| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Horizontal road towing |
| Difficulty | Standard +0.3 This is a standard M1 mechanics question involving Newton's second law applied to connected particles. Parts (a)-(c) require straightforward proportion and F=ma calculations. Parts (d)-(e) use standard equations of motion and force analysis. Part (f) asks for a simple qualitative comment. While multi-part with several steps, 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) \(M_C : M_T = 1200 : 800 = 3 : 2\) | M1 A1 | |
| \(R_C = 300 \text{ N}\) \(\therefore R_T = 200 \text{ N}\) | ||
| (b) for car and trailer, eqn. of motion is \(3000 - 500 = 2000a\) | M1 | |
| giving \(a = \frac{5}{4} \text{ ms}^{-2}\) | M1 A1 | |
| (c) for car, eqn. of motion is \(3000 - 300 - T = 1200 \times \frac{5}{4}\) | M1 | |
| giving \(T = 1200 \text{ N}\) | M1 A1 | |
| (d) total of braking + resistive forces \(= 1500\) N | ||
| \(1500 = 2000a\) so \(a = -\frac{3}{4} \text{ ms}^{-2}\) | M1 A1 | |
| \(u = 24, v = 0, a = -\frac{3}{4}\) use \(v^2 = u^2 + 2as\) | M1 | |
| \(0 = 576 - \frac{3}{2}s\) \(\therefore s = 384 \text{ m}\) | M1 A1 | |
| (e) for car \((\leftarrow)\): \(T + 1000 + 300 = 1200(\frac{3}{4})\) | M1 A1 | |
| \(T = 400 \text{ N}\) \(\therefore T = 400 \text{ N}\), pushing the car | M1 A1 | |
| (f) e.g. unlikely to be realistic, likely to decrease as speed decreases | B2 | (19) |
**(a)** $M_C : M_T = 1200 : 800 = 3 : 2$ | M1 A1 |
$R_C = 300 \text{ N}$ $\therefore R_T = 200 \text{ N}$ | |
**(b)** for car and trailer, eqn. of motion is $3000 - 500 = 2000a$ | M1 |
giving $a = \frac{5}{4} \text{ ms}^{-2}$ | M1 A1 |
**(c)** for car, eqn. of motion is $3000 - 300 - T = 1200 \times \frac{5}{4}$ | M1 |
giving $T = 1200 \text{ N}$ | M1 A1 |
**(d)** total of braking + resistive forces $= 1500$ N | |
$1500 = 2000a$ so $a = -\frac{3}{4} \text{ ms}^{-2}$ | M1 A1 |
$u = 24, v = 0, a = -\frac{3}{4}$ use $v^2 = u^2 + 2as$ | M1 |
$0 = 576 - \frac{3}{2}s$ $\therefore s = 384 \text{ m}$ | M1 A1 |
**(e)** for car $(\leftarrow)$: $T + 1000 + 300 = 1200(\frac{3}{4})$ | M1 A1 |
$T = 400 \text{ N}$ $\therefore T = 400 \text{ N}$, pushing the car | M1 A1 |
**(f)** e.g. unlikely to be realistic, likely to decrease as speed decreases | B2 | (19)7. A car of mass 1200 kg tows a trailer of mass 800 kg along a straight level road by means of a rigid towbar. The resistances to the motion of the car and the trailer are proportional to their masses.
Given that the car experiences a resistance to motion of 300 N ,
\begin{enumerate}[label=(\alph*)]
\item find the resistance to motion which the trailer experiences.
Given that the engine of the car exerts a driving force of 3 kN ,
\item find the acceleration of the system,
\item show that the tension in the towbar is 1200 N .
When the system has reached a speed of $24 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, it continues at constant speed until an electrical fault causes the engine of the car to switch off. The brakes are used to apply a constant retarding force until the system comes to rest.
Given that the retarding force of the brakes has magnitude 1 kN and assuming that the original resistances to motion of the car and the trailer remain constant,
\item calculate the distance that the system travels during the braking period,
\item find the magnitude and direction of the force exerted by the towbar on the car.
\item Comment on the assumption that the original resistances to motion of the car and the trailer remain constant throughout the motion.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q7 [19]}}