M3 June 2016 Q6
6. One end of a light inextensible string of length \(l\) is attached to a particle \(P\) of mass \(2m\). The other end of the string is attached to a fixed point \(A\). The particle is hanging freely at rest with the string vertical. The particle is then projected horizontally with speed \(\sqrt{\dfrac{7gl}{2}}\)
When the string is horizontal and \(P\) is moving upwards, the string comes into contact with a small smooth peg which is fixed at the point \(B\), where \(AB\) is horizontal and \(AB < l\). The particle then describes a complete semicircle with centre \(B\).
| Scheme | Marks |
|---|---|
| Energy to horizontal: \(\dfrac{1}{2} \times 2m \times \dfrac{7gl}{2} - \dfrac{1}{2} \times 2mv^2 = 2mgl\) | M1A1A1 |
| \(v = \sqrt{\dfrac{3gl}{2}}\) | A1 |
| (4) |
Notes
M1 Attempting an energy equation to the horizontal. Must be clear energy is being used and not \(v^2 = u^2 + 2as\). Mass can be \(m\) or \(2m\). Mixed masses are accuracy errors.
A1 Correct difference of KE terms Mass can be \(m\) or \(2m\)
A1 Correct PE and all signs correct in equation. Mass can be \(m\) or \(2m\) but mixed masses score A0.
A1 Correct speed at the horizontal (regardless of mass used).
| Scheme | Marks |
|---|---|
| Energy from horizontal to top: \(\dfrac{1}{2} \times 2m \times \dfrac{3gl}{2} - \dfrac{1}{2} \times 2mV^2 = 2mgr\) | M1A1 |
| \(V^2 = \dfrac{3gl}{2} - 2gr\) | |
| NL2 at top: \(\dfrac{2mV^2}{r} = 2mg + T\) | M1A1A1 |
| \(T \geqslant 0 \Rightarrow \dfrac{2mV^2}{r} \geqslant 2mg\) | M1 |
| \(\dfrac{3gl - 4gr}{2r} \geqslant g\) | DM1 |
| \(\dfrac{3gl}{2} - 2gr \geqslant gr\) | |
| \(r \leqslant \dfrac{1}{2}l\) | A1 |
| \(AB \geqslant \dfrac{1}{2}l\) * | A1cso |
| (9) | |
| (13 marks) |
Notes
M1 Attempt energy equation from the horizontal to the top of the new (smaller) circle with unknown radius OR from lowest point of original circle to top of the new circle. Mass can be \(m\) or \(2m\) or mixed.
A1 Correct equation. Mass can be \(m\) or \(2m\) (but same in all terms).
M1 NL2 at top of the small circle. Mass can be \(m\) or \(2m\) or mixed. Allow with \(T = 0\)
A1 Correct mass x acceleration. Mass can be \(m\) or \(2m\)
A1 Both force terms correct. Mass can be \(m\) or \(2m\) but must be the same as used in the acceleration term.
M1 Use \(T \geqslant 0\) to obtain an inequality for \(V^2\). Allow if \(T\) assumed to be zero in NL2.
DM1 Use the energy equation to eliminate \(V^2\) Dependent on first and second M marks.
These two method steps may occur in the reverse of the order shown here.
A1 Correct maximum value for \(r\)
A1cso Correct inequality for \(AB\). This is cso. Candidates who have used \(m\) in NL2 or assumed \(T = 0\) cannot be awarded this mark.
ALT: Combining lines 3 and 4 of (b):
\(\dfrac{2mV^2}{r} \geqslant 2mg\) scores M1A1A1M1 (All other marks as above.)
NB Equations to/at general positions do not gain marks until correct size of angle used.