| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | January |
| Marks | 10 |
| 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 problem requiring application of P=Fv and F=ma with resolution on an incline. Part (a) involves straightforward equilibrium at constant speed, while part (b) requires combining power equation with Newton's second law. The calculations are routine with no conceptual surprises, making it slightly easier than average for A-level. |
| Spec | 6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F = \frac{11760}{10}\) | B1 | Seen or implied |
| \(R = F - 1200g\sin\alpha\) | M1 | Motion up the slope. Allow with \(F\). Need all three terms but condone sign errors. Condone trig confusion |
| \(R = \frac{11760}{10} - 1200g\frac{1}{15}\) | A1 | Correct equation, correctly substituted |
| \(R = 390\) (392) | A1 | Max 3 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F = \frac{50000}{V}\) | B1 | Seen or implied. Could be scored after incorrect work on an energy equation to find \(F\) |
| \(F - 1200g\sin\alpha - 700 = 1200 \times 1.5\) | M1 | Equation of motion. Must be using 700. Need all four terms but condone sign errors. Condone trig confusion |
| A1 | Equation in \(V\) or \(F\) (and \(\alpha\)) with at most one error. Their \(F\) | |
| \(\frac{50000}{V} - 1200g\frac{1}{15} - 700 = 1200 \times 1.5\) | A1 | Correct equation in \(V\) |
| Solve for \(V\) | DM1 | Dependent on previous M |
| \(V = 15\) (15.2) | A1 | Max 3 s.f. |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = \frac{11760}{10}$ | B1 | Seen or implied |
| $R = F - 1200g\sin\alpha$ | M1 | Motion up the slope. Allow with $F$. Need all three terms but condone sign errors. Condone trig confusion |
| $R = \frac{11760}{10} - 1200g\frac{1}{15}$ | A1 | Correct equation, correctly substituted |
| $R = 390$ (392) | A1 | Max 3 s.f. |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = \frac{50000}{V}$ | B1 | Seen or implied. Could be scored after incorrect work on an energy equation to find $F$ |
| $F - 1200g\sin\alpha - 700 = 1200 \times 1.5$ | M1 | Equation of motion. Must be using 700. Need all four terms but condone sign errors. Condone trig confusion |
| | A1 | Equation in $V$ or $F$ (and $\alpha$) with at most one error. Their $F$ |
| $\frac{50000}{V} - 1200g\frac{1}{15} - 700 = 1200 \times 1.5$ | A1 | Correct equation in $V$ |
| Solve for $V$ | DM1 | Dependent on previous M |
| $V = 15$ (15.2) | A1 | Max 3 s.f. |
---\begin{enumerate}
\item A car of mass 1200 kg moves up a straight road. The road is inclined to the horizontal at an angle $\alpha$ where $\sin \alpha = \frac { 1 } { 15 }$. The car is moving up the road with constant speed $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the engine of the car is working at a constant rate of 11760 watts. The non-gravitational resistance to motion has a constant magnitude of $R$ newtons.\\
(a) Find the value of $R$.
\end{enumerate}
The rate of working of the car is now increased to 50 kW . At the instant when the speed of the car is $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the magnitude of the non-gravitational resistance to the motion of the car is 700 N and the acceleration of the car is $1.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(b) Find the value of $V$.
\hfill \mbox{\textit{Edexcel M2 2017 Q1 [10]}}