| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Horizontal road towing |
| Difficulty | Standard +0.3 This is a multi-part mechanics question involving Newton's second law applied to connected bodies. While it requires careful consideration of forces on different parts of the system and has several parts, each individual step uses standard M1 techniques (F=ma, resolving forces, considering separate bodies). The calculations are straightforward with no conceptual surprises, making it slightly easier than average for A-level mechanics. |
| Spec | 3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Whole train: mass \(= 150\) tonnes; Total Resistance \(= 3000\text{ N}\) | B1 | Both totals required |
| \(12000 - 3000 = 150000a\) | M1 | Correct elements must be present |
| \(a = 0.06 \text{ ms}^{-2}\) | A1 | CAO. Errors with units (e.g. not converting tonnes to kilograms) are penalised here but condoned where possible for the remainder of the question |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Truck B: \(T - 500 = 30000a\) | M1 | Correct elements must be present |
| \(T - 500 = 30000 \times 0.06\) | A1 | Allow FT for \(a\) from part (i) if units are used consistently, for all the marks in this part |
| \(T = 2300\); Between A and B, tension of 2300 N | A1 | |
| Alternative: Rest of train: \(12000 - 2500 - T = 120000a\) | M1 | Correct elements must be present |
| \(T = 12000 - 2500 - 120000 \times 0.06\) | A1 | |
| \(T = 2300\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Treating the train as a whole: \(-2000 - 5000 - 500 = 150000a\) | M1 | |
| \(a = -0.05\) | A1 | Allow FT for the remaining A marks in part (iii) from an error in \(a\) |
| \(v^2 - u^2 = 2as\) | M1 | |
| \(0^2 - 10^2 = 2 \times (-0.05) \times s\) | ||
| \(s = 1000\); Stopping distance is 1000 m | A1 | |
| B: \(T - 500 = 30000a\) | M1 | Correct elements must be present. Alternative for rest of train: \(-T - 5000 - 2000 = 120000 \times -0.05\) |
| \(T = -1000\) | A1 | The sign of 1000 must be consistent with the direction of \(T\) |
| Between A and B, thrust of 1000 N | A1 | Dependent on previous M and A marks. Accept "compression" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equilibrium parallel to the slope | M1 | Correct elements must be present and there must be an attempt to resolve the weight. Condone omission of \(g\) for this mark |
| \(150000 \times 9.8 \times \sin\alpha + 3000 = 12000\) | A1 | |
| \(\alpha = 0.35°\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| B: \(T_2 - 500 - 30000 \times 9.8 \times \sin 0.35...° = 0\) | M1 | Correct elements must be present. Condone omission of \(g\) for this mark. Do not accept 1800 N for the component of the weight without justification. Alternative for rest of train: \(12000 = T + 2500 + 120000 \times 9.8 \times \sin 0.35°\) |
| \(T_2 = 2300\); Between A and B, tension of 2300 N, as in part (ii) | A1 | This mark can only be awarded if the angle found in (iv) is correct |
## Question 3:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Whole train: mass $= 150$ tonnes; Total Resistance $= 3000\text{ N}$ | B1 | Both totals required |
| $12000 - 3000 = 150000a$ | M1 | Correct elements must be present |
| $a = 0.06 \text{ ms}^{-2}$ | A1 | CAO. Errors with units (e.g. not converting tonnes to kilograms) are penalised here but condoned where possible for the remainder of the question |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Truck B: $T - 500 = 30000a$ | M1 | Correct elements must be present |
| $T - 500 = 30000 \times 0.06$ | A1 | Allow FT for $a$ from part (i) if units are used consistently, for all the marks in this part |
| $T = 2300$; Between A and B, tension of 2300 N | A1 | |
| **Alternative:** Rest of train: $12000 - 2500 - T = 120000a$ | M1 | Correct elements must be present |
| $T = 12000 - 2500 - 120000 \times 0.06$ | A1 | |
| $T = 2300$ | A1 | |
### Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Treating the train as a whole: $-2000 - 5000 - 500 = 150000a$ | M1 | |
| $a = -0.05$ | A1 | Allow FT for the remaining A marks in part (iii) from an error in $a$ |
| $v^2 - u^2 = 2as$ | M1 | |
| $0^2 - 10^2 = 2 \times (-0.05) \times s$ | | |
| $s = 1000$; Stopping distance is 1000 m | A1 | |
| B: $T - 500 = 30000a$ | M1 | Correct elements must be present. Alternative for rest of train: $-T - 5000 - 2000 = 120000 \times -0.05$ |
| $T = -1000$ | A1 | The sign of 1000 must be consistent with the direction of $T$ |
| Between A and B, thrust of 1000 N | A1 | Dependent on previous M and A marks. Accept "compression" |
### Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| Equilibrium parallel to the slope | M1 | Correct elements must be present and there must be an attempt to resolve the weight. Condone omission of $g$ for this mark |
| $150000 \times 9.8 \times \sin\alpha + 3000 = 12000$ | A1 | |
| $\alpha = 0.35°$ | A1 | CAO |
### Part (v)
| Answer | Mark | Guidance |
|--------|------|----------|
| B: $T_2 - 500 - 30000 \times 9.8 \times \sin 0.35...° = 0$ | M1 | Correct elements must be present. Condone omission of $g$ for this mark. Do not accept 1800 N for the component of the weight without justification. Alternative for rest of train: $12000 = T + 2500 + 120000 \times 9.8 \times \sin 0.35°$ |
| $T_2 = 2300$; Between A and B, tension of 2300 N, as in part (ii) | A1 | This mark can only be awarded if the angle found in (iv) is correct |
---3 Fig. 7 illustrates a train with a locomotive, L, pulling two trucks, A and B.
The locomotive has mass 90 tonnes and is subject to a resistance force of 2000 N .\\
Each of the trucks $A$ and $B$ has mass 30 tonnes and is subject to a resistance force of $500 N$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9bff41e0-7be0-4932-ae50-a612abb3fe19-3_153_1256_457_470}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
Initially the train is travelling along a straight horizontal track. The locomotive is exerting a driving force of 12000 N .\\
(i) Find the acceleration of the train.\\
(ii) Find the tension in the coupling between trucks A and B .
When the train is travelling at $10 \mathrm {~ms} ^ { - 1 }$, a fault occurs with truck A and the resistance to its motion changes from 500 N to 5000 N .
The driver reduces the driving force to zero and allows the train to slow down under the resistance forces and come to a stop.\\
(iii) Find the distance the train travels while slowing down and coming to a stop.
Find also the force in the coupling between trucks A and B while the train is slowing down, and state whether it is a tension or a thrust.
The fault in truck A is repaired so that the resistance to its motion is again 500 N . The train continues and comes to a place where the track goes up a uniform slope at an angle of $\alpha ^ { \circ }$ to the horizontal.\\
(iv) When the train is on the slope, it travels at uniform speed. The driving force remains at 12000 N . Find the value of $\alpha$.\\
(v) Show that the force in the coupling between trucks A and B has the same value that it had in part (ii).
\hfill \mbox{\textit{OCR MEI M1 Q3 [18]}}