M1 June 2015 Q2
2. A small stone is projected vertically upwards from a point \(O\) with a speed of 19.6 m s\(^{-1}\). Modelling the stone as a particle moving freely under gravity,
| Scheme | Marks |
|---|---|
| \(0^2 = 19.6^2 - 2 \times gH\) | M1 |
| \(H = 19.6\) m (20) | A1 |
| (2) |
Notes
M1 is for a complete method (which could involve use of two suvat equations) for finding \(H\) i.e. for an equation in \(H\) only, condone sign errors
A1 for 19.6 or 20 correctly obtained (\(2g\) is A0)
| Scheme | Marks |
|---|---|
| \(14.7 = 19.6t - \tfrac{1}{2}gt^2\) | M1 A1 |
| \(t^2 - 4t + 3 = 0\) | |
| \((t - 1)(t - 3) = 0\) | DM1 |
| \(t = 1\) or 3; Answer 2 s | A1; A1 |
| (5) | |
| (7 marks) |
Notes
First M1 is for a quadratic equation in \(t\) only (where \(t\) is time at 14.7 above \(O\))
First A1 for a correct equation
Second DM1, dependent on first M1, for solving for \(t\)
Second A1 for both values of \(t\), 1 and 3.
N.B. If answer(s) are wrong or have come from an incorrect quadratic, and the quadratic formula has been used, M1 can only be awarded if there is clear evidence that the correct formula has been used. If their expression is not correct for their quadratic, allow a slip but only if we see an attempt to substitute into a stated correct formula.
Third A1 for 2 s
N.B. Obtaining \(t = 1\) at \(s = 14.7\) (above \(O\)) only, can score max M1 A1
2(b) ALT 1
| (their \(h\) − 14.7) \(= \frac{1}{2}gt^2\), \(\ t = 1\) OR \(v^2 = 19.6^2 - 2g \times 14.7 \Rightarrow v = (\pm)\,9.8\) and \(0 = 9.8 - 9.8t \Rightarrow t = 1\) | M1 A1 A1 |
| Total \(= 2 \times\) their 1 \(= 2\) s | DM 1 A1 |
(Corrected from the printed mark scheme: the “=” is missing from \(0 = 9.8 - 9.8t\).)
2(b) ALT 2/3
| \(v^2 = 19.6^2 - 2g \times 14.7\) | M1 |
| \(v = \pm 9.8\) | A1 |
| EITHER: \(-9.8 = 9.8 - gT\), \(\ T = 2\) | DM1 A1 A1 |
| OR: \(0 = 9.8t - \frac{1}{2}gt^2\), \(\ t = (0)\) or 2 | DM1 A1 A1 |