M3 June 2014 (R) Q1
1. A particle \(P\) of mass 0.25 kg is moving along the positive \(x\)-axis under the action of a single force. At time \(t\) seconds \(P\) is \(x\) metres from the origin \(O\) and is moving away from \(O\) with speed \(v\) m s\(^{-1}\) where \(\dfrac{\mathrm{d}v}{\mathrm{d}x} = 3\). It is given that \(x = 2\) and \(v = 3\) when \(t = 0\)
(a) Find the magnitude of the force acting on \(P\) when \(x = 5\) (4)
(b) Find the value of \(t\) when \(x = 5\) (4)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}v}{\mathrm{d}x} = 3 \Rightarrow v = 3x - 3\) | M1 A1 |
| \(a = 3(3x - 3)\) | DM1 |
| When \(x = 5\), \(F = 0.25 \times 3(15 - 3) = 9\) N | A1 |
| (4) |
Notes
M1 Integration
A1 correct integration
DM1 using \(a = v\,\mathrm{d}v/\mathrm{d}x\) with their \(v\)
A1 correct integration
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 3(x - 1)\) | M1 |
| \(\displaystyle\int_2^5 \frac{\mathrm{d}x}{(x - 1)} = \int_0^t 3\,\mathrm{d}t\) | A1 |
| \(\left[\ln(x - 1)\right]_2^5 = 3t\) | DM1 |
| \(t = \dfrac{1}{3}\ln 4 = 0.4620\ldots\) | A1 |
| (4) | |
| (8 marks) |
Notes
M1 using \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 3(x - 1)\)
A1 correct integrals with correct limits
DM1 Substitute the limits
A1 correct final answer