| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Collision/impulse during SHM |
| Difficulty | Challenging +1.8 This is a challenging M3 question combining circular motion, collision mechanics, and SHM in a non-trivial setup. Part (i) is routine circular motion. Part (ii) requires simultaneous equations from conservation of momentum and restitution. Parts (iii-iv) demand recognizing that small-angle SHM applies to a pendulum after collision and calculating the period—this synthesis of multiple mechanics topics with proof elements places it well above average difficulty but within reach of strong A-level students. |
| Spec | 6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([a = 0.7^2/0.4]\) | M1 | For using \(a = v^2/r\) |
| For not more than one error in \(T - 0.8g\cos 60° = 0.8 \times 0.7^2/0.4\) | A1 | |
| Above equation complete and correct | A1 | |
| Tension is 4.9N | A1 | [4 marks total] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}(0.8)v^2 = \frac{1}{2}(0.8)(0.7)^2 + 0.8g(0.4) - 0.8g(0.4)\cos 60°\) | M1 | For using the principle of conservation of energy |
| A1 | \((v = 2.1)\) | |
| \((2.1 - 0)/7 = 2u\) | M1 | For using NEL |
| Q's initial speed is \(0.15\text{ ms}^{-1}\) | A1 | [4 marks total] AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using Newton's second law transversely | |
| \((m)0.4\ddot{\theta} = -(m)g\sin\theta\) | A1 | *Allow \(m = 0.8\) (or any other numerical value) |
| \([0.4\ddot{\theta} \approx -g\theta]\) | M1 | For using \(\sin\theta \approx \theta\) |
| \([\frac{1}{2}m(0.15)^2 = mg(0.4)(1 - \cos\theta_{max}) \rightarrow \theta_{max} = 4.34°\;(0.0758\text{ rad})]\) | M1 | For using the principle of conservation of energy to find \(\theta_{max}\) |
| \(\theta_{max}\) small justifies \(0.4\ddot{\theta} \approx -g\theta\), and this implies SHM | A1 | [5 marks total] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([T = 2\pi/\sqrt{24.5} = 1.269...]\) \([\sqrt{24.5}\;t = \pi]\) | M1 | For using \(T = 2\pi/n\), or for solving either \(\sin nt = 0\) (non-zero t, considering displacement) or \(\cos nt = -1\) (considering velocity) |
| Time interval is 0.635s | A1ft | [2 marks total] From \(t = \frac{1}{2}T\) |
# Question 7:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[a = 0.7^2/0.4]$ | M1 | For using $a = v^2/r$ |
| For not more than one error in $T - 0.8g\cos 60° = 0.8 \times 0.7^2/0.4$ | A1 | |
| Above equation complete and correct | A1 | |
| Tension is 4.9N | A1 | [4 marks total] |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}(0.8)v^2 = \frac{1}{2}(0.8)(0.7)^2 + 0.8g(0.4) - 0.8g(0.4)\cos 60°$ | M1 | For using the principle of conservation of energy |
| | A1 | $(v = 2.1)$ |
| $(2.1 - 0)/7 = 2u$ | M1 | For using NEL |
| Q's initial speed is $0.15\text{ ms}^{-1}$ | A1 | [4 marks total] AG |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using Newton's second law transversely |
| $(m)0.4\ddot{\theta} = -(m)g\sin\theta$ | A1 | *Allow $m = 0.8$ (or any other numerical value) |
| $[0.4\ddot{\theta} \approx -g\theta]$ | M1 | For using $\sin\theta \approx \theta$ |
| $[\frac{1}{2}m(0.15)^2 = mg(0.4)(1 - \cos\theta_{max}) \rightarrow \theta_{max} = 4.34°\;(0.0758\text{ rad})]$ | M1 | For using the principle of conservation of energy to find $\theta_{max}$ |
| $\theta_{max}$ small justifies $0.4\ddot{\theta} \approx -g\theta$, and this implies SHM | A1 | [5 marks total] |
## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[T = 2\pi/\sqrt{24.5} = 1.269...]$ $[\sqrt{24.5}\;t = \pi]$ | M1 | For using $T = 2\pi/n$, or for solving either $\sin nt = 0$ (non-zero t, considering displacement) or $\cos nt = -1$ (considering velocity) |
| Time interval is 0.635s | A1ft | [2 marks total] From $t = \frac{1}{2}T$ |7\\
\includegraphics[max width=\textwidth, alt={}, center]{a04e6d4e-2437-4761-87ee-43e6771fbbd9-4_588_629_274_758}
A particle $P$ of mass 0.8 kg is attached to a fixed point $O$ by a light inextensible string of length 0.4 m . A particle $Q$ is suspended from $O$ by an identical string. With the string $O P$ taut and inclined at $\frac { 1 } { 3 } \pi$ radians to the vertical, $P$ is projected with speed $0.7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a direction perpendicular to the string so as to strike $Q$ directly (see diagram). The coefficient of restitution between $P$ and $Q$ is $\frac { 1 } { 7 }$.\\
(i) Calculate the tension in the string immediately after $P$ is set in motion.\\
(ii) Immediately after $P$ and $Q$ collide they have equal speeds and are moving in opposite directions. Show that $Q$ starts to move with speed $0.15 \mathrm {~ms} ^ { - 1 }$.\\
(iii) Prove that before the second collision between $P$ and $Q , Q$ is moving with approximate simple harmonic motion.\\
(iv) Hence find the time interval between the first and second collisions of $P$ and $Q$.
\hfill \mbox{\textit{OCR M3 2007 Q7 [15]}}