| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2004 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle on table with string above |
| Difficulty | Standard +0.3 This is a standard conical pendulum problem with two straightforward parts. Part (i) requires resolving forces and applying circular motion (T cos θ = mg, T sin θ = mv²/r) with given angle. Part (ii) adds a normal reaction force but follows the same method. The calculations are routine for M2 level with no conceptual surprises, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors3.03i Normal reaction force6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T\cos 60° = 0.5g\) \((T = 10)\) | B1 | |
| For applying Newton's 2nd law horizontally and using \(a = \dfrac{v^2}{r}\) (must be a component of \(T\) for M1) | M1 | |
| \(T\sin 60° = \dfrac{0.5v^2}{0.15\sin 60°}\) (for an equation in \(V^2\)) | A1 | |
| For substituting for \(T\) | M1 | |
| \(v = 1.5\) | A1 | 5 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = \dfrac{v^2}{0.15\sin 60°}\) | B1 | |
| For applying Newton's 2nd law perpendicular to the string | M1 | |
| \(0.5g\cos 30° = 0.5(a\cos 60°)\) | A1 | |
| For substituting for \(a\) | M1 | |
| \((5\cos 30° = 0.5\ v^2/0.15\tan 60°)\) (for an equation in \(V^2\)) | ||
| \(v = 1.5\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T\sin 45° = \dfrac{0.5(0.9)^2}{0.15\sin 45°}\) | B1 | |
| Tension is \(5.4\) N | B1 | 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For resolving forces vertically | M1 | |
| \(5.4\cos 45° + R = 0.5g\) | A1 ft | |
| Force is \(1.18\) N | A1 | 3 marks total |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T\cos 60° = 0.5g$ $(T = 10)$ | B1 | |
| For applying Newton's 2nd law horizontally and using $a = \dfrac{v^2}{r}$ (must be a component of $T$ for M1) | M1 | |
| $T\sin 60° = \dfrac{0.5v^2}{0.15\sin 60°}$ (for an equation in $V^2$) | A1 | |
| For substituting for $T$ | M1 | |
| $v = 1.5$ | A1 | **5 marks total** |
**Alternative Method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = \dfrac{v^2}{0.15\sin 60°}$ | B1 | |
| For applying Newton's 2nd law perpendicular to the string | M1 | |
| $0.5g\cos 30° = 0.5(a\cos 60°)$ | A1 | |
| For substituting for $a$ | M1 | |
| $(5\cos 30° = 0.5\ v^2/0.15\tan 60°)$ (for an equation in $V^2$) | | |
| $v = 1.5$ | A1 | |
### Part (ii)(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T\sin 45° = \dfrac{0.5(0.9)^2}{0.15\sin 45°}$ | B1 | |
| Tension is $5.4$ N | B1 | **2 marks total** |
### Part (ii)(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For resolving forces vertically | M1 | |
| $5.4\cos 45° + R = 0.5g$ | A1 ft | |
| Force is $1.18$ N | A1 | **3 marks total** |7 One end of a light inextensible string of length 0.15 m is attached to a fixed point which is above a smooth horizontal surface. A particle of mass 0.5 kg is attached to the other end of the string. The particle moves with constant speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a horizontal circle, with the string taut and making an angle of $\theta ^ { \circ }$ with the downward vertical.\\
(i) Given that $\theta = 60$ and that the particle is not in contact with the surface, find $v$.\\
(ii) Given instead that $\theta = 45$ and $v = 0.9$, and that the particle is in contact with the surface, find
\begin{enumerate}[label=(\alph*)]
\item the tension in the string,
\item the force exerted by the surface on the particle.
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2004 Q7 [10]}}