| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Two particles: same start time, different heights |
| Difficulty | Standard +0.3 This is a straightforward SUVAT problem requiring students to recognize that collision occurs when P reaches A and Q returns to A, giving a time constraint. Part (i) uses v² = u² + 2as and part (ii) uses s = ut + ½at². The setup is clear and the solution path is direct with standard equations, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| (complete method for total time) | M1 | For using \(v = u - gt\) to find time taken by \(Q\). Must be complete method for total time taken to return to point \(A\) |
| \(-20 = 20 - 10t \Rightarrow\) time taken is \(4\) s, or \(0 = 20 - 10t \Rightarrow\) time taken is \(2 \times 2 = 4\) s | A1 | |
| \(30 = 0 + 4a\) | M1 | For using \(v = u + at\) to find acceleration of \(P\) |
| Acceleration of \(P\) is \(7.5\) ms\(^{-2}\) | A1\(\checkmark\) | [4] ft on an incorrect positive value of the time taken |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| (suvat equation for distance \(OA\)) | M1 | For using \(v^2 = u^2 + 2as\), or \(s = \frac{(u+v)}{2}t\), or \(s = ut + \frac{1}{2}at^2\), or \(s = vt - \frac{1}{2}at^2\) to find distance \(OA\) |
| Either \(30^2 = 2 \times 7.5 \times OA\), or \(OA = \frac{(0+30)}{2} \times 4\), or \(OA = \frac{1}{2} \times 7.5 \times 4^2\), or \(OA = 30 \times 4 - \frac{1}{2} \times 7.5 \times 4^2\) \(\Rightarrow\) Distance \(OA\) is \(60\) m | A1 | [2] |
## Question 5:
### Part (i)
| Working/Answer | Mark | Guidance |
|---|---|---|
| (complete method for total time) | M1 | For using $v = u - gt$ to find time taken by $Q$. Must be complete method for total time taken to return to point $A$ |
| $-20 = 20 - 10t \Rightarrow$ time taken is $4$ s, or $0 = 20 - 10t \Rightarrow$ time taken is $2 \times 2 = 4$ s | A1 | |
| $30 = 0 + 4a$ | M1 | For using $v = u + at$ to find acceleration of $P$ |
| Acceleration of $P$ is $7.5$ ms$^{-2}$ | A1$\checkmark$ | **[4]** ft on an incorrect positive value of the time taken |
### Part (ii)
| Working/Answer | Mark | Guidance |
|---|---|---|
| (suvat equation for distance $OA$) | M1 | For using $v^2 = u^2 + 2as$, or $s = \frac{(u+v)}{2}t$, or $s = ut + \frac{1}{2}at^2$, or $s = vt - \frac{1}{2}at^2$ to find distance $OA$ |
| Either $30^2 = 2 \times 7.5 \times OA$, or $OA = \frac{(0+30)}{2} \times 4$, or $OA = \frac{1}{2} \times 7.5 \times 4^2$, or $OA = 30 \times 4 - \frac{1}{2} \times 7.5 \times 4^2$ $\Rightarrow$ Distance $OA$ is $60$ m | A1 | **[2]** |
---5 A particle $P$ starts from rest at a point $O$ on a horizontal straight line. $P$ moves along the line with constant acceleration and reaches a point $A$ on the line with a speed of $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. At the instant that $P$ leaves $O$, a particle $Q$ is projected vertically upwards from the point $A$ with a speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Subsequently $P$ and $Q$ collide at $A$. Find\\
(i) the acceleration of $P$,\\
(ii) the distance $O A$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d5f48bef-2518-4abd-b3e1-5e48ce56cf62-3_538_414_315_370}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d5f48bef-2518-4abd-b3e1-5e48ce56cf62-3_561_686_264_1080}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
Two particles $P$ and $Q$ have masses $m \mathrm {~kg}$ and $( 1 - m ) \mathrm { kg }$ respectively. The particles are attached to the ends of a light inextensible string which passes over a smooth fixed pulley. $P$ is held at rest with the string taut and both straight parts of the string vertical. $P$ and $Q$ are each at a height of $h \mathrm {~m}$ above horizontal ground (see Fig. 1). $P$ is released and $Q$ moves downwards. Subsequently $Q$ hits the ground and comes to rest. Fig. 2 shows the velocity-time graph for $P$ while $Q$ is moving downwards or is at rest on the ground.\\
\hfill \mbox{\textit{CAIE M1 2015 Q5 [6]}}