A2 October 2020 Q7

EdexcelCurrent spec11 marksCollisions: 1 Sphere Oblique

Figure 2: plan view of parallel walls AB and CD; the ball approaches AB with speed v m/s at angle α to AB, rebounds to hit CD, and then moves away at angle ½α to CD — Figure 2

Figure 2 represents the plan view of part of a horizontal floor, where \(AB\) and \(CD\) represent fixed vertical walls, with \(AB\) parallel to \(CD\).

A small ball is projected along the floor towards wall \(AB\). Immediately before hitting wall \(AB\), the ball is moving with speed \(v\ \text{m s}^{-1}\) at an angle \(\alpha\) to \(AB\), where \(0 \lt \alpha \lt \dfrac{\pi}{2}\)

The ball hits wall \(AB\) and then hits wall \(CD\).

After the impact with wall \(CD\), the ball is moving at angle \(\dfrac{1}{2}\alpha\) to \(CD\).

The coefficient of restitution between the ball and wall \(AB\) is \(\dfrac{2}{3}\)

The coefficient of restitution between the ball and wall \(CD\) is also \(\dfrac{2}{3}\)

The floor and the walls are modelled as being smooth. The ball is modelled as a particle.

(a) Show that \(\tan\left(\dfrac{1}{2}\alpha\right) = \dfrac{1}{3}\) (7)

(b) Find the percentage of the initial kinetic energy of the ball that is lost as a result of the two impacts. (4)

Diagrams: after hitting AB the ball has components v cos α parallel to the wall and (2/3) v sin α away from it; after hitting CD it has components v cos α parallel to the wall and (4/9) v sin α away from it — Mark scheme (a)

Scheme	Marks	AO
Use model to find components of velocity after the impacts:	B1 B1 B1 B1	3.1b 3.4 3.1b 3.4
\(\tan\dfrac{\alpha}{2} = \dfrac{\frac{4}{9}v\sin\alpha}{v\cos\alpha}\ \ \left(= \dfrac{4}{9}\tan\alpha\right)\)	M1	3.1b
\(t = \tan\dfrac{\alpha}{2}\ \ \Rightarrow t = \dfrac{4 \times 2t}{9(1 - t^2)}\)	M1	1.1b
\(1 - t^2 = \dfrac{8}{9}\), \(t = \dfrac{1}{3}\) *	A1*	2.2a
	(7)

Mark scheme (b)

Scheme	Marks	AO
\(\tan\alpha = \dfrac{\frac{2}{3}}{1 - \frac{1}{9}} = \dfrac{3}{4}\)	B1	1.1b
change in KE \(\dfrac{1}{2}mv^2 - \dfrac{1}{2}m\left(v^2\cos^2\alpha + \left(\dfrac{4}{9}v\right)^2\sin^2\alpha\right)\)	M1	3.1b
% of KE lost \(= 100\left(1 - \dfrac{\frac{1}{2}mv^2\left(\frac{16}{25} + \frac{16}{81} \times \frac{9}{25}\right)}{\frac{1}{2}mv^2}\right)\)	M1	1.1b
\(= 28.888\ldots\ (\%)\)	A1	1.1b
	(4)
	(11 marks)

Notes

B1: Correct use of \(t = \dfrac{1}{3}\) Must be seen / used in part (b)

M1: Dimensionally correct expression for change in KE
NB note that they may not show component parallel to the wall

M1: Dimensionally correct expression for the percentage of KE lost.

A1: Accept 29 (%) or better Accept \(\dfrac{260}{9}\)