| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Up and down hill: two equations |
| Difficulty | Standard +0.3 This is a standard M2 work-energy-power problem requiring application of P=Fv and F=ma in two scenarios (up and down the incline). It involves resolving forces parallel to the incline and solving simultaneous equations, but follows a well-practiced template with no novel insight required. Slightly easier than average due to straightforward setup and arithmetic. |
| Spec | 6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Driving force \(= \frac{3P}{12}\) | B1 | Use of \(P = Fv\). Allow for \(\frac{P}{12}\) in second equation if not awarded here |
| Motion up the hill: \(F - R - W\sin\theta = 0\) | M1 | Need all terms. Condone sign errors and sin/cos confusion |
| \(\frac{3P}{12} - R - \frac{9000}{15} = 0\), \(\left(\frac{3P}{12} - R = 600\right)\) | A1 | Correct substituted equation. Any equivalent form |
| Motion down the hill: \(F + W\sin\theta - R = \frac{9000}{9.8} \times \frac{9.8}{20}\) | M1 | Need all terms. Condone sign errors and sin/cos confusion |
| \(\frac{P}{12} + \frac{9000}{15} - R = 450\), \(\left(\frac{P}{12} - R = -150\right)\) | A1 | Substituted equation with at most one error. Any equivalent form |
| (correct second equation) | A1 | Correct substituted equation. Any equivalent form |
| Solve for \(P\) or \(R\) | DM1 | Dependent on both preceding M marks |
| \(\left(\frac{2P}{12} = 750\right) \Rightarrow P = 4500\) | A1 | One correct |
| \(R = 525\ (530)\) | A1 | Both correct |
| Answer | Marks | Guidance |
|---|---|---|
| SC1: Misread mass = 9000 kg | B1, M1A0, M1A1ftA0, M1A1ftA1ft | Total 7/9 |
| Gives equations \(\frac{P}{4} = R + 5880\), \(\frac{P}{12} = R - 1470\) | ||
| Solutions: \(P = 44100\), \(R = 5145\) | ||
| SC2: Use of mass = weight = 9000 | B1, M1A1, M1A1A0, M1A0A0 | Total 6/9 |
| Gives equations \(\frac{P}{4} = R + 600\), \(\frac{P}{12} = R + 3810\) | ||
| Solutions: \(P = -19260\), \(R = -5415\) |
# Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Driving force $= \frac{3P}{12}$ | B1 | Use of $P = Fv$. Allow for $\frac{P}{12}$ in second equation if not awarded here |
| Motion up the hill: $F - R - W\sin\theta = 0$ | M1 | Need all terms. Condone sign errors and sin/cos confusion |
| $\frac{3P}{12} - R - \frac{9000}{15} = 0$, $\left(\frac{3P}{12} - R = 600\right)$ | A1 | Correct substituted equation. Any equivalent form |
| Motion down the hill: $F + W\sin\theta - R = \frac{9000}{9.8} \times \frac{9.8}{20}$ | M1 | Need all terms. Condone sign errors and sin/cos confusion |
| $\frac{P}{12} + \frac{9000}{15} - R = 450$, $\left(\frac{P}{12} - R = -150\right)$ | A1 | Substituted equation with at most one error. Any equivalent form |
| (correct second equation) | A1 | Correct substituted equation. Any equivalent form |
| Solve for $P$ or $R$ | DM1 | Dependent on both preceding M marks |
| $\left(\frac{2P}{12} = 750\right) \Rightarrow P = 4500$ | A1 | One correct |
| $R = 525\ (530)$ | A1 | Both correct |
**Special Cases:**
| SC1: Misread mass = 9000 kg | B1, M1A0, M1A1ftA0, M1A1ftA1ft | Total 7/9 |
|---|---|---|
| Gives equations $\frac{P}{4} = R + 5880$, $\frac{P}{12} = R - 1470$ | | |
| Solutions: $P = 44100$, $R = 5145$ | | |
| SC2: Use of mass = weight = 9000 | B1, M1A1, M1A1A0, M1A0A0 | Total 6/9 |
|---|---|---|
| Gives equations $\frac{P}{4} = R + 600$, $\frac{P}{12} = R + 3810$ | | |
| Solutions: $P = -19260$, $R = -5415$ | | |
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\item A truck of weight 9000 N is travelling up a hill on a straight road that is inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 1 } { 15 }$
\end{enumerate}
When the truck travels up the hill with the engine working at $3 P$ watts, the truck is moving at a constant speed of $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
Later on, the truck travels down the hill along the same road, with the engine working at $P$ watts. At the instant when the speed of the truck is $12 \mathrm {~ms} ^ { - 1 }$, the acceleration of the truck is $\frac { g } { 20 }$
The resistance to motion of the truck from non-gravitational forces is a constant force of magnitude $R$ newtons in all circumstances.
Find (i) the value of $P$,\\
(ii) the value of $R$.\\
WIHW SIHI NIT INHM ION OC\\
WIIV SIHI NI III IM I ON OC\\
VAYV SIHI NI JLIUM ION OO
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\hfill \mbox{\textit{Edexcel M2 2020 Q2 [9]}}