M3 June 2016 Q1
1. A particle \(P\) of mass 0.5 kg is moving along the positive \(x\)-axis under the action of a resultant force. The force acts along the \(x\)-axis. At time \(t\) seconds, \(P\) is \(x\) metres from the origin \(O\) and is moving away from \(O\) in the positive \(x\) direction with speed \(\dfrac{12}{x + 3}\) m s\(^{-1}\)
Given that \(x = 4\) when \(t = 2\)
| Scheme | Marks |
|---|---|
| \(v = \dfrac{12}{x + 3}\) | |
| \(\dfrac{\mathrm{d}v}{\mathrm{d}x} = -\dfrac{12}{(x + 3)^2}\) | M1 |
| \(F = 0.5v\dfrac{\mathrm{d}v}{\mathrm{d}x} = 0.5 \times \dfrac{12}{x + 3} \times -\dfrac{12}{(x + 3)^2}\) | DM1A1 |
| \(x = 3 \quad |F| = 0.5 \times \dfrac{12}{6} \times \dfrac{12}{6^2} = \dfrac{1}{3}\) N | A1 |
| (4) |
Notes
M1 Attempt differentiation of \(v = \dfrac{12}{x + 3}\) or \(\dfrac{1}{2}v^2 = \dfrac{72}{(x + 3)^2}\) wrt \(x\) \((x + 3)^{-2}\) or \((x + 3)^{-3}\) (oe) must be seen. Both sides of the equation must be differentiated wrt \(x\)
DM1 Use NL2 with accel \(v\dfrac{\mathrm{d}v}{\mathrm{d}x}\) as obtained above. Must include mass. Dependent on the first M mark.
A1 Correct expression for \(F\) with correct mass and correct acceleration seen here or before use in NL2
A1 Use \(x = 3\) to obtain the correct magnitude, \(\dfrac{1}{3}\), 0.33 or better Must be positive
| Scheme | Marks |
|---|---|
| \(\displaystyle\int (x + 3)\,\mathrm{d}x = \int 12\,\mathrm{d}t\) | |
| \(\dfrac{1}{2}x^2 + 3x = 12t\ \ (+c)\) | M1A1 |
| \(x = 4,\ t = 2 \quad 8 + 12 = 24 + c \quad c = -4\) | DM1 |
| \(x = 10 \quad 50 + 30 = 12t - 4\) | DM1 |
| \(t = 7\) | A1cao |
| (5) | |
| (9 marks) |
Notes
M1 Use \(v = \dfrac{\mathrm{d}x}{\mathrm{d}t}\) and attempt the integration
A1 Correct integration constant of integration not needed
DM1 Use given values to obtain a value for \(c\). Dependent on first M mark
DM1 Use \(x = 10\) to obtain a linear equation for \(t\). Dependent on the first but not the second M mark
A1 cao \(t = 7\)
ALT (b)
| Definite integration: | |
| \(\displaystyle\int_2^T 12\,\mathrm{d}t = \int_4^{10} (x + 3)\,\mathrm{d}x\) | |
| \(12(T - 2) = \left[\dfrac{x^2}{2} + 3x\right]_4^{10}\) | M1A1 as main scheme - limits not needed DM1 Correct limits shown |
| \(12(T - 2) = 80 - 20,\ \ T = 7\) | DM1 Substitute limits, A1 \(T = 7\) |