| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Inclined road towing |
| Difficulty | Standard +0.2 This is a standard M1 connected particles problem with straightforward application of F=ma on an incline. Part (a) is a 'show that' requiring basic resolution of forces, part (b) uses standard tension calculation, part (c) applies a SUVAT equation, and part (d) requires comparing driving forces on level vs inclined road. All techniques are routine M1 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03d Newton's second law: 2D vectors3.03f Weight: W=mg3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| (a) for car + caravan, eqn. of motion is \(3000 - 900 - 2100\sin\alpha = 2100a\); \(2100 - 1470 = 2100a \therefore a = 0.3 \text{ ms}^{-2}\) | M2 A1 M1 A1 | |
| (b) for caravan, \(T - 500 - 850\sin\alpha = 850 \times 0.3\) | M1 | |
| \(T - 500 - 595 = 255 \therefore T = 1350 \text{ N}\) | M1 A1 | |
| (c) \(u = 0\), \(a = 0.3\), \(s = 540\) use \(v^2 = u^2 + 2as\) | M1 | |
| \(v^2 = 0 + 2(0.3)(540) = 324 \therefore v = 18 \text{ ms}^{-1}\) | M1 A1 | |
| (d) \(D - 900 = 0 \therefore D = 900 \text{ N}\) | M1 A1 | |
| % reduction \(= \frac{3000 - 900}{3000} \times 100 = 70\%\) | M1 A1 | (15) |
**(a)** for car + caravan, eqn. of motion is $3000 - 900 - 2100\sin\alpha = 2100a$; $2100 - 1470 = 2100a \therefore a = 0.3 \text{ ms}^{-2}$ | M2 A1 M1 A1 |
**(b)** for caravan, $T - 500 - 850\sin\alpha = 850 \times 0.3$ | M1 |
$T - 500 - 595 = 255 \therefore T = 1350 \text{ N}$ | M1 A1 |
**(c)** $u = 0$, $a = 0.3$, $s = 540$ use $v^2 = u^2 + 2as$ | M1 |
$v^2 = 0 + 2(0.3)(540) = 324 \therefore v = 18 \text{ ms}^{-1}$ | M1 A1 |
**(d)** $D - 900 = 0 \therefore D = 900 \text{ N}$ | M1 A1 |
% reduction $= \frac{3000 - 900}{3000} \times 100 = 70\%$ | M1 A1 | (15)
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**Total: (75)**7. A car of mass 1250 kg tows a caravan of mass 850 kg up a hill inclined at an angle $\alpha$ to the horizontal where $\sin \alpha = \frac { 1 } { 14 }$. The total resistance to motion experienced by the car is 400 N , and by the caravan is 500 N .
Given that the driving force of the engine is 3 kN ,
\begin{enumerate}[label=(\alph*)]
\item show that the acceleration of the system is $0.3 \mathrm {~m} \mathrm {~s} ^ { - 2 }$,
\item find the tension in the towbar linking the car and the caravan.
Starting from rest, the car accelerates uniformly for 540 m until it reaches a speed of $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the top of the hill.
\item Find v.
At the top of the hill the road becomes level and the driver maintains the speed at which the car and caravan reached the top of the hill.
\item Assuming that the resistance to motion on each part of the system is unchanged, find the percentage reduction in the driving force of the engine required to achieve this.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q7 [15]}}