Edexcel M1 — Question 4 11 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNewton's laws and connected particles
TypeLift with passenger or load
DifficultyStandard +0.3 This is a standard M1 lift problem requiring application of Newton's second law in multiple parts. Parts (a) and (b) involve routine F=ma calculations with weight and normal force, while parts (c) and (d) apply impulse-momentum theorem. All techniques are textbook standard with no novel insight required, making it slightly easier than average.
Spec3.03d Newton's second law: 2D vectors6.03f Impulse-momentum: relation

4. A lift of mass 70 kg is supported by a cable which remains taut at all times. A man of mass 90 kg gets into the lift and it begins to descend vertically from rest with constant acceleration \(0.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). Calculate, giving your answers correct to 3 significant figures,
  1. the magnitude of the force which the lift exerts on the man,
  2. the tension in the cable. Prior to slowing down, the lift is moving at \(2 \mathrm {~ms} ^ { - 1 }\). It then uniformly decelerates until it is brought to rest.
  3. Find the impulse exerted by the cable on the lift in bringing the lift to rest.
  4. Given that it takes 2 seconds to come to rest, use your answer to part (c) to calculate the magnitude of the force exerted by the cable on the lift in bringing the lift to rest.
    (2 marks)

AnswerMarks Guidance
(a) resolve ↓ for man: \(90g - R = 90(0.5)\) so \(R = 837 \text{ N}\)M2 A1
(b) resolve ↓ for lift: \(R + 70g - T = 70(0.5)\) \(837 + 686 - T = 35\) so \(T = 1488 \text{ N}\)M2 A1, A1
(c) impulse \(= \Delta\) mom. \(= 160(0-2) = 320 \text{ Ns}\)M1 A1
(d) \(Ft = 320\), so \(F = \frac{320}{2} = 160 \text{ N}\)M1 A1 (11)
| (a) resolve ↓ for man: $90g - R = 90(0.5)$ so $R = 837 \text{ N}$ | M2 A1 | |
| (b) resolve ↓ for lift: $R + 70g - T = 70(0.5)$ $837 + 686 - T = 35$ so $T = 1488 \text{ N}$ | M2 A1, A1 | |
| (c) impulse $= \Delta$ mom. $= 160(0-2) = 320 \text{ Ns}$ | M1 A1 | |
| (d) $Ft = 320$, so $F = \frac{320}{2} = 160 \text{ N}$ | M1 A1 | (11) |
4. A lift of mass 70 kg is supported by a cable which remains taut at all times. A man of mass 90 kg gets into the lift and it begins to descend vertically from rest with constant acceleration $0.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.

Calculate, giving your answers correct to 3 significant figures,
\begin{enumerate}[label=(\alph*)]
\item the magnitude of the force which the lift exerts on the man,
\item the tension in the cable.

Prior to slowing down, the lift is moving at $2 \mathrm {~ms} ^ { - 1 }$. It then uniformly decelerates until it is brought to rest.
\item Find the impulse exerted by the cable on the lift in bringing the lift to rest.
\item Given that it takes 2 seconds to come to rest, use your answer to part (c) to calculate the magnitude of the force exerted by the cable on the lift in bringing the lift to rest.\\
(2 marks)
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1  Q4 [11]}}