| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Power from force and speed |
| Difficulty | Moderate -0.3 Part (i) is a direct application of P=Fv at constant speed (routine 2-mark calculation). Part (ii) requires setting up an energy equation with KE loss, PE gain, and work against resistance—standard M1 energy method with clear structure, though the multi-step calculation and managing units across 5 marks elevates it slightly above pure recall. Overall slightly easier than average due to straightforward application of standard techniques. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| 4(i) | P=3000 × 30 | M1 |
| P=90000 W = 90kW | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 4(ii) | PE gained = 25000gh | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | B1 | For either correct |
| Answer | Marks | Guidance |
|---|---|---|
| Initial KE = Final KE + 25000gdsin2 + 3000d | M1 | For a 4 term work-energy equation, correct dimensions |
| A1 | Correct work-energy equation involving h or d | |
| h=10.2m (10.2318…) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Mark |
Question 4:
--- 4(i) ---
4(i) | P=3000 × 30 | M1 | Use of P=Fv with F=resistance
P=90000 W = 90kW | A1
2
--- 4(ii) ---
4(ii) | PE gained = 25000gh | B1 | Correct expression for PE Allow PE=25000gdsin2
1
Initial KE= ×25000×302 [= 11 250 000]
2
1
Final KE= ×25000×252 [= 7 812 500]
2 | B1 | For either correct
[KE loss = 3 437 500]
3000h
Initial KE = Final KE + 25000gh +
sin2
OR
Initial KE = Final KE + 25000gdsin2 + 3000d | M1 | For a 4 term work-energy equation, correct dimensions
A1 | Correct work-energy equation involving h or d
h=10.2m (10.2318…) | A1
5
Question | Answer | Mark | GuidanceA lorry of mass 25 000 kg travels along a straight horizontal road. There is a constant force of 3000 N resisting the motion.
\begin{enumerate}[label=(\roman*)]
\item Find the power required to maintain a constant speed of 30 m s$^{-1}$. [2]
\end{enumerate}
The lorry comes to a straight hill inclined at 2° to the horizontal. The driver switches off the engine of the lorry at the point $A$ which is at the foot of the hill. Point $B$ is further up the hill. The speeds of the lorry at $A$ and $B$ are 30 m s$^{-1}$ and 25 m s$^{-1}$ respectively. The resistance force is still 3000 N.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Use an energy method to find the height of $B$ above the level of $A$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2019 Q4 [7]}}