| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | November |
| Marks | 8 |
| 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 M1 mechanics question involving power, tension, and forces on connected particles. Part (i) uses P=Fv with total resistance; part (ii) requires considering forces on the trailer only; part (iii) applies F=ma with power and incline components. All techniques are routine for M1 level with straightforward multi-step application, making it slightly easier than average A-level difficulty. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([\text{Power} = 400 \times 25]\) | M1 | For using \(P = Fv\) where \(F =\) resistance \(= 400 \text{ N}\) |
| Power \(= 10000 \text{ W}\) | A1 [2] | Allow \(10 \text{ kW}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Tension \(= 100 \text{ N}\) | B1 [1] | Considering the trailer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| New driving force \(= 25000/20 = 1250 \text{ N}\) | B1 | Driving force \(= P/v\) at the instant when \(v = 20\) |
| \([\text{DF} - 300 - T - 3000\ g\sin4 = 3000a]\) or \([T - 100 - 500\ g\sin4 = 500a]\) or \([\text{DF} - 400 - 3500\ g\sin4 = 3500a]\) | M1 | For using Newton's second law applied either to the van or to the trailer or to the system of van and trailer |
| M1 | For using N2 applied to one of the other cases | |
| \([a = -0.4547 \text{ may be seen}]\) | M1 | Solving or using substitution to find \(T\) |
| \(T = 221 \text{ N}\) | A1 [5] | Allow \(T = 1550/7 \text{ N}\) |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[\text{Power} = 400 \times 25]$ | M1 | For using $P = Fv$ where $F =$ resistance $= 400 \text{ N}$ |
| Power $= 10000 \text{ W}$ | A1 [2] | Allow $10 \text{ kW}$ |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Tension $= 100 \text{ N}$ | B1 [1] | Considering the trailer |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| New driving force $= 25000/20 = 1250 \text{ N}$ | B1 | Driving force $= P/v$ at the instant when $v = 20$ |
| $[\text{DF} - 300 - T - 3000\ g\sin4 = 3000a]$ or $[T - 100 - 500\ g\sin4 = 500a]$ or $[\text{DF} - 400 - 3500\ g\sin4 = 3500a]$ | M1 | For using Newton's second law applied **either** to the van **or** to the trailer **or** to the system of van and trailer |
| | M1 | For using N2 applied to one of the other cases |
| $[a = -0.4547 \text{ may be seen}]$ | M1 | Solving or using substitution to find $T$ |
| $T = 221 \text{ N}$ | A1 [5] | Allow $T = 1550/7 \text{ N}$ |6 A van of mass 3000 kg is pulling a trailer of mass 500 kg along a straight horizontal road at a constant speed of $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The system of the van and the trailer is modelled as two particles connected by a light inextensible cable. There is a constant resistance to motion of 300 N on the van and 100 N on the trailer.\\
(i) Find the power of the van's engine.\\
(ii) Write down the tension in the cable.
The van reaches the bottom of a hill inclined at $4 ^ { \circ }$ to the horizontal with speed $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The power of the van's engine is increased to 25000 W .\\
(iii) Assuming that the resistance forces remain the same, find the new tension in the cable at the instant when the speed of the van up the hill is $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\hfill \mbox{\textit{CAIE M1 2016 Q6 [8]}}