| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Particle on slope then horizontal |
| Difficulty | Moderate -0.3 This is a straightforward application of work-energy principles with clearly defined parameters. Part (i) requires simple multiplication of force and distance, while part (ii) involves a standard energy conservation calculation accounting for KE, PE, and work against resistance. The multi-step nature and need to track energy changes across two ramps adds slight complexity, but the method is routine for M1 students and requires no novel insight. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(640 \times 18\) | M1 | For use of work done \(= F \times d\) |
| Work done \(= 11\,520\) J | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| KE at start \(= \frac{1}{2} \times 840 \times 14^2 = 82\,320\) J | B1 | |
| PE gained \(= 840g \times 8\sin 30 - 840g \times 10\sin 20 = 4870\) J | B1 | |
| \(\frac{1}{2} \times 840 \times v^2 = 82\,320 - 11\,520 - 4870\) | M1 | For using work–energy equation with 4 terms and solving for \(v\) |
| \(v = 12.5\) m s\(^{-1}\) | A1 |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $640 \times 18$ | M1 | For use of work done $= F \times d$ |
| Work done $= 11\,520$ J | A1 | |
---
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| KE at start $= \frac{1}{2} \times 840 \times 14^2 = 82\,320$ J | B1 | |
| PE gained $= 840g \times 8\sin 30 - 840g \times 10\sin 20 = 4870$ J | B1 | |
| $\frac{1}{2} \times 840 \times v^2 = 82\,320 - 11\,520 - 4870$ | M1 | For using work–energy equation with 4 terms and solving for $v$ |
| $v = 12.5$ m s$^{-1}$ | A1 | |
---3 A roller-coaster car (including passengers) has a mass of 840 kg . The roller-coaster ride includes a section where the car climbs a straight ramp of length 8 m inclined at $30 ^ { \circ }$ above the horizontal. The car then immediately descends another ramp of length 10 m inclined at $20 ^ { \circ }$ below the horizontal. The resistance to motion acting on the car is 640 N throughout the motion.\\
(i) Find the total work done against the resistance force as the car ascends the first ramp and descends the second ramp.\\
(ii) The speed of the car at the bottom of the first ramp is $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Use an energy method to find the speed of the car when it reaches the bottom of the second ramp.\\
\hfill \mbox{\textit{CAIE M1 2017 Q3 [6]}}