| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Particle on smooth curved surface |
| Difficulty | Moderate -0.3 Part (a) is a straightforward application of conservation of energy on a smooth surface (PE = KE), requiring only substitution into standard formulas. Part (b) involves using the work-energy principle with given numerical values to find mass—slightly more involved but still a standard textbook exercise with clear setup and direct calculation. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(mg \times 1.8 = \frac{1}{2}mv^2\) | M1 | Use of conservation of energy, 2 terms. Must NOT use constant acceleration equations. Use of equations such as \(v^2 = u^2 + 2as\) scores M0 A0. |
| Speed of block at \(B = v = 6\) ms\(^{-1}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt the work-energy equation | M1 | In the form: \(\pm\) KE lost \(= \pm\) PE gain \(\pm\) WD against Resistance |
| \(\frac{1}{2} \times m \times 6^2 = 4.5 + mg \times 1.2\) | A1 | If using motion from \(A\) to final point: \(mg \times 1.8 = mg \times 1.2 + 4.5\) |
| Mass of the block \(= m = 0.75\) kg | A1 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $mg \times 1.8 = \frac{1}{2}mv^2$ | M1 | Use of conservation of energy, 2 terms. Must NOT use constant acceleration equations. Use of equations such as $v^2 = u^2 + 2as$ scores **M0 A0**. |
| Speed of block at $B = v = 6$ ms$^{-1}$ | A1 | AG |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt the work-energy equation | M1 | In the form: $\pm$ KE lost $= \pm$ PE gain $\pm$ WD against Resistance |
| $\frac{1}{2} \times m \times 6^2 = 4.5 + mg \times 1.2$ | A1 | If using motion from $A$ to final point: $mg \times 1.8 = mg \times 1.2 + 4.5$ |
| Mass of the block $= m = 0.75$ kg | A1 | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{083d3e44-1e42-461f-aa8d-a1a22047a47e-04_416_792_260_674}
The diagram shows a semi-circular track $A B C$ of radius 1.8 m which is fixed in a vertical plane. The points $A$ and $C$ are at the same horizontal level and the point $B$ is at the bottom of the track. The section $A B$ is smooth and the section $B C$ is rough. A small block is released from rest at $A$.
\begin{enumerate}[label=(\alph*)]
\item Show that the speed of the block at $B$ is $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
The block comes to instantaneous rest for the first time at a height of 1.2 m above the level of $B$. The work done against the resistance force during the motion of the block from $B$ to this point is 4.5 J .
\item Find the mass of the block.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q3 [5]}}