| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2022 |
| Session | October |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Engine power on road constant/variable speed |
| Difficulty | Standard +0.3 This is a standard M2 power-resistance-incline problem requiring P=Fv to find driving force, then F=ma with component of weight down the slope minus resistance. It's slightly easier than average because the setup is straightforward with given values, requiring only routine application of well-practiced techniques with no conceptual surprises. |
| Spec | 3.03d Newton's second law: 2D vectors6.02l Power and velocity: P = Fv |
| VIAV SIHI NI IIIIIM ION OC | VIIIV SIHI NI III IM I O N OC | VIIV SIHI NI IIIIM I I ON OC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(P = Fv\) | M1 | Seen or implied e.g. \(F = \dfrac{15000}{16}\ (=937.5)\). Condone 15 in place of 15000 or extra zeros on 15000 |
| Equation of motion | M1 | Need all terms. Condone sign errors and sin/cos confusion. Dimensionally consistent |
| \(F + 900g\sin\theta - 400 = 900a\) | A1 | Unsimplified equation in \(P\) or their \(F\) with at most one error |
| \(\dfrac{15000}{16} + 900g \times \dfrac{1}{12} - 400 = 900a\) | A1 | Correct unsimplified equation with \(F\) and \(\sin\theta\) substituted |
| \(a = 1.41\ (1.4)\ (\text{ms}^{-2})\) | A1 | 3sf or 2sf |
| Total: (5) |
# Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $P = Fv$ | M1 | Seen or implied e.g. $F = \dfrac{15000}{16}\ (=937.5)$. Condone 15 in place of 15000 or extra zeros on 15000 |
| Equation of motion | M1 | Need all terms. Condone sign errors and sin/cos confusion. Dimensionally consistent |
| $F + 900g\sin\theta - 400 = 900a$ | A1 | Unsimplified equation in $P$ or their $F$ with at most one error |
| $\dfrac{15000}{16} + 900g \times \dfrac{1}{12} - 400 = 900a$ | A1 | Correct unsimplified equation with $F$ and $\sin\theta$ substituted |
| $a = 1.41\ (1.4)\ (\text{ms}^{-2})$ | A1 | 3sf or 2sf |
| **Total: (5)** | | |
---2. A car of mass 900 kg is moving down a straight road which is inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 1 } { 12 }$
The engine of the car is working at a constant rate of 15 kW .\\
The resistance to the motion of the car is modelled as a constant force of magnitude 400 N .
Find the acceleration of the car at the instant when it is moving at $16 \mathrm {~ms} ^ { - 1 }$\\
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VIAV SIHI NI IIIIIM ION OC & VIIIV SIHI NI III IM I O N OC & VIIV SIHI NI IIIIM I I ON OC \\
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\section*{Qu}
\hfill \mbox{\textit{Edexcel M2 2022 Q2 [5]}}