| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | November |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Distance from velocity-time graph |
| Difficulty | Easy -1.2 This is a straightforward SUVAT kinematics problem requiring basic application of standard equations of motion. The question involves reading values from a velocity-time graph and applying formulas for acceleration and displacement, which are routine skills for M1 students with no novel problem-solving required. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.03a Force: vector nature and diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using the gradient property for acceleration | |
| Acceleration is \(4 \text{ ms}^{-2}\) | A1 | |
| M1 | For applying Newton's 2nd law to both particles or using \((M+m)a = (M-m)g\) and \(m + M = 1\) | |
| \(T - mg = 4m\) and \((1-m)g - T = 4(1-m)\) or \(4 = (1 - m - m)g\) | A1 | |
| \(P\) has mass \(0.3\) kg and \(Q\) has mass \(0.7\) kg | A1 | [5 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For using area property of graph or \(h = \frac{1}{2}at^2\) to obtain \(h = 2\) | B1 | [1 mark] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance travelled upwards by \(P = \frac{1}{2} \times 1.4 \times 4\) | B1 | |
| Height is \(4.8\) m | B1 | [2 marks] |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using the gradient property for acceleration |
| Acceleration is $4 \text{ ms}^{-2}$ | A1 | |
| | M1 | For applying Newton's 2nd law to both particles or using $(M+m)a = (M-m)g$ and $m + M = 1$ |
| $T - mg = 4m$ **and** $(1-m)g - T = 4(1-m)$ **or** $4 = (1 - m - m)g$ | A1 | |
| $P$ has mass $0.3$ kg and $Q$ has mass $0.7$ kg | A1 | **[5 marks]** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For using area property of graph or $h = \frac{1}{2}at^2$ to obtain $h = 2$ | B1 | **[1 mark]** |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance travelled upwards by $P = \frac{1}{2} \times 1.4 \times 4$ | B1 | |
| Height is $4.8$ m | B1 | **[2 marks]** |
---6
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9c7e8624-c4cd-4a8e-83d9-f92d0bd6f95b-3_462_218_1343_287}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9c7e8624-c4cd-4a8e-83d9-f92d0bd6f95b-3_563_1143_1238_712}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
Particles $P$ and $Q$ have a total mass of 1 kg . The particles are attached to opposite ends of a light inextensible string which passes over a smooth fixed pulley. $P$ is held at rest and $Q$ hangs freely, with both straight parts of the string vertical. Both particles are at a height of $h \mathrm {~m}$ above the floor (see Fig. 1). $P$ is released from rest and the particles start to move with the string taut. Fig. 2 shows the velocity-time graphs for $P$ 's motion and for $Q$ 's motion, where the positive direction for velocity is vertically upwards. Find\\
(i) the magnitude of the acceleration with which the particles start to move and the mass of each of the particles,\\
(ii) the value of $h$,\\
(iii) the greatest height above the floor reached by particle $P$.
\hfill \mbox{\textit{CAIE M1 2014 Q6}}