| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Work-energy over time interval |
| Difficulty | Standard +0.3 This is a straightforward application of the power-force-velocity relationship (P=Fv) and the work-energy theorem. Part (i) requires simple algebraic manipulation of given ratios to find speed, while part (ii) applies the work-energy principle with given values. Both parts are routine calculations with no conceptual challenges beyond standard M1 content. |
| Spec | 6.02a Work done: concept and definition6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([v_B = 1.2 \times 28 \div 0.96]\) | M1 | For using \(P = Fv\) and factors 1.2 and 0.96 and an equation in \(v_B\) only |
| Speed of the train at \(B\) is \(35 \text{ ms}^{-1}\) | A1 | 2 marks total; AG |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| KE increase \(= 100000(35^2 - 28^2)\) | B1 | |
| WD by engine \(= 44.1 \times 10^6 + 2.3 \times 10^6 \text{ J}\) | M1 | For using WD by engine \(=\) KE increase \(+\) WD against resistance |
| Work done is \(46400 \text{ kJ}\) or \(46.4 \times 10^6 \text{ J}\) | A1 | 3 marks total; or \(46\,400\,000 \text{ J}\) |
## Question 3:
### Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| $[v_B = 1.2 \times 28 \div 0.96]$ | M1 | For using $P = Fv$ and factors 1.2 and 0.96 and an equation in $v_B$ only |
| Speed of the train at $B$ is $35 \text{ ms}^{-1}$ | A1 | 2 marks total; AG |
### Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| KE increase $= 100000(35^2 - 28^2)$ | B1 | |
| WD by engine $= 44.1 \times 10^6 + 2.3 \times 10^6 \text{ J}$ | M1 | For using WD by engine $=$ KE increase $+$ WD against resistance |
| Work done is $46400 \text{ kJ}$ or $46.4 \times 10^6 \text{ J}$ | A1 | 3 marks total; or $46\,400\,000 \text{ J}$ |
---3 A train of mass 200000 kg moves on a horizontal straight track. It passes through a point $A$ with speed $28 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and later it passes through a point $B$. The power of the train's engine at $B$ is 1.2 times the power of the train's engine at $A$. The driving force of the train's engine at $B$ is 0.96 times the driving force of the train's engine at $A$.\\
(i) Show that the speed of the train at $B$ is $35 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(ii) For the motion from $A$ to $B$, find the work done by the train's engine given that the work done against the resistance to the train's motion is $2.3 \times 10 ^ { 6 } \mathrm {~J}$.
\hfill \mbox{\textit{CAIE M1 2014 Q3 [5]}}