| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle on table with string above |
| Difficulty | Moderate -0.5 This appears to be a standard circular motion problem involving a conical pendulum or particle on a table with string through a hole. These are routine M2 exercises requiring resolution of forces and application of F=mrω². The setup is typical and the solution method is well-practiced, making it easier than average but not trivial. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(0.6\,dv/dt = 0.6g - 3v\) | B1 | Newton's Second Law |
| \(\int 1/(10-5v)\,dv = \int dt\) | M1 | \(0.6\int 1/(0.6g - 3v)\,dv = \int dt\) |
| \(-\frac{1}{5}\ln(10 - 5v) = t(+c)\) | A1 | \(\frac{0.6}{-3}\ln(0.6g - 3v) = t(+c)\) |
| Finds \(c\) or uses limits twice | M1 | |
| \(t = 0.738 \text{ s}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(0.6v\,dv/dx = -3v\) | B1 | Newton's Second Law |
| \(\int 0.2\,dv = -\int dx\) | M1 | Integration with use of limits or finding \(c\) |
| \(x = 0.39 \text{ m}\) | A1 |
## Question 6:
### Part (i):
| Working | Marks | Guidance |
|---------|-------|----------|
| $0.6\,dv/dt = 0.6g - 3v$ | B1 | Newton's Second Law |
| $\int 1/(10-5v)\,dv = \int dt$ | M1 | $0.6\int 1/(0.6g - 3v)\,dv = \int dt$ |
| $-\frac{1}{5}\ln(10 - 5v) = t(+c)$ | A1 | $\frac{0.6}{-3}\ln(0.6g - 3v) = t(+c)$ |
| Finds $c$ or uses limits twice | M1 | |
| $t = 0.738 \text{ s}$ | A1 | |
**Total: 5 marks**
### Part (ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| $0.6v\,dv/dx = -3v$ | B1 | Newton's Second Law |
| $\int 0.2\,dv = -\int dx$ | M1 | Integration with use of limits or finding $c$ |
| $x = 0.39 \text{ m}$ | A1 | |
**Total: 3 marks**
---6 A particle $P$ of mass 0.6 kg is released from rest at a point above ground level and falls vertically. The motion of $P$ is opposed by a force of magnitude $3 v \mathrm {~N}$, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the speed of $P$. Immediately before $P$ reaches the ground, $v = 1.95$.\\
(i) Calculate the time after its release when $P$ reaches the ground.\\
$P$ is now projected horizontally with speed $1.95 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ across a smooth horizontal surface. The motion of $P$ is again opposed by a force of magnitude $3 v \mathrm {~N}$, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the speed of $P$.\\
(ii) Calculate the distance $P$ travels after projection before coming to rest.
\hfill \mbox{\textit{CAIE M2 2014 Q6}}