| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Energy method with work done |
| Difficulty | Standard +0.8 This M1 question requires energy methods with variable resistance and power, demanding students to work with work-done integrals and apply energy conservation across multiple stages. The need to handle R=420/v + constant, find intermediate speeds, and compare resistance changes across two segments requires sophisticated problem-solving beyond standard textbook exercises, though it remains within M1 scope. |
| Spec | 3.02d Constant acceleration: SUVAT formulae6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = (5^2 - 3^2) \div (2 \times 500) = 0.016\) | B1 | |
| M1 | For using Newton's 2nd law | |
| \(DF + 90g \times 0.05 - R = 90 \times 0.016\) | A1 | |
| \(R = \frac{420}{v} - 90(0.016 - 0.5)\) | M1 | For using \(DF = P/v\) |
| \(R = \frac{420}{v} + 43.56\) | A1 | AG — Total: 5 marks |
| SR for assuming constant \(R\) and \(DF\) (max 2/5): PE loss \(= 90g(500)(0.05)\) and KE gain \(= \frac{1}{2}(90)(5^2-3^2)\) | B1 | |
| \(WD_{DF}\) + PE loss = KE gain + \(WD_R \rightarrow R = 420/v + 43.56\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v_M^2 = 3^2 + 2\times0.016\times250 \rightarrow\) speed at mid-point is \(4.12\ \text{ms}^{-1}\) | B1 | |
| Decrease in \(R\) from top to mid-way \(= 420[(1\div3) - (1\div\sqrt{17})]\) or Decrease in \(R\) from midway to b'm \(= 420[(1\div\sqrt{17}) - (1\div5)]\) | M1 | For finding the difference in \(R\) for either top to midway or midway to bottom |
| 38.1 and 17.9 | A1 | Total: 3 marks |
## Question 5(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = (5^2 - 3^2) \div (2 \times 500) = 0.016$ | B1 | |
| | M1 | For using Newton's 2nd law |
| $DF + 90g \times 0.05 - R = 90 \times 0.016$ | A1 | |
| $R = \frac{420}{v} - 90(0.016 - 0.5)$ | M1 | For using $DF = P/v$ |
| $R = \frac{420}{v} + 43.56$ | A1 | AG — Total: 5 marks |
| **SR** for assuming constant $R$ and $DF$ (max 2/5): PE loss $= 90g(500)(0.05)$ and KE gain $= \frac{1}{2}(90)(5^2-3^2)$ | B1 | |
| $WD_{DF}$ + PE loss = KE gain + $WD_R \rightarrow R = 420/v + 43.56$ | B1 | |
## Question 5(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v_M^2 = 3^2 + 2\times0.016\times250 \rightarrow$ speed at mid-point is $4.12\ \text{ms}^{-1}$ | B1 | |
| Decrease in $R$ from top to mid-way $= 420[(1\div3) - (1\div\sqrt{17})]$ or Decrease in $R$ from midway to b'm $= 420[(1\div\sqrt{17}) - (1\div5)]$ | M1 | For finding the difference in $R$ for either top to midway or midway to bottom |
| 38.1 and 17.9 | A1 | Total: 3 marks |
---5 A cyclist and his bicycle have a total mass of 90 kg . The cyclist starts to move with speed $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from the top of a straight hill, of length 500 m , which is inclined at an angle of $\sin ^ { - 1 } 0.05$ to the horizontal. The cyclist moves with constant acceleration until he reaches the bottom of the hill with speed $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The cyclist generates 420 W of power while moving down the hill. The resistance to the motion of the cyclist and his bicycle, $R \mathrm {~N}$, and the cyclist's speed, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, both vary.\\
(i) Show that $R = \frac { 420 } { v } + 43.56$.\\
(ii) Find the cyclist's speed at the mid-point of the hill. Hence find the decrease in the value of $R$ when the cyclist moves from the top of the hill to the mid-point of the hill, and when the cyclist moves from the mid-point of the hill to the bottom of the hill.
\hfill \mbox{\textit{CAIE M1 2015 Q5 [8]}}