| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Particle moving through liquid or resistance |
| Difficulty | Standard +0.3 This is a standard two-stage SUVAT problem with clearly defined phases (air then liquid). Parts (i) and (ii) require straightforward application of kinematic equations with given values. Part (iii) adds a simple force balance (weight - resistance = ma), but all values are provided. The problem is slightly above average due to multiple parts and the need to recognize two different motion regimes, but requires no novel insight or complex problem-solving. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.03c Newton's second law: F=ma one dimension |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | M1 | For using the area property for displacements or for using \(s = (u + v)t/2\) (for (a) or (b)) |
| (a) Height is \(2.45\) m | A1 | 3 |
| (b) Depth is \(3\) m | A1 | |
| (ii) | M1 | For using the idea that gradient represents acceleration or for using \(v = u + at\) |
| Deceleration is \(4\) ms\(^{-2}\) | A1 | 2 |
| (iii) | A1 ft | For using Newton's second law (3 terms needed) |
| \(0.7 - mg = 4m\) | A1 | 3 |
| Mass is \(0.05\) kg |
**(i)** | M1 | For using the area property for displacements or for using $s = (u + v)t/2$ (for (a) or (b)) |
(a) Height is $2.45$ m | A1 | 3 |
(b) Depth is $3$ m | A1 | |
**(ii)** | M1 | For using the idea that gradient represents acceleration or for using $v = u + at$ |
Deceleration is $4$ ms$^{-2}$ | A1 | 2 |
**(iii)** | A1 ft | For using Newton's second law (3 terms needed) |
$0.7 - mg = 4m$ | A1 | 3 |
Mass is $0.05$ kg | |
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\includegraphics[max width=\textwidth, alt={}, center]{b5873699-d207-4cad-9518-1321dc429c15-3_568_1084_269_532}
The diagram shows the velocity-time graph for the motion of a small stone which falls vertically from rest at a point $A$ above the surface of liquid in a container. The downward velocity of the stone $t \mathrm {~s}$ after leaving $A$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The stone hits the surface of the liquid with velocity $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when $t = 0.7$. It reaches the bottom of the container with velocity $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when $t = 1.2$.\\
(i) Find
\begin{enumerate}[label=(\alph*)]
\item the height of $A$ above the surface of the liquid,
\item the depth of liquid in the container.\\
(ii) Find the deceleration of the stone while it is moving in the liquid.\\
(iii) Given that the resistance to motion of the stone while it is moving in the liquid has magnitude 0.7 N , find the mass of the stone.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2006 Q4 [8]}}