M4 June 2015 Q3
3.
Two smooth uniform spheres \(A\) and \(B\) with equal radii have masses \(m\) and \(2m\) respectively. The spheres are moving in opposite directions on a smooth horizontal surface and collide obliquely. Immediately before the collision, \(A\) has speed \(3u\) with its direction of motion at an angle \(\theta\) to the line of centres, and \(B\) has speed \(u\) with its direction of motion at an angle \(\theta\) to the line of centres, as shown in Figure 1. The coefficient of restitution between the spheres is \(\dfrac{1}{8}\)
Immediately after the collision, the speed of \(A\) is twice the speed of \(B\).
Find the size of the angle \(\theta\). (12)
| Scheme | Marks |
|---|---|
| After collision \(u\sin\theta\) and \(3u\sin\theta\) perpendicular to \(l\) of \(c\) | B1 |
| CLM : \(r + 2s = 3u\cos\theta - 2u\cos\theta\ (= u\cos\theta)\) | M1 A1 |
| Impact: \(s - r = e \times 4u\cos\theta\left(= \dfrac{u\cos\theta}{2}\right)\) | M1 A1 |
| \(\Rightarrow r = 0,\ s = \dfrac{u\cos\theta}{2}\) | DM1 A1 |
| After the collision: \((3u\sin\theta)^2 + r^2 = 4\left((u\sin\theta)^2 + s^2\right)\) | M1 A1 |
| \(9u^2\sin^2\theta = 4u^2\sin^2\theta + 4.\dfrac{u^2}{4}\cos^2\theta\) | A1 |
| \(\tan^2\theta = \dfrac{1}{5},\quad \theta = 24.1(^\circ)\ \ (0.421\text{ radians})\) | DM1 A1 |
| (12) | |
| (12 marks) |
Notes
M1 Requires all four terms but condone sign errors.
A1 Correct unsimplified equation
M1 Must be the right way round, but condone sign errors
A1 Correct unsimplified equation
DM1 Solve the simultaneous equations to find the horizontal components of velocities. Dependent on the two preceding M marks
A1 Both correct
M1 Use \(v_A = 2v_B\). Condone 2 on the wrong side?
A1 Correct unsimplified equation (their \(r\) and \(s\))
A1 Obtain an equation in \(\theta\)
DM1 Solve for \(\theta\). Dependent on the previous M1
A1 Correct to 3 sf or better
3 alt
| For those who prefer everything with trig: | |
| \(v_A\sin\alpha = 3u\sin\theta,\quad v_B\sin\beta = u\sin\theta\) | B1 |
| \(m.3u\cos\theta - 2m.u\cos\theta = mv_A\cos\alpha + 2mv_B\cos\beta\) | M1 |
| \(\left(u\cos\theta = v_A\cos\alpha + 2v_B\cos\beta\right)\) | A1 |
| \(\dfrac{1}{8} \times (3u\cos\theta + u\cos\theta) = v_B\cos\beta - v_A\cos\alpha\) | M1 |
| \(\left(\dfrac{u}{2}\cos\theta = v_B\cos\beta - v_A\cos\alpha\right)\) | A 1 |
| \(\dfrac{u}{2}\cos\theta = v_B\cos\beta,\quad 0 = v_A\cos\alpha\ (\Rightarrow \sin\alpha = 1)\) | DM1 A1 |
| \(v_A\sin\alpha = v_A = 2v_B = 3u\sin\theta\) | M1 |
| \(v_B\sin\beta = u\sin\theta \Rightarrow \dfrac{3u\sin\theta}{2}\sin\beta = u\sin\theta\) | A1 |
| \(\sin\beta = \dfrac{2}{3}\) | A1 |
| \(2v_B = 3u\sin\theta\ \ \&\ \ \dfrac{u}{2}\cos\theta = v_B\cos\beta\) \(\Rightarrow 6\tan\theta = \dfrac{2}{\cos\beta}\left(= 2 \times \dfrac{3}{\sqrt{5}}\right)\) | M1 |
| \(\tan\theta = \dfrac{1}{\sqrt{5}},\quad \theta = 24.1(^\circ)\ \ (0.421\text{ radians})\) | A1 |
B1 Perpendicular to the l.o.c.
M1 CLM
M1 Impact law
DM1 Simultaneous equations
M1 Use \(v_A = 2v_B\) to find \(\beta\)
M1 Solve for \(\theta\)