M4 June 2015 Q7

EdexcelOld spec13 marksCollisions: 1 Sphere Oblique

Figure 4: plan view of walls AB and BC at 120 degrees; ball approaches AB with speed u at 60 degrees to AB, rebounds to hit BC and leaves with speed w — Figure 4

Figure 4 represents the plan view of part of a smooth horizontal floor, where \(AB\) and \(BC\) are smooth vertical walls. The angle between \(AB\) and \(BC\) is 120\(^\circ\). A ball is projected along the floor towards \(AB\) with speed \(u\) m s\(^{-1}\) on a path at an angle of 60\(^\circ\) to \(AB\). The ball hits \(AB\) and then hits \(BC\). The ball is modelled as a particle. The coefficient of restitution between the ball and each wall is \(\dfrac{1}{2}\)

(a) Show that the speed of the ball immediately after it has hit \(AB\) is \(\dfrac{\sqrt{7}}{4}u\). (6)

The speed of the ball immediately after it has hit \(BC\) is \(w\) m s\(^{-1}\)

(b) Find \(w\) in terms of \(u\). (7)

Path of ball: speed u at 60 degrees to AB, speed v at angle theta after AB, speed w at angle alpha after BC — Mark scheme (a)

Scheme	Marks
Resolve parallel to barrier - condone sin/cos confusion	M1
\(u\cos 60 = v\cos\theta\)	A1
Resolve perpendicular to the barrier - condone sin/cos confusion	M1
\(eu\sin 60 = v\sin\theta\)	A1
\(v^2 = u^2\cos^2 60 + e^2u^2\sin^2 60 = \dfrac{u^2}{4} + \dfrac{3u^2}{16} = \dfrac{7u^2}{16}\)	M1
\(v = \dfrac{\sqrt{7}}{4}u\)	A1
	(6)

Mark scheme (b)

Scheme	Marks
Angle of approach with \(BC = 19.1^\circ\)	B1
\(v\cos 19.1 = w\cos\phi\)	M1
\(\dfrac{1}{2}v\sin 19.1 = w\sin\phi\)	M1
	A1
Form equation in \(v\) and \(\phi\)	M1
\(w^2 = v^2\left(\dfrac{1}{4}\sin^2 19.1 + \cos^2 19.1\right)\)	A1
\(0.634u\)	A1
	(7)
	(13 marks)

Notes

M1 Components parallel to \(BC\)

M1 Components perpendicular to \(BC\)

A1 Equations correct for their 19.1

M1 Square and add or divide to find \(\tan\phi\)

A1 \((\phi = 9.83^\circ)\) Follow their 19,1?

7balt

\(\tan\theta = \dfrac{1}{2}\tan 60\)	B1
\(\tan\alpha = \dfrac{1}{2}\tan(60 - \theta) \quad \left(= \dfrac{1}{2}\left(\dfrac{\sqrt{3} - \frac{1}{2}\sqrt{3}}{1 + \sqrt{3}.\frac{1}{2}\sqrt{3}}\right) = \dfrac{\sqrt{3}}{10}\right)\)	M1 A1
\(v\cos(60 - \theta) = w\cos\alpha\)	M1
\(v\left(\dfrac{1}{2}.\dfrac{2}{\sqrt{7}} + \dfrac{\sqrt{3}}{2}.\dfrac{\sqrt{3}}{\sqrt{7}}\right) = w\dfrac{10}{\sqrt{103}}\left(= v\dfrac{5}{2\sqrt{7}}\right)\)	M1 A1
\(w = \dfrac{\sqrt{103}}{4\sqrt{7}}v = \dfrac{\sqrt{103}}{4\sqrt{7}}.\dfrac{\sqrt{7}}{4}u = \dfrac{\sqrt{103}}{16}u \qquad (0.634u)\)	A1