| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2002 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Multi-stage motion with all parameters given |
| Difficulty | Moderate -0.3 This is a standard M1 kinematics and dynamics question requiring routine application of SUVAT equations, area under velocity-time graphs, and Newton's second law. While it has multiple parts (5 total), each part follows textbook methods with no novel problem-solving required. The calculations are straightforward: finding time from v=u+at, using area of trapezium for distance, and applying F=ma with weight. Slightly easier than average due to its highly structured nature and explicit guidance ('Hence, or otherwise'). |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| shape | B1 | |
| \((3, 2.5)\) | B1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Area \(= 27 = \frac{1}{2} \times 1.5 \times 3 + 3T + \frac{1}{2} \times 2.5 \times 3\) | M1 A1 | |
| \(\Rightarrow T = 7 \text{ s}\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| shape \(0 \leq t \leq 8.5\) | B1 | |
| shape \(t > 8.5\) | B1 | |
| \((2, 7 \text{ (ft)}, 2.5)\) | B1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(T - 200g = 200 \times 2\) | M1 A1 | |
| \(\Rightarrow T = 2360 \text{ N}\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(R - 80g = -80 \times 1.2\) | M1 A1 | |
| \(\Rightarrow R = 688 \text{ N}\) | A1 | (3) |
## (a)
shape | B1 |
$(3, 2.5)$ | B1 | (2) |
## (b)
Area $= 27 = \frac{1}{2} \times 1.5 \times 3 + 3T + \frac{1}{2} \times 2.5 \times 3$ | M1 A1 |
$\Rightarrow T = 7 \text{ s}$ | A1 | (3) |
## (c)
shape $0 \leq t \leq 8.5$ | B1 |
shape $t > 8.5$ | B1 |
$(2, 7 \text{ (ft)}, 2.5)$ | B1 | (3) |
## (d)
(System)
$T - 200g = 200 \times 2$ | M1 A1 |
$\Rightarrow T = 2360 \text{ N}$ | A1 | (3) |
## (e)
(Man)
$R - 80g = -80 \times 1.2$ | M1 A1 |
$\Rightarrow R = 688 \text{ N}$ | A1 | (3) |
**Total: (14 marks)**
---A man travels in a lift to the top of a tall office block. The lift starts from rest on the ground floor and moves vertically. It comes to rest again at the top floor, having moved a vertical distance of $27 \text{ m}$. The lift initially accelerates with a constant acceleration of $2 \text{ m s}^{-1}$ until it reaches a speed of $3 \text{ m s}^{-1}$. It then moves with a constant speed of $3 \text{ m s}^{-1}$ for $T$ seconds. Finally it decelerates with a constant deceleration for $2.5 \text{ s}$ before coming to rest at the top floor.
\begin{enumerate}[label=(\alph*)]
\item Sketch a speed-time graph for the motion of the lift. [2]
\item Hence, or otherwise, find the value of $T$. [3]
\item Sketch an acceleration-time graph for the motion of the lift. [3]
\end{enumerate}
The mass of the man is $80 \text{ kg}$ and the mass of the lift is $120 \text{ kg}$. The lift is pulled up by means of a vertical cable attached to the top of the lift. By modelling the cable as light and inextensible, find
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item the tension in the cable when the lift is accelerating, [3]
\item the magnitude of the force exerted by the lift on the man during the last $2.5 \text{ s}$ of the motion. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2002 Q6 [14]}}