M2 January 2008 Q6
6.
[In this question, the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are in a vertical plane, \(\mathbf{i}\) being horizontal and \(\mathbf{j}\) being vertical.]
A particle \(P\) is projected from the point \(A\) which has position vector \(47.5\mathbf{j}\) metres with respect to a fixed origin \(O\). The velocity of projection of \(P\) is \((2u\mathbf{i} + 5u\mathbf{j})\) m s\(^{-1}\). The particle moves freely under gravity passing through the point \(B\) with position vector \(30\mathbf{i}\) metres, as shown in Figure 3.
(a) Show that the time taken for \(P\) to move from \(A\) to \(B\) is 5 s. (6)
(b) Find the value of \(u\). (2)
(c) Find the speed of \(P\) at \(B\). (5)
| Scheme | Marks |
|---|---|
| \(\rightarrow\) \(30 = 2ut\) | B1 |
| \(\uparrow\) \(-47.5 = 5ut - 4.9t^2\) | M1 A1 |
| \(-47.5 = 75 - 4.9t^2\) eliminating \(u\) or \(t\) | DM1 |
| \(t^2 = \dfrac{75 + 47.5}{4.9}\ (= 25)\) | DM1 |
| \(t = 5\) * cso | A1 |
| (6) |
| Scheme | Marks |
|---|---|
| \(30 = 2ut \ \Rightarrow\ 30 = 10u \ \Rightarrow\ u = 3\) | M1 A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\uparrow\) \(\dot{y} = 5u - 9.8t = -34\) M1 requires both \(\dot{x}\) and \(\dot{y}\) | M1 A1 |
| \(\rightarrow\) \(\dot{x} = 2u = 6\) | A1 |
| \(v^2 = 6^2 + (-34)^2\) | DM1 |
| \(v \approx 34.5\) (m s\(^{-1}\)) accept 35 | A1 |
| (5) | |
| (13 marks) |
Alternative to (c)
| \(\tfrac{1}{2}mv_B^2 - \tfrac{1}{2}mv_A^2 = m \times g \times 47.5\) with \(v_A^2 = 6^2 + 15^2 = 261\) | M1 A(2,1,0) |
| \(v_B^2 = 261 + 2 \times 9.8 \times 47.5\ \ (= 1192)\) | DM1 |
| \(v_B \approx 34.5\) (m s\(^{-1}\)) accept 35 | A1 |
(5)
BEWARE : Watch out for incorrect use of \(v^2 = u^2 + 2as\)