M2 January 2007 Q7
7.
A particle \(P\) is projected from a point \(A\) with speed \(u\) m s\(^{-1}\) at an angle of elevation \(\theta\), where \(\cos\theta = \tfrac{4}{5}\). The point \(B\), on horizontal ground, is vertically below \(A\) and \(AB = 45\) m. After projection, \(P\) moves freely under gravity passing through a point \(C\), 30 m above the ground, before striking the ground at the point \(D\), as shown in Figure 3.
Given that \(P\) passes through \(C\) with speed 24.5 m s\(^{-1}\),
(a) using conservation of energy, or otherwise, show that \(u = 17.5\), (4)
(b) find the size of the angle which the velocity of \(P\) makes with the horizontal as \(P\) passes through \(C\), (3)
(c) find the distance \(BD\). (7)
| Scheme | Marks |
|---|---|
| Energy \(\dfrac{1}{2}m(24.5^2 - u^2) = mg \times 15\) | M1 A1=A1 |
| \(u^2 = 24.5^2 - 30g = 306.25\) | |
| \(u = \sqrt{306.25} = 17.5\ \ *\) cso | A1 |
| (4) |
Alternative for (a)
| \(\rightarrow u_x = u\cos\theta = 0.8u,\ \uparrow u_y = u\sin\theta = 0.6u\) | |
| \(v_y^2 = 0.36u^2 + 2 \times 9.8 \times 15 = 0.36u^2 + 294\) | |
| \(24.5^2 = u_x^2 + v_y^2 = 0.64u^2, +0.36u^2 + 294\) | M1 A1,A1 |
| \(u^2 = 306.25\ \ \Rightarrow\ \ u = 17.5\ \ *\) cso | A1 (4) |
| Scheme | Marks |
|---|---|
| \(\rightarrow\) \(u_x = u\cos\theta = 17.5 \times 0.8 = 14\) | B1 |
| \(\psi = \arccos\dfrac{14}{24.5} \approx 55^\circ\) accept 55.2\(^\circ\) (0.96 rads, or 0.963 rads) | M1 A1 |
| (3) |
Alternative for (b)
| \(\rightarrow\ u_x = u\cos\theta = 17.5 \times 0.8 = 14\) | B1 |
| \(\uparrow\ v_y^2 = u^2\sin^2\theta + 2 \times 9.8 \times 15 = 404.25\) | |
| \(\psi = \arctan\dfrac{\sqrt{404.25}}{14} \approx 55^\circ\) accept 55.2\(^\circ\) | M1 A1 (3) |
| Scheme | Marks |
|---|---|
| \(\uparrow\) \(u_y = u\sin\theta = 17.5 \times 0.6 = 10.5\) | B1 |
| \(s = ut + \dfrac{1}{2}at^2\ \ \Rightarrow\ \ -45 = 10.5t - 4.9t^2\) | M1 A1 |
| leading to \(t = 4.3\), awrt \(t = 4.3\) or \(t = 4\tfrac{2}{7}\) | A1 |
| \(\rightarrow\ \ BD = 14 \times 4\tfrac{2}{7}\) (14 x \(t\)) ft their \(t\) | M1 A1ft |
| \(= 60\) (m) only | A1 |
| (7) | |
| (14 marks) |
Alternative for (c)
| Use of \(y = x\tan\theta - \dfrac{g\sec^2\theta}{2u^2}x^2\) | M1 |
| \(-45 = \dfrac{3}{4}x,\ -\dfrac{g}{2 \times 17.5^2} \times \dfrac{25}{16}x^2\) | B1,A1 |
| \(x^2 - 30x - 1800 = 0\) o.e. | A1 |
| Factors or quadratic formula | M1 A1ft |
| BD = 60 (m) | A1 |