1 here.
Alternative for (b)
WD: \(573\dfrac{1}{3}s = \dfrac{1}{2} \times 800 \times 15^2\)
\(s = 157\)
Use of \(v^2 = u^2 + 2as\) M1 for getting as far as an equation in \(a\).
\(a = 0.72\) A1
finish as above.
2nd Alternative for (b)
\(Ft\) = Change in momentum:
M1 Using the correct \(F\)
M1 Use of the method to form an equation
A1 Equation correct unsimplified but fully substituted
A1 \(T \approx 21\)