M3 June 2016 Q7
7. A particle \(P\) of mass 0.5 kg is attached to one end of a light elastic spring, of natural length 1.2 m and modulus of elasticity 15 N. The other end of the spring is attached to a fixed point \(A\) on a smooth horizontal table. The particle is placed on the table at the point \(B\) where \(AB = 1.2\) m. The particle is pulled away from \(B\) to the point \(C\), where \(ABC\) is a straight line and \(BC = 0.8\) m, and is then released from rest.
The point \(D\) is the midpoint of \(AB\).
When \(P\) first comes to instantaneous rest a particle \(Q\) of mass 0.3 kg is placed at \(B\). When \(P\) reaches \(B\) again, \(P\) strikes and adheres to \(Q\) to form a single particle \(R\).
| Scheme | Marks |
|---|---|
| \(T = -m\ddot{x}\) | M1 |
| \(\dfrac{15x}{1.2} = -0.5\ddot{x}\) | M1A1 |
| (i) \(\ddot{x} = -25x \qquad \therefore\) SHM | A1cso |
| (ii) Period \(= \dfrac{2\pi}{5}\) (= 1.256 Accept 1.3 or better) | B1ft |
| (5) |
Notes
M1 Using NL2 with \(T\) for tension, acceleration \(a\) or \(\ddot{x}\)
M1 Using HL to obtain an equation connecting \(\ddot{x}\) or \(a\) and \(x\)
A1 A correct equation - any equivalent to that shown - must have \(\ddot{x}\) now.
(i)A1 \(\ddot{x} = -25x\) and stating SHM
These 4 marks are available without substituting for any or all of \(m\), \(\lambda\) or \(l\)
(ii)B1ft Period \(= \dfrac{2\pi}{\text{their numerical }\omega}\) or decimal equivalent
| Scheme | Marks |
|---|---|
| \(v^2 = \omega^2\left(a^2 - x^2\right)\) | |
| \(x = 0 \quad v = 5 \times 0.8 \quad v = 4\) m s\(^{-1}\) | M1A1 |
| (2) |
Notes
M1 Using \(v^2 = \omega^2\left(a^2 - x^2\right)\) with \(x = 0\) (or just \(v = a\omega\)) and \(a = 0.8\), their \(\omega\)
OR: using \(v = -a\omega\sin\omega t\) with \(t = \dfrac{1}{4} \times\) their period, \(a = 0.8\), their \(\omega\)
A1 Correct value for \(v\)
ALT: Using energy: \(\dfrac{1}{2} \times 0.5v^2 = \dfrac{15 \times 0.8^2}{2 \times 1.2}\) M1 \(v = 4\) A1
| Scheme | Marks |
|---|---|
| \(x = a\cos\omega t\) | |
| \(x = -0.6 \qquad -0.6 = 0.8\cos 5t\) | M1A1ft |
| \(t = \dfrac{1}{5}\cos^{-1}\left(-\dfrac{6}{8}\right) = 0.4837\ldots\) s accept 0.48 or better | A1cso |
| (3) |
Notes
M1 Using \(x = a\cos\omega t\) with their \(\omega\) and \(a = 0.8\), \(x = \pm 0.6\) ( or any other complete method)
Use of \(x = a\sin\omega t\) requires further work to complete the method.
A1ft Correct equation follow through their \(\omega\)
A1cso \(t = 0.48\) or better.
| Scheme | Marks |
|---|---|
| \(T = -m\ddot{y}\) | |
| \(\dfrac{15y}{1.2} = -0.8\ddot{y}\) | M1A1 |
| \(\ddot{y} = -15.625y \quad\) or \(\quad \ddot{y} = -\dfrac{125}{8}y \quad \therefore\) SHM | A1 |
| (3) |
Notes
M1 Using NL2 with tension at the new extension and increased mass (any variable inc \(x\) for extension), acceleration in differential form or just \(a\)
A1 Correct equation, any equivalent form, acceleration in differential form or \(a\)
A1 \(\ddot{y} = -15.625y\) and stating SHM (unless already penalised in (a)) must have differential form for acceleration
| Scheme | Marks |
|---|---|
| Con of mom: \(0.5 \times 4 = (0.5 + 0.3)V\) | M1 |
| \(V = \dfrac{2}{0.8} = 2.5\) m s\(^{-1}\) | A1 |
| \((2.5)^2 = 15.625a^2\) | DM1 |
| \(a^2 = \dfrac{2.5^2}{15.625}\) | |
| \(a = 0.6324\ldots\) m accept 0.63 or better or \(a = \dfrac{\sqrt{10}}{5}\) oe | A1 cso |
| (4) | |
| (17 marks) |
Notes
M1 Using the conservation of momentum equation with their speed of \(P\) at \(B\) (see ans to (b)
A1 Correct speed for \(R\) at \(B\) (no ft)
Momentum equation often seen in (d). Marks can be awarded if the result is seen in (e)
DM1 Using \(v^2 = \omega^2\left(a^2 - y^2\right)\) with \(y = 0\) (or just \(v = a\omega\)) with their speed for \(R\) and their \(\omega\)
Dependent on the first M mark of (e)
A1cso \(a = 0.63\) or better or exact answer