| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Work-energy method on incline |
| Difficulty | Standard +0.3 This is a straightforward work-energy problem with standard mechanics techniques. Part (a) applies conservation of energy with given work done (routine calculation with KE, PE, and work). Part (b) uses F=ma with constant acceleration and resolving forces on the trailer. Both parts are direct applications of standard M1 methods with no novel insight required, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Increase in KE \(= \frac{1}{2} \times 1900 \times 30^2 - \frac{1}{2} \times 1900 \times 20^2\) \([= 475000\ \text{J}]\) | B1 | May be implied by energy equation |
| Loss of PE \(= 1900 \times g \times s\sin 5\) \([= 1655.95s\ \text{J}]\) | B1 | May be implied by energy equation |
| \(1900 \times g \times s\sin 5 + 150\,000 = \frac{1}{2} \times 1900 \times 30^2 - \frac{1}{2} \times 1900 \times 20^2\) | M1 | For attempt at work/energy equation |
| A1 | Correct | |
| \(s = 196\) m | A1 | |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(30^2 = 20^2 + 2a \times 200\) | M1 | Use of \(v^2 = u^2 + 2as\) |
| \(a = 1.25\ \text{m s}^{-2}\) | A1 | |
| \(T - 100 + 500g\sin 5 = 500a\) | M1 | For applying Newton's second law to the trailer |
| \(T = 289\ \text{N}\) | A1 | |
| 4 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Increase in KE $= \frac{1}{2} \times 1900 \times 30^2 - \frac{1}{2} \times 1900 \times 20^2$ $[= 475000\ \text{J}]$ | B1 | May be implied by energy equation |
| Loss of PE $= 1900 \times g \times s\sin 5$ $[= 1655.95s\ \text{J}]$ | B1 | May be implied by energy equation |
| $1900 \times g \times s\sin 5 + 150\,000 = \frac{1}{2} \times 1900 \times 30^2 - \frac{1}{2} \times 1900 \times 20^2$ | M1 | For attempt at work/energy equation |
| | A1 | Correct |
| $s = 196$ m | A1 | |
| | **5** | |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $30^2 = 20^2 + 2a \times 200$ | M1 | Use of $v^2 = u^2 + 2as$ |
| $a = 1.25\ \text{m s}^{-2}$ | A1 | |
| $T - 100 + 500g\sin 5 = 500a$ | M1 | For applying Newton's second law to the trailer |
| $T = 289\ \text{N}$ | A1 | |
| | **4** | |
---5 A car of mass 1400 kg is towing a trailer of mass 500 kg down a straight hill inclined at an angle of $5 ^ { \circ }$ to the horizontal. The car and trailer are connected by a light rigid tow-bar. At the top of the hill the speed of the car and trailer is $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and at the bottom of the hill their speed is $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item It is given that as the car and trailer descend the hill, the engine of the car does 150000 J of work, and there are no resistance forces.
Find the length of the hill.
\item It is given instead that there is a resistance force of 100 N on the trailer, the length of the hill is 200 m , and the acceleration of the car and trailer is constant.
Find the tension in the tow-bar between the car and trailer.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q5 [9]}}