M1 June 2012 Q5
5. A particle \(P\) is projected vertically upwards from a point \(A\) with speed \(u\) m s\(^{-1}\). The point \(A\) is 17.5 m above horizontal ground. The particle \(P\) moves freely under gravity until it reaches the ground with speed 28 m s\(^{-1}\).
At time \(t\) seconds after projection, \(P\) is 19 m above \(A\).
The ground is soft and, after \(P\) reaches the ground, \(P\) sinks vertically downwards into the ground before coming to rest. The mass of \(P\) is 4 kg and the ground is assumed to exert a constant resistive force of magnitude 5000 N on \(P\).
| Scheme | Marks |
|---|---|
| \(v^2 = u^2 + 2as\ \Rightarrow\ 28^2 = u^2 + 2 \times 9.8 \times 17.5\) | M1 A1 |
| Leading to \(u = 21\) \(\ast\) cso | A1 |
| (3) |
Notes
First M1 for a complete method for finding \(u\) e.g.
\(28^2 = u^2 + 2g \times 17.5\)
or \(28^2 = u^2 + 2(-g) \times (-17.5)\)
or \(28^2 = 2gs \Rightarrow s = 40\) then \(0^2 = u^2 + 2(-g) \times (22.5)\)
condone sign errors
First A1 for a correct equation(s) with \(g = 9.8\)
Second A1 for “\(u = 21\)” PRINTED ANSWER
N.B. Allow a verification method, but they must state, as a conclusion, that “\(u = 21\)”, to score the final A1.
| Scheme | Marks |
|---|---|
| \(s = ut + \dfrac{1}{2}at^2\ \Rightarrow\ 19 = 21t - 4.9t^2\) | M1 A1 |
| \(4.9t^2 - 21t + 19 = 0\) \(t = \dfrac{21 \pm \sqrt{21^2 - 4 \times 4.9 \times 19}}{9.8}\) \(t = 2.99\) or 3.0 \(t = 1.30\) or 1.3 | DM1 A1 A1 |
| (5) |
Notes
First M1 for a complete method for finding at least one \(t\) value i.e. for producing an equation in \(t\) only. (condone sign errors but not missing terms)
First A1 for a correct quadratic equation in \(t\) only or TWO correct linear equations in \(t\) only.
Second DM1, dependent on first M1, for attempt to solve the quadratic or one of the linear equations.
Second A1 for 3.0 or 3 or 2.99
Third A1 for 1.3 or 1.30
| Scheme | Marks |
|---|---|
| N2L \(4g - 5000 = 4a\) \((a = -1240.2)\) | M1 A1 |
| \(v^2 = u^2 + 2as\ \Rightarrow\ 0^2 = 28^2 - 2 \times 1240.2 \times s\) Leading to \(s = 0.316\) (m) or 0.32 | M1 A1 |
| (4) | |
| (12 marks) |
Notes
First M1 for resolving vertically with usual rules.
First A1 for a correct equation
Second M1 for use of \(v^2 = u^2 + 2as\), with \(v = 0\), \(u = 28\) or \(u = 0\) and \(v = 28\) and their \(a\), (or any other complete method which produces an equation in \(s\), which could be negative)
M0 if they haven’t calculated a value of \(a\).
Second A1 for 0.32 or 0.316. (must be positive since it’s a distance)
OR
| Work-Energy: \(\tfrac{1}{2} \times 4 \times 28^2 + 4gs = 5000s\) | M1 A1 |
| \(s = 0.316\) or 0.32 | M1 A1 |