| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find power at constant speed |
| Difficulty | Moderate -0.3 This is a straightforward M2 power question requiring standard formulas (P=Fv, F=ma, resolving forces on incline). Part (a) uses P=Fv to find driving force then F=ma. Part (b) applies force balance at constant speed with component of weight down the slope. Both parts are direct applications of well-practiced techniques with no problem-solving insight required, making it slightly easier than average. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(T = \dfrac{30000}{24}\) | M1 | Use of \(P = Fv\) |
| \((= 1250 \text{ N})\) | A1 | Seen or implied |
| \(1250 - 650 = 1200a\) | M1 | Equation of motion. All terms required & must be dimensionally correct. Allow for their 1250. Condone sign errors |
| \(a = \dfrac{600}{1200} = 0.5 \text{ m s}^{-2}\) | A1 | |
| Total: (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(T' - 650 - 1200g\sin\alpha = 0\) | M1 | No acceleration. Condone sign errors and sin/cos confusion. |
| \(T' = 650 + 1200g \times \dfrac{1}{12}\) \((= 1630)\) | A1 | Correct unsimplified substituted expression for the driving force. |
| Rate of working \(= \left(650 + 1200g \times \dfrac{1}{12}\right)\times 24\) W | dM1 | Their driving force \(\times 24\). Dependent on the previous M1 |
| \((= 39.12) = 39\) (kW) | A1 | Or 39.1 (kW) Max 3 s.f. The Q asks for kW |
| Total: (4) [8] |
# Question 2(a):
| Working/Answer | Marks | Notes |
|---|---|---|
| $T = \dfrac{30000}{24}$ | M1 | Use of $P = Fv$ |
| $(= 1250 \text{ N})$ | A1 | Seen or implied |
| $1250 - 650 = 1200a$ | M1 | Equation of motion. All terms required & must be dimensionally correct. Allow for their 1250. Condone sign errors |
| $a = \dfrac{600}{1200} = 0.5 \text{ m s}^{-2}$ | A1 | |
| **Total: (4)** | | |
# Question 2(b):
| Working/Answer | Marks | Notes |
|---|---|---|
| $T' - 650 - 1200g\sin\alpha = 0$ | M1 | No acceleration. Condone sign errors and sin/cos confusion. |
| $T' = 650 + 1200g \times \dfrac{1}{12}$ $(= 1630)$ | A1 | Correct unsimplified substituted expression for the driving force. |
| Rate of working $= \left(650 + 1200g \times \dfrac{1}{12}\right)\times 24$ W | dM1 | Their driving force $\times 24$. Dependent on the previous M1 |
| $(= 39.12) = 39$ (kW) | A1 | Or 39.1 (kW) Max 3 s.f. The Q asks for kW |
| **Total: (4) [8]** | | |
---2. A van of mass 1200 kg is travelling along a straight horizontal road. The resistance to the motion of the van has a constant magnitude of 650 N and the van's engine is working at a rate of 30 kW .
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of the van when its speed is $24 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
The van now travels up a straight road which is inclined at angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 1 } { 12 }$. The resistance to the motion of the van from non-gravitational forces has a constant magnitude of 650 N . The van moves up the road at a constant speed of $24 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
\item Find, in kW , the rate at which the van's engine is now working.
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\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2017 Q2 [8]}}