| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Towing system: inclined road |
| Difficulty | Moderate -0.3 This is a standard mechanics question testing power-force-velocity relationships and Newton's second law on an incline. Part (a) is direct substitution (P=Fv), part (b) requires resolving forces and using P=Fv, and part (c) involves connected particles with two equations. All techniques are routine for M1 level with no novel insight required, making it slightly easier than average. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| 3(a) | Power =140028 | B1 |
| Power = 39.3kW | B1 |
| Answer | Marks |
|---|---|
| 3(b) | 43500 |
| Answer | Marks | Guidance |
|---|---|---|
| v | B1 | oe |
| Answer | Marks | Guidance |
|---|---|---|
| or DF =1400+1250gsin6.89=2899.544602 | M1 | 3 terms, no need for DF in terms of v. |
| Answer | Marks | Guidance |
|---|---|---|
| Speed =15 m s–1 | A1 | Awrt 15.0 |
| Answer | Marks |
|---|---|
| 3(c) | Attempt at N2L on either car, trailer or the system |
| Answer | Marks | Guidance |
|---|---|---|
| System: 5000−1400−300−1250g0.12−600g0.12=( 1250+600 )a | M1 | Allow sign errors, sin/cos mix. |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | For any 2 equations correct. | |
| Solve for a or T | M1 | From equation(s) with at most 1 term. |
| Answer | Marks | Guidance |
|---|---|---|
| 185 37 | A1 | Awrt 0.584 and 1370. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 3:
--- 3(a) ---
3(a) | Power =140028 | B1
Power = 39.3kW | B1
2
--- 3(b) ---
3(b) | 43500
DF =
v | B1 | oe
Attempt to resolve parallel to the hill
DF =1400+1250g0.12=2900
or DF =1400+1250gsin6.89=2899.544602 | M1 | 3 terms, no need for DF in terms of v.
Allow sign errors, sin/cos mix.
Allow use of 6.89º or 6.9º.
Speed =15 m s–1 | A1 | Awrt 15.0
3
--- 3(c) ---
3(c) | Attempt at N2L on either car, trailer or the system
Car: 5000−1400−1250g0.12−T =1250a
Trailer: T −300−600g0.12=600a
System: 5000−1400−300−1250g0.12−600g0.12=( 1250+600 )a | M1 | Allow sign errors, sin/cos mix.
Correct number of relevant terms.
Allow use of 6.89º or 6.9º.
Allow with g missing.
A1 | For any 2 equations correct.
Solve for a or T | M1 | From equation(s) with at most 1 term.
missing/extra in total. Allow with g missing.
108 50700
Acceleration = =0.584 ms-2, Tension = =1370 N
185 37 | A1 | Awrt 0.584 and 1370.
a=0.583787838, T =1370.27027.
4
Question | Answer | Marks | GuidanceA constant resistance of magnitude 1400 N acts on a car of mass 1250 kg.
\begin{enumerate}[label=(\alph*)]
\item The car is moving along a straight level road at a constant speed of 28 m s$^{-1}$.
Find, in kW, the rate at which the engine of the car is working. [2]
\item The car now travels at a constant speed up a hill inclined at an angle of $\theta$ to the horizontal, where $\sin \theta = 0.12$, with the engine working at 43.5 kW.
Find this speed. [3]
\item On another occasion, the car pulls a trailer of mass 600 kg up the same hill. The system of the car and the trailer is modelled as particles connected by a light inextensible cable. The car's engine produces a driving force of 5000 N and the resistance to the motion of the trailer is 300 N. The resistance to the motion of the car remains 1400 N.
Find the acceleration of the system and the tension in the cable. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2022 Q3 [9]}}