M1 June 2014 (R) Q4
4. At time \(t = 0\), a particle is projected vertically upwards with speed \(u\) from a point \(A\). The particle moves freely under gravity. At time \(T\) the particle is at its maximum height \(H\) above \(A\).
The point \(A\) is at a height \(3H\) above the ground.
| Scheme | Marks |
|---|---|
| Max ht \(v = 0\). \(\ v = u - gt \Rightarrow T = \dfrac{u}{g}\) | M1A1 |
| (2) |
Notes
In this question, condone sign errors in a suvat equation for the M mark, but a missing term is M0 or an incorrect term is M0. An incorrect suvat formula is M0
Allow use of symmetry of motion.
e.g. in (a), using \(v = u + at\), either \(0 = u - gT\) or \(u = 0 + gT\)
M1 for use of suvat to obtain an equation in \(T\), \(u\) and \(g\) only.
A1 for \(T = u/g\) correctly obtained.
| Scheme | Marks |
|---|---|
| Max ht \(H = ut + \dfrac{1}{2}at^2 = \dfrac{u^2}{g} - \dfrac{u^2}{2g} = \dfrac{u^2}{2g}\) * Given answer* | M1A1 |
| Or use of \(v^2 = u^2 + 2as\) | |
| (2) |
Notes
In this question, condone sign errors in a suvat equation for the M mark, but a missing term is M0 or an incorrect term is M0. An incorrect suvat formula is M0
Allow use of symmetry of motion.
M1 for use of suvat to obtain an equation in \(H\), \(u\) and \(g\) only.
A1 for \(H = u^2/2g\) correctly obtained (given answer)
| Scheme | Marks |
|---|---|
| \(-3 \times \dfrac{u^2}{2g} = ut - \dfrac{1}{2}gt^2\) | M1 |
| \(-3u^2 = 2ugt - g^2t^2\) | |
| \(g^2t^2 - 2ugt - 3u^2 = 0,\quad gt = \dfrac{2u \pm \sqrt{4u^2 + 12u^2}}{2}\) | DM1 A1 |
| \(t = \dfrac{3u}{g} = 3T\) | A1 |
| (4) | |
| (8 marks) |
Notes
In this question, condone sign errors in a suvat equation for the M mark, but a missing term is M0 or an incorrect term is M0. An incorrect suvat formula is M0
Allow use of symmetry of motion.
Question 4(c) Watch out for t / T confusion (N.B. if only \(T\)’s used, M0DM0)
First M1 for a complete method to find the total time in terms of \(u\), \(g\), \(H\) or \(T\):-
either: \(3H = -ut + \frac{1}{2}gt^2\)
or: \(4H = \frac{1}{2}gt^2\) and \(t + T\)
or: \(v^2 = u^2 + 6gH\) and \(v = -u + gt\), with \(v\) eliminated
Second M1, dependent on first M1, for producing an expression, in terms of \(u\), \(g\), \(H\) or \(T\), for the total time, by solving a quadratic
First A1 for any correct expression for the total time in terms of \(u\), \(g\), \(H\) or \(T\).
Second A1 for \(3T\) cso
(c) alt
| \(-4H = -\dfrac{1}{2}gt^2\) | M1 |
| Total time \(= T + \sqrt{\dfrac{8H}{g}} = T + \sqrt{\dfrac{8u^2}{2g^2}}\) | DM1A1 |
| \(= T + 2T = 3T\) | A1 |