Edexcel M1 2022 October — Question 4 6 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2022
SessionOctober
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNewton's laws and connected particles
TypeLift with passenger or load
DifficultyModerate -0.8 This is a straightforward application of Newton's second law to connected particles with constant acceleration. Part (a) requires a single F=ma equation for the lift (T - mg = ma), and part (b) similarly applies F=ma to Alan (R - mg = ma using the acceleration from part (a)). Both parts involve routine substitution into standard formulas with no conceptual challenges or problem-solving insight required.
Spec3.03c Newton's second law: F=ma one dimension3.03f Weight: W=mg

\includegraphics{figure_3} Two children, Alan and Bhavana, are standing on the horizontal floor of a lift, as shown in Figure 3. The lift has mass 250 kg. The lift is raised vertically upwards with constant acceleration by a vertical cable which is attached to the top of the lift. The cable is modelled as being light and inextensible. While the lift is accelerating upwards, the tension in the cable is 3616 N. As the lift accelerates upwards, the floor of the lift exerts a force of magnitude 565 N on Alan and a force of magnitude 226 N on Bhavana. Air resistance is modelled as being negligible and Alan and Bhavana are modelled as particles.
  1. By considering the forces acting on the lift only, find the acceleration of the lift. [3]
  2. Find the mass of Alan. [3]

Question 4:

AnswerMarks Guidance
4(a)3616−250g−565−226=250a M1 A1
a=1.5 (m s−2)A1
(3)
AnswerMarks
4(a)M1 Equation in a only, correct no. of terms, condone sign errors
A1 Correct equation
A1 oe

AnswerMarks Guidance
4(b)5 6 5 − m g = m  1 .5 M1A1ft
m = 50 (kg)A1
(3)
(6)
Notes for question 4
AnswerMarks
4(b)M1 Equation in m (mass of A) only, correct terms, condone sign errors
A1ft Correct equation ft on their a
A1 cao
Question 4:
--- 4(a) ---
4(a) | 3616−250g−565−226=250a | M1 A1
a=1.5 (m s−2) | A1
(3)
4(a) | M1 Equation in a only, correct no. of terms, condone sign errors
A1 Correct equation
A1 oe
--- 4(b) ---
4(b) | 5 6 5 − m g = m  1 .5 | M1A1ft
m = 50 (kg) | A1
(3)
(6)
Notes for question 4
4(b) | M1 Equation in m (mass of A) only, correct terms, condone sign errors
A1ft Correct equation ft on their a
A1 cao
\includegraphics{figure_3}

Two children, Alan and Bhavana, are standing on the horizontal floor of a lift, as shown in Figure 3.

The lift has mass 250 kg. The lift is raised vertically upwards with constant acceleration by a vertical cable which is attached to the top of the lift. The cable is modelled as being light and inextensible. While the lift is accelerating upwards, the tension in the cable is 3616 N.

As the lift accelerates upwards, the floor of the lift exerts a force of magnitude 565 N on Alan and a force of magnitude 226 N on Bhavana.

Air resistance is modelled as being negligible and Alan and Bhavana are modelled as particles.

\begin{enumerate}[label=(\alph*)]
\item By considering the forces acting on the lift only, find the acceleration of the lift. [3]

\item Find the mass of Alan. [3]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2022 Q4 [6]}}