M2 January 2013 Q6
6.
A ball is thrown from a point \(O\), which is 6 m above horizontal ground. The ball is projected with speed \(u\) m s\(^{-1}\) at an angle \(\theta\) above the horizontal. There is a thin vertical post which is 4 m high and 8 m horizontally away from the vertical through \(O\), as shown in Figure 2. The ball passes just above the top of the post 2 s after projection. The ball is modelled as a particle.
The ball hits the ground \(T\) seconds after projection.
Immediately before the ball hits the ground the direction of motion of the ball makes an angle \(\alpha\) with the horizontal.
| Scheme | Marks |
|---|---|
| \(2 = -2u\sin\theta + \dfrac{1}{2}g \times 4\) \(\left(-2 = u\sin\theta t - \dfrac{1}{2}gt^2\right)\) | M1 |
| \(u\sin\theta = g - 1\) | A1 |
| \(2u\cos\theta = 8\ \ \ (u\cos\theta = 4)\) \((u\cos\theta t = 8)\) | B1 |
| \(\tan\theta = \dfrac{g - 1}{4} = 2.2\ \ *\) | M1 A1 |
| (5) |
Notes
M1 Vertical distance. Condone sign errors. Must have used \(t = 2\), but could be using \(u_y = u\sin\theta\)
A1 All correct
B1 Horizontal distance. Accept \(u_x = 4\) o.e.
M1 Divide to obtain expression for \(\tan\theta\)
A1 Given answer It is acceptable to quote and use the equation for the projectile path. Incorrect equation is 0/5.
| Scheme | Marks |
|---|---|
| \(u\cos\theta = 4\) | M1 |
| \(u = \dfrac{4}{\cos\theta} = 9.66\ldots = 9.7\) | A1 |
| (2) |
Notes
M1 Use the horizontal distance and \(\theta\) to find \(u\)
A1 9.67 or 9.7 NB \(\theta = 65.6^\circ\) leading to 9.68 is an accuracy penalty.
OR use components from (a) and Pythagoras.
| Scheme | Marks |
|---|---|
| \(6 = (1 - g)T + \dfrac{1}{2} \times 9.8T^2\) \(4.9T^2 - 8.8T - 6 = 0\) | M1 |
| \(T = \dfrac{8.8 \pm \sqrt{[(-)8.8]^2 + 24 \times 4.9}}{9.8}\) | DM1 |
| \(T = 2.323\ldots = 2.32\) or 2.3 | A1 |
| (3) |
Notes
M1 Equation for vertical distance \(= \pm 6\) to give a quadratic in \(T\). Allow their \(u_y\)
DM1 Solve a 3 term quadratic
A1 2.3 or 2.32 only
| Scheme | Marks |
|---|---|
| \(v^2 = 8.8^2 + 2g \times 6\) or \(v = -8.8 + gT\) | M1 |
| \(v = 13.96\ldots\) | A1 |
| Horiz speed \(= 4\) \(\tan\alpha = \dfrac{v}{4}\) | DM1 A1 |
| \(\alpha = 74.01\ldots = 74^\circ\) | A1 |
| (5) | |
| (15 marks) |
Notes
M1 Use suvat to find vertical speed
A1 Correct equation their \(u_y\), \(T\)
DM1 Correct trig. with their vertical speed to find the required angle.
A1 Correct equation
A1 \(74^\circ\) or \(74.0^\circ\). Allow 106.
Alternative
| \(\dfrac{1}{2}m(9.6664)^2 + 6mg = \dfrac{1}{2}mv^2\) | M1 |
| \(v = 14.52719\ldots\) | A1 |
| \(\cos\alpha = \dfrac{4}{14.5}\) | DM1 A1 |
| \(\alpha = 74.01\ldots = 74^\circ\) | A1 |
M1 Conservation of energy to find speed
DM1 Correct method for \(\alpha\)
A1 Allow 106