M3 June 2017 Q7
7.
The fixed points \(A\) and \(B\) are 4 m apart on a smooth horizontal floor. One end of a light elastic string, of natural length 1.8 m and modulus of elasticity 45 N, is attached to a particle \(P\) and the other end is attached to \(A\). One end of another light elastic string, of natural length 1.2 m and modulus of elasticity 20 N, is attached to \(P\) and the other end is attached to \(B\). The particle \(P\) rests in equilibrium at the point \(O\), where \(AOB\) is a straight line, as shown in Figure 6.
The point \(C\) lies on the straight line \(AOB\) with \(AC = 2.7\) m. The mass of \(P\) is 0.6 kg. The particle \(P\) is held at \(C\) and then released from rest.
The point \(D\) lies on the straight line \(AOB\) with \(AD = 1.8\) m. When \(P\) reaches \(D\) the string \(PB\) breaks.
| Scheme | Marks |
|---|---|
| \(T = \dfrac{45e}{1.8} = \dfrac{20(1 - e)}{1.2} \qquad\) or \(\qquad \dfrac{45(1 - e^{\prime})}{1.8} = \dfrac{20e^{\prime}}{1.2}\) | M1A1 |
| \(e = 0.4 \qquad\) or \(\qquad e^{\prime} = 0.6\) | A1 |
| \(AO = 2.2\) m * | A1cso |
| (4) |
Notes
M1 Use Hooke's law to obtain the tension in each string and equate the 2 tensions. Either extension can be used as the unknown. ALT: Use both extensions and use \(e + e^{\prime} = 1\) later
A1 Correct equation.
A1 Correct result for their choice of unknown.
A1cso Correct completion to the given answer with no errors seen.
ALT:
| \(AO = y\) | |
| \(\dfrac{45(y - 1.8)}{1.8} = \dfrac{20(2.8 - y)}{1.2}\) | M1A1A1 |
| \(54y - 97.2 = 100.8 - 36y\) | |
| \(y = 2.2\) m * | A1cso |
| Scheme | Marks |
|---|---|
| \(0.6\ddot{x} = \dfrac{20(0.6 - x)}{1.2} - \dfrac{45(0.4 + x)}{1.8}\) | M1A1A1 |
| \(0.6\ddot{x} = -\dfrac{125}{3}x\) | |
| \(\ddot{x} = -\dfrac{625}{9}x \quad \therefore\) SHM | dM1A1cso |
| (5) |
Notes
M1 Form an equation of motion with the difference of 2 tensions (from applying Hooke's law) which must have different extensions which both include a variable. Acceleration can be \(a\) or \(\ddot{x}\)
If the variable is measured from \(A\) or \(B\) a substitution is required to obtain the necessary SHM equation. Do not award this mark until this substitution is attempted.
A1A1 Deduct 1 for each error. Difference of tensions the wrong way round counts as one error. Acceleration can be \(a\) or \(\ddot{x}\) but if \(a\) is used it must be in the same direction as \(\ddot{x}\). The numbers in the extensions must be as shown (as SHM cannot be assumed before it is shown and so the numbers must come from the situation and not just chosen so that the constant terms cancel out).
dM1 Solve the equation to \(\pm\ddot{x} = \ldots\) Acceleration \(a\) scores M0 now.
A1cso Correct equation (must be \(\ddot{x} = \ldots\) now) and the conclusion.
| Scheme | Marks |
|---|---|
| \(\omega^2 = \dfrac{625}{9}\) oe | B1 |
| Time from \(C\) to \(D\): \(-0.4 = 0.5\cos\dfrac{25}{3}t\) | M1A1ft |
| \(t = \dfrac{3}{25}\cos^{-1}(-0.8)\) | A1 |
| Speed at \(D\): \(v^2 = \dfrac{625}{9}\left(0.5^2 - 0.4^2\right)\) or use \(v = -a\omega\sin\omega t\) | M1 |
| \(v = \dfrac{25}{3} \times 0.3 = 2.5\) | A1 |
| Time from \(D\) to \(A\) \(= \dfrac{1.8}{2.5}\) | M1 |
| Total time: \(= \dfrac{1.8}{2.5} + \dfrac{3}{25}\cos^{-1}(-0.8) = 1.01977\ldots\) (s) (accept 1.0 or better) | A1 cao |
| (8) | |
| (17 marks) |
Notes
B1 Correct value for \(\omega^2\) or \(\omega\) seen explicitly or used
M1 Obtain the time from \(C\) to \(D\). \(x = a\cos\omega t\) or \(x = a\sin\omega t\) can be used as long as the method is complete. Amplitude to be 0.5, \(x = \pm 0.4\) with their \(\omega\) which must have come from an equation of the form \(\ddot{x} = \pm\omega^2x\)
Sometimes done in two parts: Time \(C\) to \(O\) \(\left(\dfrac{3\pi}{50}\text{ or } 0.1884\ldots\right)\) and \(O\) to \(D\) \((0.111\ldots)\) added now or later.
A1ft Correct equation. Follow through their \(\omega\)
A1 Correct time, as shown or 0.2997...seen now or later
M1 Find the speed of \(P\) as it reaches \(D\) with their \(\omega\).
A1 Correct speed at \(D\)
M1 Use their speed at \(D\) to find the time from \(D\) to \(A\)
A1cao Add the 2 (or 3) times to obtain the required time.