M3 June 2018 Q7
7. A particle \(P\) of mass 0.5 kg is attached to one end of a light elastic string. The string has natural length \(l\) metres and modulus of elasticity 29.4 N. The other end of the string is attached to a fixed point \(A\). The particle hangs freely in equilibrium at the point \(B\), where \(B\) is vertically below \(A\) and \(AB = 1.4\) m.
The point \(C\) is vertically below \(A\) and \(AC = 1.8\) m. The particle is pulled down to \(C\) and released from rest.
The particle first comes to instantaneous rest at the point \(D\).
| Scheme | Marks |
|---|---|
| \(0.5g = \dfrac{29.4 \times (1.4 - l)}{l} \qquad\) OR \(\quad 0.5g = \dfrac{29.4 \times x}{l} \quad x = \dfrac{l}{6} \quad \dfrac{7l}{6} = 1.4\) | M1A1 |
| \(l = 1.2\) * | A1cso |
| (3) |
Notes
M1 Use Hooke's Law to find the extension at \(B\)
A1 Correct equation
A1 Obtain given value for \(l\) with no errors seen
| Scheme | Marks |
|---|---|
| \(0.5g - T = 0.5\ddot{x}\) | |
| \(0.5g - \dfrac{29.4(x + 0.2)}{1.2} = 0.5\ddot{x}\) | M1A1 |
| \(\ddot{x} = -\dfrac{29.4}{1.2 \times 0.5}x \qquad \ddot{x} = -49x \qquad \therefore\) SHM | dM1A1 |
| (4) |
Notes
M1 Attempt an equation of motion, using Hooke's law for the tension when extension is \(x + 0.2\) \(m\) or 0.5 allowed Acceleration can be \(\ddot{x}\) or \(a\)
A1 Fully correct equation \(m\) or 0.5 allowed Acceleration can be \(\ddot{x}\) or \(a\) but if \(a\) used the direction must be consistent with the direction for \(\ddot{x}\)
dM1 Re-arrange to the form \(\ddot{x} = (\pm)\omega^2x\) Must be \(\ddot{x}\) now and probably will have 0.5 for mass.
Depends on the first M mark
A1 Correct equation and conclusion stated.
| Scheme | Marks |
|---|---|
| \(v^2 = 49\left(0.4^2 - (\pm 0.2)^2\right)\) | M1A1ft |
| \(v = 2.42487\ldots = 2.4\) or 2.42 ms\(^{-1}\) | A1 |
| (3) |
Notes
M1 Use \(v^2 = \omega^2\left(a^2 - x^2\right)\) with their \(\omega^2\), obtained from a "correct" equation ie \(\ddot{x}\) or \(a = -\omega^2x\) and \(x = \pm 0.2\) amp \(= 0.4\)
A1ft Correct equation, follow through their \(\omega\)
A1 Correct speed at instant the string becomes slack.
Must be 2 or 3 significant figures as value used for \(g\) in (a)
NB Can be solved using energy.
M1 Energy equation with an EPE term, a (final) KE term and a GPE term, all with the correct dimensions.
A1 All terms correct (No follow through on this method)
A1 Correct speed at instant the string becomes slack.
Must be 2 or 3 significant figures as value used for \(g\) in (a)
| Scheme | Marks |
|---|---|
| Motion under gravity: \(0 = \) "\(7\sqrt{0.4^2 - 0.2^2}\)" \(-\ gt\) | M1A1ft |
| \(t = \left(7\sqrt{0.4^2 - 0.2^2}\right) \div g = 0.24743\ldots\) | A1 |
| SHM: \(-0.2 = 0.4\cos 7t\) | M1 |
| \(t = \dfrac{1}{7}\cos^{-1}(-0.5) = 0.29919\ldots\) | dM1A1 |
| Total time \(= 0.29919\ldots + 0.24743\ldots = 0.5466\ldots = 0.55\) or 0.547 s | A1 cao |
| (7) | |
| (17 marks) |
Notes
M1 Use SUVAT for the motion under gravity with their speed from (c) to find the time from \(D\) to the string becoming taut again. The exact method chosen must be complete.
A1ft Correct numbers used, follow through their speed from (c)
A1 Correct time, shown explicitly or implied by correct final answer. Need not be 2 or 3 sf as not a demanded answer.
M1 \((\pm)0.2 = 0.4\cos\)"7"\(t\) or \(0.4\sin\)"7"\(t\) with their \(\omega\) from a “correct” equation (see (c))
dM1 Solve for the time to \(C\). Must be radians. Can be implied by a correct final answer.
The method here must be complete. Depends on the previous M mark.
A1 Correct time. Can be implied by a correct final answer.
A1 Add the times for the 2 parts of the motion to obtain the correct total time. Must be 2 or 3 significant figures.