| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Constant speed up/down incline |
| Difficulty | Standard +0.3 This is a straightforward M2 mechanics question requiring standard application of power = force × velocity and Newton's second law with constant acceleration. Part (a) involves resolving forces in equilibrium and calculating power; part (b) uses F=ma with given forces. Both parts follow routine procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.03d Newton's second law: 2D vectors6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| \(W = 573\frac{1}{3} \times 15 = 8600 = 8.6\) kW | M1, A1, M1, A1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| (b) N2L: \(800 \times 9.8 \times \frac{1}{24} - 900 = 800a\) | M1 | \(a = -\frac{43}{60}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(T \approx 21\) | M1, A1cso | accept 20.9; 8 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Use of \(v^2 = u^2 + 2as\) \(a = 0.72\) | M1 for getting as far as an equation in a, A1 | finish as above |
(a) $F + 800g \sin \alpha = 900$
$F = 573\frac{1}{3}$
$W = 573\frac{1}{3} \times 15 = 8600 = 8.6$ kW | M1, A1, M1, A1 | 4 marks
**NB.** Going up hill is an error, not a Misread
(b) N2L: $800 \times 9.8 \times \frac{1}{24} - 900 = 800a$ | M1 | $a = -\frac{43}{60}$ | A1 | awrt $-0.72$
$0 = 15 - \frac{43}{60}T$
$T \approx 21$ | M1, A1cso | accept 20.9; 8 marks total
**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):**
$\text{WD: } 573\frac{1}{3} \times t = \frac{1}{2} \times 800 \times 15^2$
$s = 157$
Use of $v^2 = u^2 + 2as$ $a = 0.72$ | M1 for getting as far as an equation in a, 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 $T \approx 21$
---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 $\sin \alpha = \frac{3}{4}$. The resistance to motion from non-gravitational forces is modelled as a constant force of magnitude 900 N.
\begin{enumerate}[label=(\alph*)]
\item Find, in kW, the rate of working of the engine of the car. [4]
\end{enumerate}
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.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the value of $T$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2007 Q2 [8]}}