| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | March |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Vehicle pulling trailer - tension |
| Difficulty | Moderate -0.3 This is a straightforward mechanics problem requiring standard application of P=Fv and F=ma. Part (i) is simple algebra from power equation at constant speed. Part (ii) involves calculating driving force from new power, applying Newton's second law to the system, then considering the trailer separately to find tension. All steps are routine textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| 4(i) | Driving force = 6000/20 [= 300 N] | B1 |
| R = 300 – 80 = 220 | B1ft | Net force on system = 300 – R – 220 = 0 ft on DF found |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 4(ii) | [New driving force DF = 12500/25 = 500 N |
| Answer | Marks | Guidance |
|---|---|---|
| System: DF – 80 – R = 1800a] | M1 | Any one equation from the following: |
| Answer | Marks | Guidance |
|---|---|---|
| Two correct equations | A1ft | Correct DF = 500 must be used. ft on R value found |
| M1 | EITHER solve two dimensionally correct simultaneous |
| Answer | Marks | Guidance |
|---|---|---|
| a = 0.111 m s–2 | A1 | Allow a = 1/9 |
| T = 113 N (= 113.3333...) | A1 | Allow T = 340/3 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 4:
--- 4(i) ---
4(i) | Driving force = 6000/20 [= 300 N] | B1 | Using F = P/v
R = 300 – 80 = 220 | B1ft | Net force on system = 300 – R – 220 = 0 ft on DF found
2
Question | Answer | Marks | Guidance
--- 4(ii) ---
4(ii) | [New driving force DF = 12500/25 = 500 N
Car: DF – T – R = 1500a
Trailer: T – 80 = 300a
System: DF – 80 – R = 1800a] | M1 | Any one equation from the following:
Apply Newton’s 2nd law to the car
Apply Newton’s 2nd law to the trailer
Apply Newton’s 2nd law to the system of car and trailer.
Two correct equations | A1ft | Correct DF = 500 must be used. ft on R value found
M1 | EITHER solve two dimensionally correct simultaneous
equations in a and T to find a or T OR solve the system
equation to find a
a = 0.111 m s–2 | A1 | Allow a = 1/9
T = 113 N (= 113.3333...) | A1 | Allow T = 340/3
5
Question | Answer | Marks | GuidanceA car of mass $1500$ kg is pulling a trailer of mass $300$ kg along a straight horizontal road at a constant speed of $20$ m s$^{-1}$. The system of the car and trailer is modelled as two particles, connected by a light rigid horizontal rod. The power of the car's engine is $6000$ W. There are constant resistances to motion of $R$ N on the car and $80$ N on the trailer.
\begin{enumerate}[label=(\roman*)]
\item Find the value of $R$. [2]
\item The power of the car's engine is increased to $12\,500$ W. The resistance forces do not change. Find the acceleration of the car and trailer and the tension in the rod at an instant when the speed of the car is $25$ m s$^{-1}$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2019 Q4 [7]}}