CAIE M1 2020 Specimen — Question 5 9 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2020
SessionSpecimen
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNewton's laws and connected particles
TypeCar towing trailer, inclined road
DifficultyStandard +0.3 This is a standard connected particles problem on an inclined plane requiring resolution of forces, Newton's second law applied to a two-body system, and basic kinematics. While it involves multiple steps and careful bookkeeping of forces (driving force, resistances, weight components, tension), it follows a completely routine template taught in M1 courses with no novel insight required. Slightly easier than average due to being a textbook application.
Spec3.03o Advanced connected particles: and pulleys3.03q Dynamics: motion under force in plane

5 A car of mass 1200 kg is pulling a trailer of mass 800 kg up a hill inclined at an angle of \(\sin ^ { - 1 } ( 0.1 )\) to the horizontal. The car and the trailer are connected by a light rigid tow-bar which is parallel to the road. The driving force of the car's engine is 2500 N and the resistances to the car and trailer are 300 N and 100 N respectively.
  1. Find the acceleration of the system and the tension in the tow-bar.
  2. When the car and trailer are travelling at a speed of \(30 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the driving force becomes zero. Find the time, in seconds, before the system comes to rest and the force in the tow-bar during this time.

AnswerMarks Guidance
5(a)\(2500 = 2000g \times 0.1 - 400 = 2000a\) \(\sigma = 0.05\text{m}^2\) \(T = 300 - 1200 \times 0.05\) M1 M1 M1 A1 A1
- \(T - 30 = 1.2\)
- \(T = 100 - 800 \times 1.2\)
Accept \((Tmust) = 600N\)
AnswerMarks Guidance
5(b)For Newton's 2nd law \(-2000g \times 0.1 - 400 = 2000a\) \(T = -25\) \(T = 100 - 800 \times (-1.2)\) M1 M1 A1 M1 A1
**5(a)** | $2500 = 2000g \times 0.1 - 400 = 2000a$ $\sigma = 0.05\text{m}^2$ $T = 300 - 1200 \times 0.05$ | M1 M1 M1 A1 A1 | For Newton's 2nd law either car or trailer to:
- $T - 30 = 1.2$
- $T = 100 - 800 \times 1.2$
Accept $(Tmust) = 600N$

**5(b)** | For Newton's 2nd law $-2000g \times 0.1 - 400 = 2000a$ $T = -25$ $T = 100 - 800 \times (-1.2)$ | M1 M1 A1 M1 A1 | Accept (Tmust) = 60N

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5 A car of mass 1200 kg is pulling a trailer of mass 800 kg up a hill inclined at an angle of $\sin ^ { - 1 } ( 0.1 )$ to the horizontal. The car and the trailer are connected by a light rigid tow-bar which is parallel to the road. The driving force of the car's engine is 2500 N and the resistances to the car and trailer are 300 N and 100 N respectively.\\
(a) Find the acceleration of the system and the tension in the tow-bar.\\
(b) When the car and trailer are travelling at a speed of $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the driving force becomes zero.

Find the time, in seconds, before the system comes to rest and the force in the tow-bar during this time.\\

\hfill \mbox{\textit{CAIE M1 2020 Q5 [9]}}