| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - smooth inclined plane (no resistance) |
| Difficulty | Standard +0.3 This is a straightforward energy conservation problem with two standard parts: (i) uses basic PE + KE conservation with no resistance, and (ii) includes work done against resistance. Both require routine application of energy equations with no novel problem-solving insight, making it slightly easier than average for A-level mechanics. |
| Spec | 6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{PE loss} = mg \times 100\sin20\) | B1 | |
| \([\frac{1}{2}mv^2 - \frac{1}{2}m \times 5^2 = mg \times 100\sin20]\) | M1 | Using KE gain \(=\) PE loss |
| \(v = 26.6 \text{ ms}^{-1}\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = g\sin20\ [= 3.42]\) | B1 | |
| \([v^2 = 5^2 + 2 \times a \times 100]\) | M1 | Using \(v^2 = u^2 + 2as\) |
| \(v = 26.6 \text{ ms}^{-1}\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{KE} = \pm(0.5m \times 441 - 0.5m \times 25)\ [= \pm 208m]\) | B1 | |
| \([mg \times 100\sin20 = 8500 + 208m]\) | M1 | For using PE loss \(=\) WD against Friction \(+\) KE gain |
| Mass \(m = 63.4 \text{ kg}\) | A1 [3] |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{PE loss} = mg \times 100\sin20$ | B1 | |
| $[\frac{1}{2}mv^2 - \frac{1}{2}m \times 5^2 = mg \times 100\sin20]$ | M1 | Using KE gain $=$ PE loss |
| $v = 26.6 \text{ ms}^{-1}$ | A1 [3] | |
**Alternative method for 4(i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = g\sin20\ [= 3.42]$ | B1 | |
| $[v^2 = 5^2 + 2 \times a \times 100]$ | M1 | Using $v^2 = u^2 + 2as$ |
| $v = 26.6 \text{ ms}^{-1}$ | A1 [3] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{KE} = \pm(0.5m \times 441 - 0.5m \times 25)\ [= \pm 208m]$ | B1 | |
| $[mg \times 100\sin20 = 8500 + 208m]$ | M1 | For using PE loss $=$ WD against Friction $+$ KE gain |
| Mass $m = 63.4 \text{ kg}$ | A1 [3] | |
---4 A girl on a sledge starts, with a speed of $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, at the top of a slope of length 100 m which is at an angle of $20 ^ { \circ }$ to the horizontal. The sledge slides directly down the slope.\\
(i) Given that there is no resistance to the sledge's motion, find the speed of the sledge at the bottom of the slope.\\
(ii) It is given instead that the sledge experiences a resistance to motion such that the total work done against the resistance is 8500 J , and the speed of the sledge at the bottom of the slope is $21 \mathrm {~ms} ^ { - 1 }$. Find the total mass of the girl and the sledge.
\hfill \mbox{\textit{CAIE M1 2016 Q4 [6]}}