M3 June 2013 Q4
4. A particle \(P\) of mass 2 kg is attached to one end of a light elastic string of natural length 1.2 m. The other end of the string is attached to a fixed point \(O\) on a rough horizontal plane. The coefficient of friction between \(P\) and the plane is \(\dfrac{2}{5}\). The particle is held at rest at a point \(B\) on the plane, where \(OB = 1.5\) m. When \(P\) is at \(B\), the tension in the string is 20 N. The particle is released from rest.
The particle comes to rest at the point \(C\).
| Scheme | Marks |
|---|---|
| \(T = \dfrac{\lambda x}{l}\) | |
| \(20 = \dfrac{\lambda \times 0.3}{1.2}\) | M1A1 |
| \(\lambda = 80\) N | A1 |
| Initial EPE \(= \dfrac{\lambda x^2}{2l} = \dfrac{80 \times 0.3^2}{2.4}\ \ (= 3\text{ J})\) | B1 |
| \(\dfrac{80 \times 0.3^2}{2.4} - 0.4 \times 2g \times 0.3 = \dfrac{1}{2} \times 2v^2\) | M1A1ft |
| \(v^2 = 0.648\) | |
| \(v = 0.80\) or 0.805 m s\(^{-1}\) | A1 |
| (7) |
Notes
M1 for attempting Hooke's Law, formula must be correct, either explicitly or by correct substitution.
A1 for \(20 = \dfrac{\lambda \times 0.3}{1.2}\)
A1 for obtaining \(\lambda = 80\)
B1 for the initial EPE \(\dfrac{"\lambda" \times 0.3^2}{2.4}\ \ (= 3\text{ J})\) their value for \(\lambda\) allowed. May only be seen in the eqaution.
M1 for a work-energy equation with one EPE term, one KE term and work done against friction (Award if second EPE/KE terms included provided these become 0). The EPE must be dimensionally correct, but need not be fully correct (eg denominator 1.2 instead of 2.4)
A1ft for a completely correct equation follow through their EPE
A1 cao for \(v = 0.80\) or 0.805 must be 2 or 3 sf
NB: This is damped harmonic motion (due to friction) so all SHM attempts lose the last 4 marks.
| Scheme | Marks |
|---|---|
| Comes to rest \(0.4 \times 2g \times y = 3\) | M1 |
| \(y = \dfrac{3}{0.4 \times 2 \times 9.8} = 0.38\) or 0.383 m | A1 |
| (2) | |
| (9 marks) |
Notes
M1 for any complete method leading to a value for either \(BC\). If the distance travelled after the string becomes slack is found the work must be completed by adding 0.3 Their EPE found in (a) used in energy methods.
MS method is energy from \(B\) to \(C\) ie work done against friction = loss of EPE.
OR Energy from point where the string becomes slack to \(C\) ie work done against friction = loss of KE and completed for the required distance
OR NL2 to obtain the acceleration \(\left(-\dfrac{2g}{5}\right)\) while the string is slack and \(v^2 = u^2 + 2as\) to find the distance and completed for the required distance
A1cso for \(BC = 0.38\) or 0.383 (m) must be 2 or 3 sf
Alternatives:
| Energy from string going slack to rest: | |
| \(\dfrac{1}{2} \times 2 \times 0.648 = 0.4 \times 2g \times x\) | |
| \(x = 0.08265\ldots\) | |
| \(y = 0.3 + 0.08265\ldots = 0.38\) or 0.383 | M1 Complete method A1 |
(Corrected from the printed mark scheme: \(x\) is printed as \(0.8265\ldots\).)
| NL2 to obtain the accel when string is slack \(\left(-\dfrac{2g}{5}\right)\) and \(v^2 = u^2 + 2as\) | |
| \(0 = 0.648 + 2 \times \left(-\dfrac{2g}{5}\right)s\) | |
| \(BC = \dfrac{0.648 \times 5}{4g} + 0.3 = 0.38\) or 0.383 | M1A1 |