| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find power at constant speed |
| Difficulty | Standard +0.3 This is a standard M1 power question requiring P=Fv and F=ma applications. Part (a) uses direct formula application, part (b) involves finding new driving force then deceleration (routine two-step), and part (c) requires resolving forces on an incline with variable resistance and solving a quadratic. All techniques are standard M1 fare with no novel insight required, making it slightly easier than average overall. |
| Spec | 3.03f Weight: W=mg3.03v Motion on rough surface: including inclined planes6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks |
|---|---|
| 5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or the |
| Answer | Marks | Guidance |
|---|---|---|
| 5(a)(i) | DF = 750 | B1 |
| Answer | Marks |
|---|---|
| =24kW | B1 FT |
| Answer | Marks |
|---|---|
| 5(a)(ii) | 16000=DF×32 |
| DF =500 | M1 |
| 500−750=1250×a | M1 |
| a=[−]0.2 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 5(b) | DF =1000+8v+1250×10×0.096 | M1 |
| 2200+8v | A1 | |
| 60000=(2200+8v)v | M1 | |
| 8v2 +2200v−60000=0 | A1 | |
| v=25 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or the
method required, award all marks earned and deduct just 1 mark for the misread.
--- 5(a)(i) ---
5(a)(i) | DF = 750 | B1
Power =their(750)×32
=24kW | B1 FT
2
--- 5(a)(ii) ---
5(a)(ii) | 16000=DF×32
DF =500 | M1
500−750=1250×a | M1
a=[−]0.2 | A1
3
--- 5(b) ---
5(b) | DF =1000+8v+1250×10×0.096 | M1
2200+8v | A1
60000=(2200+8v)v | M1
8v2 +2200v−60000=0 | A1
v=25 | A1
5
Question | Answer | MarksA car of mass 1250 kg is moving on a straight road.
\begin{enumerate}[label=(\alph*)]
\item On a horizontal section of the road, the car has a constant speed of $32 \text{ ms}^{-1}$ and there is a constant force of 750 N resisting the motion.
\begin{enumerate}[label=(\roman*)]
\item Calculate, in kW, the power developed by the engine of the car. [2]
\item Given that this power is suddenly decreased by 8 kW, find the instantaneous deceleration of the car. [3]
\end{enumerate}
\item On a section of the road inclined at $\sin^{-1} 0.096$ to the horizontal, the resistance to the motion of the car is $(1000 + 8v)$ N when the speed of the car is $v \text{ ms}^{-1}$. The car travels up this section of the road at constant speed with the engine working at 60 kW.
Find this constant speed. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2020 Q5 [10]}}