AS June 2022 Q2

EdexcelAS paperCurrent spec8 marksCollisions: 2 Spheres Direct Impulse & Momentum

2. Two particles, \(A\) and \(B\), have masses \(m\) and \(3m\) respectively. The particles are moving in opposite directions along the same straight line on a smooth horizontal plane when they collide directly.

Immediately before they collide, \(A\) is moving with speed \(2u\) and \(B\) is moving with speed \(u\).

The direction of motion of each particle is reversed by the collision.

In the collision, the magnitude of the impulse exerted on \(A\) by \(B\) is \(\dfrac{9mu}{2}\)

(a) Find the value of the coefficient of restitution between \(A\) and \(B\). (7)

(b) Hence, write down the total loss in kinetic energy due to the collision, giving a reason for your answer. (1)

Mark scheme (a)

Scheme	Marks	AO
\[\begin{array}{cccc} & 2u \rightarrow & \leftarrow u & \\ \dfrac{9mu}{2} \longleftarrow & m & 3m & \longrightarrow \dfrac{9mu}{2} \\ & v \leftarrow & \rightarrow w & \end{array}\]
Use of Impulse-momentum principle for \(A\) or \(B\)	M1	3.4
\(A\): \(\dfrac{9mu}{2} = m(v - -2u)\) or \(B\): \(\dfrac{9mu}{2} = 3m(w - -u)\)	A1	1.1b
Use of Impulse-momentum principle for \(B\) or \(A\) or CLM	M1	3.4
\(\dfrac{9mu}{2} = 3m(w - -u)\) or \(\dfrac{9mu}{2} = m(v - -2u)\) or \(2mu - 3mu = -mv + 3mw\)	A1	1.1b
\(v = \dfrac{5u}{2}\) and \(w = \dfrac{u}{2}\)	A1	1.1b
\(e = \dfrac{\dfrac{5u}{2} + \dfrac{u}{2}}{2u + u}\)	M1	3.1a
\(e = 1\)	A1cso	1.1b
	(7)

Notes

N.B. Ignore diagrams if it helps the candidate.
Equations need to be consistent, where appropriate, to earn A marks.

M1: Use of Impulse-momentum principle for \(A\) or \(B\), condone sign errors but M0 if dimensionally incorrect e.g. if \(m\) missing

A1: Correct unsimplified equation

M1: Use of Impulse-momentum principle for other particle or CLM, condone sign errors but M0 if dimensionally incorrect e.g. if \(m\) missing from impulse
For CLM, allow consistent missing \(m\)’s or extra \(g\)’s.

A1: Correct unsimplified equation

A1: Cao for both. Allow one or both negative if correct for their symbols.

M1: Use of NEL to obtain \(e = \ldots\), condone sign errors in numerator but must be terms in \(u\) only AND must be \((2u + u)\) in denominator.
M0 if inverted

A1: cso

ALTERNATIVE

Scheme	Marks	AO
NEL is written down before \(v\) and \(w\) are found: \(v + w = 3ue\)	3^rd M1
Use of Impulse-momentum principle for \(A\) or \(B\)	1^st M1
\(A\): \(\dfrac{9mu}{2} = m(v - -2u)\) or \(B\): \(\dfrac{9mu}{2} = 3m(w - -u)\)	1^st A1
Use of Impulse-momentum principle for \(B\) or \(A\) or CLM	2^nd M1
\(\dfrac{9mu}{2} = 3m(w - -u)\) or \(\dfrac{9mu}{2} = m(v - -2u)\) or \(2mu - 3mu = -mv + 3mw\)	2^nd A1
An equation (not an identity) in \(u\) and \(e\) only is produced	3^rd A1
\(e = 1\)	A1cso

Mark scheme (b)

Scheme	Marks	AO
Perfectly elastic (or the coefficient of restitution is 1) so no loss in kinetic energy. Allow a direct evaluation of the KE loss i.e. \(\dfrac{1}{2}m(2u)^2 + \dfrac{1}{2} \times 3mu^2 - \left(\dfrac{1}{2}m\left(\dfrac{5u}{2}\right)^2 + \dfrac{1}{2} \times 3m\left(\dfrac{u}{2}\right)^2\right) = 0\) B0 if incorrect extras	DB1	2.4
	(1)
	(8 marks)

Notes

N.B. Ignore diagrams if it helps the candidate.
Equations need to be consistent, where appropriate, to earn A marks.

DB1: Dependent on \(e = 1\) correctly obtained in (a)
A correct statement e.g. zero, 0 etc and a correct reason