M2 January 2010 Q8
8. [In this question \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors in a horizontal and upward vertical direction respectively]
A particle \(P\) is projected from a fixed point \(O\) on horizontal ground with velocity \(u(\mathbf{i} + c\mathbf{j})\) m s\(^{-1}\), where \(c\) and \(u\) are positive constants. The particle moves freely under gravity until it strikes the ground at \(A\), where it immediately comes to rest. Relative to \(O\), the position vector of a point on the path of \(P\) is \((x\mathbf{i} + y\mathbf{j})\) m.
Given that \(u = 7\), \(OA = R\) m and the maximum vertical height of \(P\) above the ground is \(H\) m,
Given also that when \(P\) is at the point \(Q\), the velocity of \(P\) is at right angles to its initial velocity,
| Scheme | Marks |
|---|---|
| \(x = ut\) | B1 |
| \(y = cut - 4.9t^2\) | M1 A1 |
| eliminating \(t\) and simplifying to give \(\ y = cx - \dfrac{4.9x^2}{u^2}\ **\) | DM1 A1 |
| (5) |
| Scheme | Marks |
|---|---|
| (i) \(\ 0 = cx - \dfrac{4.9x^2}{u^2}\) | M1 |
| \(0 = x\left(c - \dfrac{4.9x}{u^2}\right) \Rightarrow R = \dfrac{u^2c}{4.9} = 10c\) | M1 A1 |
| (ii) When \(x = 5c,\ \ y = H\) | M1 |
| \(= 5c^2 - \dfrac{(5c)^2}{10} = 2.5c^2\) | M1 A1 |
| (6) |
| Scheme | Marks |
|---|---|
| \(\dfrac{dy}{dx} = c - \dfrac{9.8x}{u^2} = c - \dfrac{x}{5}\) | M1 A1 |
| When \(x = 0,\ \dfrac{dy}{dx} = c\) | B1 |
| So, \(\ c - \dfrac{x}{5} = \dfrac{-1}{c}\) | DM1 A1 |
| \(x = 5\left(c + \dfrac{1}{c}\right)\) | A1 |
| (6) | |
| (17 marks) |
Alternative to 8(c)
| \(\tan\theta = \dfrac{u}{cu} = \dfrac{1}{c} = \dfrac{v}{u}\) | B1 |
| \(\Rightarrow v = \dfrac{u}{c} = \dfrac{7}{c}\) | M1 A1 |
| \(v = u + at\ ;\quad -\dfrac{7}{c} = 7c - 9.8t\) | M1 |
| \(t = \dfrac{7}{9.8}\left(c + \dfrac{1}{c}\right)\) | A1 |
| \(x = ut = 7t\ ;\quad x = 5\left(c + \dfrac{1}{c}\right)\) | A1 |