| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Particle on smooth curved surface |
| Difficulty | Moderate -0.3 This is a straightforward application of conservation of energy on a smooth surface with three standard parts: finding maximum speed at lowest point, KE at a given height, and minimum speed for reverse journey. All parts use the same principle (PE ↔ KE) with no complex problem-solving or novel insight required, making it slightly easier than average. |
| Spec | 6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Loss of PE = \(mg \times 2.45\) | B1 | |
| \([\frac{1}{2}mv^2 = 24.5m]\) | M1 | For using KE gain = PE loss |
| Greatest speed is 7ms\(^{-1}\) | A1 | 3 |
| (ii) \(KE = 0.5g(2.45 - 1.2)\) or \(KE = \frac{1}{4} - 0.5x^2 - 0.5gx1.2\) | B1ft | |
| Kinetic energy is 6.25J | B1 | 2 |
| (iii) \([\frac{1}{2} \times 0.5v^2 = 6.25]\) | M1 | For using KE found in (ii)=\(\frac{1}{2}mv^2\) or \(\frac{1}{2}m1^2 - 1.2mg = \frac{1}{2}mv^2\) |
| Least value is 5 | A1 | 2 |
(i) Loss of PE = $mg \times 2.45$ | B1 |
$[\frac{1}{2}mv^2 = 24.5m]$ | M1 | For using KE gain = PE loss
Greatest speed is 7ms$^{-1}$ | A1 | 3
(ii) $KE = 0.5g(2.45 - 1.2)$ or $KE = \frac{1}{4} - 0.5x^2 - 0.5gx1.2$ | B1ft |
Kinetic energy is 6.25J | B1 | 2 | SR(max 1 mark out of 2): For use of $v^2 = 2x10(2.45 - 1.2)$ to obtain $v = 5$ and then $KE = \frac{1}{2} \times 0.5 \times 5^2 = 6.25$ | B1
(iii) $[\frac{1}{2} \times 0.5v^2 = 6.25]$ | M1 | For using KE found in (ii)=$\frac{1}{2}mv^2$ or $\frac{1}{2}m1^2 - 1.2mg = \frac{1}{2}mv^2$
Least value is 5 | A1 | 25\\
\includegraphics[max width=\textwidth, alt={}, center]{d0fa61eb-f320-427e-8883-de224d293933-3_515_789_995_676}
The diagram shows the vertical cross-section $L M N$ of a fixed smooth surface. $M$ is the lowest point of the cross-section. $L$ is 2.45 m above the level of $M$, and $N$ is 1.2 m above the level of $M$. A particle of mass 0.5 kg is released from rest at $L$ and moves on the surface until it leaves it at $N$. Find\\
(i) the greatest speed of the particle,\\
(ii) the kinetic energy of the particle at $N$.
The particle is now projected from $N$, with speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, along the surface towards $M$.\\
(iii) Find the least value of $v$ for which the particle will reach $L$.
\hfill \mbox{\textit{CAIE M1 2006 Q5 [7]}}