| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Collision/impulse during circular motion |
| Difficulty | Challenging +1.2 This is a multi-part circular motion problem combining energy conservation, collision mechanics, and tension analysis. Part (i) is straightforward energy conservation between two points. Parts (ii)-(iv) involve collision with coalescence and finding the mass ratio, requiring systematic application of standard techniques (conservation of momentum, energy methods, and tension calculation) but no novel insights. The problem is longer than average and requires careful bookkeeping across multiple parts, placing it moderately above typical A-level difficulty. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}m\left(\frac{17ag}{3} - 5ag\right) = mga(\cos\beta - \cos\alpha)\) | M1, A1 | Energy equation A to B |
| \(\cos\alpha = \frac{1}{3}\) | E1 | Correctly shown |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When string goes slack at A: Use \(T=0\) | M1 | May be implied |
| \(Mg\cos\alpha = \frac{Mu^2}{a}\) and so \(u = \sqrt{\frac{ag}{3}}\) | E1 | Correctly shown (N2L at A). Condone use of \(m\) or \(km\) for \(M\ (=(k+1)m)\) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{1}{2}m(v^2 - 5ag) = mg\frac{5a}{3}\) | M1* | Energy equation B (or A) to lowest point |
| \(v^2 = \frac{25ag}{3}\) | A1 | |
| \(\frac{1}{2}M(V^2 - u^2) = Mg\frac{4a}{3}\) | M1* | Energy equation: lowest point to when string goes slack at A. *Condone use of \(m\) or \(km\) for \(M\) \((= (k+1)m)\)* |
| \(V^2 = \frac{8ag}{3} + \frac{ag}{3} = 3ag\) | A1 | *The first 5 marks can be earned in (iv). SC B2 for \(V^2 = 3ag\) seen (or clearly implied) after M0* |
| \((k+1)mV = mv\) | M1* A1 | PCLM at collision |
| \((k+1)\sqrt{3ag} = \sqrt{\frac{25ag}{3}}\) | M1 | Dep M3* Substitute into momentum equation |
| \((k+1)^2 = \frac{25}{9}\) | ||
| \(k = \frac{2}{3}\) | A1 | Must be positive |
| [9] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(T_1 - mg = \frac{mv^2}{a}\) | M1 | N2L at lowest point before collision |
| \(T_1 = \frac{28}{3}mg\) | A1 | |
| \(T_2 - (k+1)mg = \frac{m(k+1)V^2}{a}\) | M1 | N2L at lowest point immediately after collision. *Must use \((k+1)m\) in both terms* |
| \(T_2 = \frac{20}{3}mg\) | ||
| Change in tension is \(\frac{8}{3}mg\) | A1 | CAO |
| [4] |
# Question 4:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}m\left(\frac{17ag}{3} - 5ag\right) = mga(\cos\beta - \cos\alpha)$ | M1, A1 | Energy equation A to B |
| $\cos\alpha = \frac{1}{3}$ | E1 | Correctly shown |
| **[3]** | | |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| When string goes slack at A: Use $T=0$ | M1 | May be implied |
| $Mg\cos\alpha = \frac{Mu^2}{a}$ and so $u = \sqrt{\frac{ag}{3}}$ | E1 | Correctly shown (N2L at A). Condone use of $m$ or $km$ for $M\ (=(k+1)m)$ |
| **[2]** | | |
## Question (iii):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{1}{2}m(v^2 - 5ag) = mg\frac{5a}{3}$ | M1* | Energy equation B (or A) to lowest point |
| $v^2 = \frac{25ag}{3}$ | A1 | |
| $\frac{1}{2}M(V^2 - u^2) = Mg\frac{4a}{3}$ | M1* | Energy equation: lowest point to when string goes slack at A. *Condone use of $m$ or $km$ for $M$ $(= (k+1)m)$* |
| $V^2 = \frac{8ag}{3} + \frac{ag}{3} = 3ag$ | A1 | *The first 5 marks can be earned in (iv). SC B2 for $V^2 = 3ag$ seen (or clearly implied) after M0* |
| $(k+1)mV = mv$ | M1* A1 | PCLM at collision |
| $(k+1)\sqrt{3ag} = \sqrt{\frac{25ag}{3}}$ | M1 | Dep M3* Substitute into momentum equation |
| $(k+1)^2 = \frac{25}{9}$ | | |
| $k = \frac{2}{3}$ | A1 | Must be positive |
| | **[9]** | |
---
## Question (iv):
| Working | Mark | Guidance |
|---------|------|----------|
| $T_1 - mg = \frac{mv^2}{a}$ | M1 | N2L at lowest point before collision |
| $T_1 = \frac{28}{3}mg$ | A1 | |
| $T_2 - (k+1)mg = \frac{m(k+1)V^2}{a}$ | M1 | N2L at lowest point immediately after collision. *Must use $(k+1)m$ in both terms* |
| $T_2 = \frac{20}{3}mg$ | | |
| Change in tension is $\frac{8}{3}mg$ | A1 | CAO |
| | **[4]** | |4 A particle P of mass $m$ is attached to one end of a light inextensible string of length $a$. The other end of the string is attached to a fixed point O . Particle P is projected so that it moves in complete vertical circles with centre O ; there is no air resistance. A and B are two points on the circle, situated on opposite sides of the vertical through O . The lines OA and OB make angles $\alpha$ and $\beta$ with the upward vertical as shown in Fig. 4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{68cbb8bb-2898-4812-a221-6ea5363b0812-5_414_399_434_833}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
The speed of P at A is $\sqrt { \frac { 17 a g } { 3 } }$. The speed of P at B is $\sqrt { 5 a g }$ and $\cos \beta = \frac { 2 } { 3 }$.\\
(i) Show that $\cos \alpha = \frac { 1 } { 3 }$.
On one occasion, when P is at its lowest point and moving in a clockwise direction, it collides with a stationary particle Q . The two particles coalesce and the combined particle continues to move in the same vertical circle. When this combined particle reaches the point A , the string becomes slack.\\
(ii) Show that when the string becomes slack, the speed of the combined particle is $\sqrt { \frac { a g } { 3 } }$.
The mass of the particle Q is $k m$.\\
(iii) Find the value of $k$.\\
(iv) Find, in terms of $m$ and $g$, the instantaneous change in the tension in the string as a result of the collision.
\hfill \mbox{\textit{OCR MEI M3 2016 Q4 [18]}}