M3 June 2013 Q2
2. A particle \(P\) of mass 0.5 kg is moving along the positive \(x\)-axis in the positive \(x\)-direction. The only force on \(P\) is a force of magnitude \(\left(2t + \dfrac{1}{2}\right)\) N acting in the direction of \(x\) increasing, where \(t\) seconds is the time after \(P\) leaves the origin \(O\). When \(t = 0\), \(P\) is at rest at \(O\).
The particle passes through the point \(A\) with speed 6 m s\(^{-1}\).
| Scheme | Marks |
|---|---|
| \(\left(2t + \dfrac{1}{2}\right) = 0.5\dfrac{\mathrm{d}v}{\mathrm{d}t}\) | M1 |
| \(\displaystyle\int (4t + 1)\,\mathrm{d}t = \int \mathrm{d}v\) | |
| \(2t^2 + t = v + c\) | M1dep \(c\) not needed |
| \(t = 0\ \ v = 0\ \ \ \ \ \ c = 0\) | |
| \(v = 2t^2 + t\) m s\(^{-1}\) | A1 inc the value for \(c\) |
| (3) |
Notes
M1 for NL2 with acceleration in the form \(\dfrac{\mathrm{d}v}{\mathrm{d}t}\), seen explicitly or implied by the integration mass can be 0.5 or \(m\)
M1dep for integrating with respect to \(t\) - constant not needed
A1cso for showing that \(c = 0\) and giving the final result \(v = 2t^2 + t\) Must see \(t = 0\), \(v = 0\) as a minimum
By definite integration:
M1 as above
M1dep for integrating, ignore limits
A1 for substituting the limits 0 and \(v\) and 0 and \(t\) and obtaining \(v = 2t^2 + t\)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 2t^2 + t\) | |
| \(x = \dfrac{2}{3}t^3 + \dfrac{1}{2}t^2 + k\) | M1 |
| \(t = 0\ \ x = 0\ \ \ \ \ k = 0\) | |
| \(x = \dfrac{2}{3}t^3 + \dfrac{1}{2}t^2\) | A1 |
| \(v = 6\ \ \ 6 = 2t^2 + t\ \ \ 2t^2 + t - 6 = 0\) | |
| \((2t - 3)(t + 2) = 0\ \ \ \ t = \dfrac{3}{2}\) | M1A1 |
| \(x = \dfrac{2}{3} \times \left(\dfrac{3}{2}\right)^3 + \dfrac{1}{2}\left(\dfrac{3}{2}\right)^2\) | M1dep |
| \(x = \dfrac{27}{8}\) (oe 3.4, 3.375, 3.38) m | A1 cso |
| (6) | |
| (9 marks) |
Notes
M1 for integrating their \(v\) with respect to \(t\) constant not needed
A1 for showing that \(k = 0\) If no constant shown this mark is lost.
M1 for setting \(v = 6\) using their answer from (a) and attempting to solve the resulting quadratic equation, any valid method. If solved by calculator, both solutions must be shown.
A1 for \(t = \dfrac{3}{2}\) negative solution need not be shown with an algebraic solution
M1dep for using their (positive) value for \(t\) to obtain \(x = \ldots\). If two positive values were obtained, then allow M1 for substituting either value. Dependent on the first M1 of (b) but not the second.
A1cso for \(x = \dfrac{27}{8}\) (oe eg 3.375, 3.38) (All marks for (b) must have been awarded)
By definite integration:
M1 for integrating their \(v\) with respect to \(t\) limits not needed
A1 for correct integration with lower limits 0.
M1 for setting \(v = 6\) using their answer from (a) and attempting to solve the resulting quadratic equation, any valid method. If solved by calculator, both solutions must be shown.
A1 for \(t = \dfrac{3}{2}\) negative solution need not be shown with an algebraic solution
M1dep for substituting their limits into their integrated \(v\) (sub should be shown). Dependent on the first M1 of (b) but not the second
A1cso for \(x = \dfrac{27}{8}\) (oe eg 3.375, 3.38)