| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | March |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Vehicle on slope with resistance |
| Difficulty | Standard +0.3 This is a straightforward mechanics problem requiring application of work-energy principles and resolution of forces on a slope. Part (a) uses work done equals change in gravitational PE at constant speed (single equation to solve for m). Part (b) requires resolving forces on the trailer alone. Both parts are standard M1 techniques with no conceptual surprises, making it slightly easier than average. |
| Spec | 3.03f Weight: W=mg6.02a Work done: concept and definition |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| PE lost in 50 m \(= (m+300) \, g \times 50\sin 3\) | B1 | |
| \((m+300) \, g \times 50\sin 3 - 40\,000 = 0\) | M1 | Use of the work-energy equation |
| \(m = 1230\) to 3 sf | A1 | \(m = 1228.6\) |
| Alternative: Resistance force \(R = \frac{40000}{50}\) \([= 800\text{ N}]\) | B1 | |
| \((m+300)\,g\sin 3 - R = 0\) | M1 | Apply Newton's second law to the system, 3 terms |
| \(m = 1230\) to 3 sf | A1 | \(m = 1228.6\) |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(T + 300\,g\sin 3 - 200 = 0\) (Trailer) or \(mg\sin 3 = T + 600\) (Car) | M1 | Apply Newton's 2nd law either to the trailer or to the car using \(a=0\), three terms in either case |
| \(T = 43[.0]\) N to 3 sf | A1 | |
| 2 |
## Question 3:
### Part 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| PE lost in 50 m $= (m+300) \, g \times 50\sin 3$ | B1 | |
| $(m+300) \, g \times 50\sin 3 - 40\,000 = 0$ | M1 | Use of the work-energy equation |
| $m = 1230$ to 3 sf | A1 | $m = 1228.6$ |
| **Alternative:** Resistance force $R = \frac{40000}{50}$ $[= 800\text{ N}]$ | B1 | |
| $(m+300)\,g\sin 3 - R = 0$ | M1 | Apply Newton's second law to the system, 3 terms |
| $m = 1230$ to 3 sf | A1 | $m = 1228.6$ |
| | **3** | |
### Part 3(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $T + 300\,g\sin 3 - 200 = 0$ (Trailer) or $mg\sin 3 = T + 600$ (Car) | M1 | Apply Newton's 2nd law either to the trailer or to the car using $a=0$, three terms in either case |
| $T = 43[.0]$ N to 3 sf | A1 | |
| | **2** | |
---3 A car of mass $m \mathrm {~kg}$ is towing a trailer of mass 300 kg down a straight hill inclined at $3 ^ { \circ }$ to the horizontal at a constant speed. There are resistance forces on the car and on the trailer, and the total work done against the resistance forces in a distance of 50 m is 40000 J . The engine of the car is doing no work and the tow-bar is light and rigid.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $m$.\\
The resistance force on the trailer is 200 N .
\item Find the tension in the tow-bar between the car and the trailer.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2022 Q3 [5]}}