A2 June 2019 Q5

EdexcelCurrent spec11 marksCollisions: 2 Spheres Direct

5. A particle \(P\) of mass \(3m\) and a particle \(Q\) of mass \(2m\) are moving along the same straight line on a smooth horizontal plane. The particles are moving in opposite directions towards each other and collide directly.

Immediately before the collision the speed of \(P\) is \(u\) and the speed of \(Q\) is \(2u\).

Immediately after the collision \(P\) and \(Q\) are moving in opposite directions.

The coefficient of restitution between \(P\) and \(Q\) is \(e\).

(a) Find the range of possible values of \(e\), justifying your answer. (8)

Given that \(Q\) loses 75% of its kinetic energy as a result of the collision,

(b) find the value of \(e\). (3)

Diagram: P (3m) moving right at u and Q (2m) moving left at 2u; afterwards P moves left at v and Q moves right at w — Mark scheme (a)

Scheme	Marks	AO

Use of CLM	M1	3.1a
\(3mu - 4mu = 2mw - 3mv\) \((-u = -3v + 2w)\)	A1	1.1b
Use of impact law	M1	3.4
\(w + v = 3ue\)	A1	1.1b
Correct strategy to form equation in \(w\) and find critical value of \(e \in (0, 1)\) \(\big(5w = u(9e - 1)\big)\)	M1	3.1a
\(w \gt 0: e \gt \dfrac{1}{9}\)	A1	1.1b
Complete strategy to justify the range of values of \(e\) \(\big(5v = u(1 + 6e)\big)\) \(v \gt 0\): true for all \(e\)	M1	3.1a
Therefore \(\dfrac{1}{9} \lt e \leqslant 1\)	A1	2.2a
	(8)

Mark scheme (b)

Scheme	Marks	AO
Final KE \(=\) 25% of initial KE	M1	3.1a
\(\dfrac{1}{2} \times 2m \times \dfrac{u^2(9e - 1)^2}{25} = \dfrac{1}{4} \times \dfrac{1}{2} \times 2m \times 4u^2\) \(\left(\text{or } w = \dfrac{1}{2} \times 2u\right)\)	A1ft	1.1b
\(\Rightarrow (9e - 1)^2 = 25\), \(e = \dfrac{2}{3}\) only	A1	1.1b
	(3)
	(11 marks)

Notes

M1: Use KE to form equation in \(e\). 25% should be used correctly
Condone if mass cancelled throughout

A1ft: Correct unsimplified equation – follow their \(w\)

A1: Or equivalent. Correct conclusion
ISW after correct answer.