| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Towing system: inclined road |
| Difficulty | Standard +0.3 This is a standard M2 mechanics problem involving power, forces on an incline, and work-energy principle. Part (a) requires P=Fv to find driving force then F=ma with resistance and weight components. Part (b) isolates the trailer with standard force resolution. Part (c) applies work-energy principle with given values. All techniques are routine M2 content with straightforward multi-step application, making it slightly easier than average A-level difficulty. |
| Spec | 3.03o Advanced connected particles: and pulleys6.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = \frac{60000}{10} = 6000\) | B1 | Correct application of \(P = Fv\) seen or implied |
| \(F - 1200g\sin\alpha - 400g\sin\alpha - 1000 - 200 = 1600a\) | M1 | Use of \(F = ma\) parallel to the slope for the car and trailer. Must have all the terms, but condone sign errors |
| A1 | At most one error (with \(F\) or their \(F\)) | |
| A1 | Correct equation (with \(F\) or their \(F\)) | |
| \(a = 2.3 \text{ (m s}^{-2})\) | A1 | only |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| \(T - 400g\sin\alpha - 200 = 400 \times 2.3\) | M1 | Use of \(F = ma\) parallel to the slope for the trailer |
| A1 ft | At most one error (their \(a\)) | |
| A1 ft | All correct (their \(a\)) | |
| \(T = 1400\) | A1 | only |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| \(6000 - 1200g\sin\alpha - 1000 - T = 1200 \times 2.3\) | M1 | Use of \(F = ma\) parallel to the slope for the car |
| A1 ft | At most one error (their \(a\)) | |
| A1 ft | All correct (their \(a\)) | |
| \(T = 1400\) | A1 | only |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = 6000\) | B1 | Simultaneous equations in \(T\) and \(a\) |
| \(T - 400g\sin\alpha - 200 = 400 \times a\) | ||
| \(6000 - 1200g\sin\alpha - 1000 - T = 1200 \times a\) | ||
| \(6000 - 1600g\sin\alpha - 1200 = 1600a\) | ||
| \(a = 2.3 \text{ (m s}^{-2})\) | M1A1A1 | Add to eliminate \(T\) |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| \(-800a = 2T + 800g\sin\alpha + 800 - 6000\) | M1A1A1 | Subtract and / or substitute to eliminate \(a\) |
| \(2T = 5200 - 800g\sin\alpha - 800 \times 2.3\) | ||
| \(T = 1400\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(200d = \frac{1}{2} \times 400 \times 12^2 - 400gd\sin\alpha\) | M1 | Use of work-energy. Must have all three terms. Do not accept duplication of terms, but condone sign errors. Equation in only one unknown, but could be vertical distance |
| A1 | At most one error in the equation | |
| A1 | All correct in one unknown | |
| DM1 | Solve for \(d\) – dependent on M for work-energy equation. only | |
| \(d = 60 \text{ (m)}\) | A1 | only |
| (5) | ||
| 14 | ||
| For vertical distance \(\left(= \frac{60}{14} = 4.29\right)\) allow 3/5 |
**(a)**
| $F = \frac{60000}{10} = 6000$ | B1 | Correct application of $P = Fv$ seen or implied |
| $F - 1200g\sin\alpha - 400g\sin\alpha - 1000 - 200 = 1600a$ | M1 | Use of $F = ma$ parallel to the slope for the car and trailer. Must have all the terms, but condone sign errors |
| | A1 | At most one error (with $F$ or their $F$) |
| | A1 | Correct equation (with $F$ or their $F$) |
| $a = 2.3 \text{ (m s}^{-2})$ | A1 | only |
| | (5) | |
**(b)**
| $T - 400g\sin\alpha - 200 = 400 \times 2.3$ | M1 | Use of $F = ma$ parallel to the slope for the trailer |
| | A1 ft | At most one error (their $a$) |
| | A1 ft | All correct (their $a$) |
| $T = 1400$ | A1 | only |
| | (4) | |
**OR**
| $6000 - 1200g\sin\alpha - 1000 - T = 1200 \times 2.3$ | M1 | Use of $F = ma$ parallel to the slope for the car |
| | A1 ft | At most one error (their $a$) |
| | A1 ft | All correct (their $a$) |
| $T = 1400$ | A1 | only |
| | (4) | |
**OR (a)**
| $F = 6000$ | B1 | Simultaneous equations in $T$ and $a$ |
| $T - 400g\sin\alpha - 200 = 400 \times a$ | | |
| $6000 - 1200g\sin\alpha - 1000 - T = 1200 \times a$ | | |
| $6000 - 1600g\sin\alpha - 1200 = 1600a$ | | |
| $a = 2.3 \text{ (m s}^{-2})$ | M1A1A1 | Add to eliminate $T$ |
| | (5) | |
**(b)**
| $-800a = 2T + 800g\sin\alpha + 800 - 6000$ | M1A1A1 | Subtract and / or substitute to eliminate $a$ |
| $2T = 5200 - 800g\sin\alpha - 800 \times 2.3$ | | |
| $T = 1400$ | A1 | |
**(c)**
| $200d = \frac{1}{2} \times 400 \times 12^2 - 400gd\sin\alpha$ | M1 | Use of work-energy. Must have all three terms. Do not accept duplication of terms, but condone sign errors. Equation in only one unknown, but could be vertical distance |
| | A1 | At most one error in the equation |
| | A1 | All correct in one unknown |
| | DM1 | Solve for $d$ – dependent on M for work-energy equation. only |
| $d = 60 \text{ (m)}$ | A1 | only |
| | (5) |
| | 14 | |
| | | For vertical distance $\left(= \frac{60}{14} = 4.29\right)$ allow 3/5 |
---6. A car of mass 1200 kg pulls a trailer of mass 400 kg up a straight road which is inclined to the horizontal at an angle $\alpha$, where $\sin \alpha = \frac { 1 } { 14 }$. The trailer is attached to the car by a light inextensible towbar which is parallel to the road. The car's engine works at a constant rate of 60 kW . The non-gravitational resistances to motion are constant and of magnitude 1000 N on the car and 200 N on the trailer.
At a given instant, the car is moving at $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find
\begin{enumerate}[label=(\alph*)]
\item the acceleration of the car at this instant,
\item the tension in the towbar at this instant.
The towbar breaks when the car is moving at $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item Find, using the work-energy principle, the further distance that the trailer travels before coming instantaneously to rest.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2012 Q6 [14]}}