M1 June 2008 Q2
2. At time \(t = 0\), a particle is projected vertically upwards with speed \(u\) m s\(^{-1}\) from a point 10 m above the ground. At time \(T\) seconds, the particle hits the ground with speed 17.5 m s\(^{-1}\). Find
(a) the value of \(u\), (3)
(b) the value of \(T\). (4)
| Scheme | Marks |
|---|---|
| \(v^2 = u^2 + 2as\ \ \Rightarrow\ \ 17.5^2 = u^2 + 2 \times 9.8 \times 10\) | M1 A1 |
| Leading to \(u = 10.5\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(v = u + at\ \ \Rightarrow\ \ 17.5 = -10.5 + 9.8T\) | M1 A1 f.t. |
| \(T = 2\dfrac{6}{7}\ \ (\text{s})\) | DM1 A1 |
| (4) | |
| (7 marks) |
Alternatives for (b)
| \(s = \left(\dfrac{u + v}{2}\right)T\ \Rightarrow\ 10 = \left(\dfrac{17.5 + -10.5}{2}\right)T\) | |
| \(\dfrac{20}{7} = T\) | M1A1 f.t. DM1A1 (4) |
OR
| \(s = ut + \tfrac{1}{2}at^2\ \ \Rightarrow\ \ -10 = 10.5t - 4.9t^2\) | M1 A1 f.t. |
| Leading to \(T = 2\dfrac{6}{7},\ \left(-\dfrac{5}{7}\right)\) Rejecting negative | DM1 A1 (4) |
(b) can be done independently of (a)
| \(s = vt - \tfrac{1}{2}at^2\ \ \Rightarrow\ \ -10 = -17.5t + 4.9t^2\) | M1 A1 |
| Leading to \(T = 2\dfrac{6}{7},\ \dfrac{5}{7}\) | DM1 |
| For final A1, second solution has to be rejected. \(\dfrac{5}{7}\) leads to a negative \(u\). | A1 (4) |