AS October 2020 Q3

EdexcelAS paperCurrent spec12 marksCollisions: 2 Spheres Direct Impulse & Momentum

3. Three particles \(A\), \(B\) and \(C\) are at rest on a smooth horizontal plane. The particles lie along a straight line with \(B\) between \(A\) and \(C\).

Particle \(B\) has mass \(4m\) and particle \(C\) has mass \(km\), where \(k\) is a positive constant. Particle \(B\) is projected with speed \(u\) along the plane towards \(C\) and they collide directly.

The coefficient of restitution between \(B\) and \(C\) is \(\dfrac{1}{4}\)

(a) Find the range of values of \(k\) for which there would be no further collisions. (8)

The magnitude of the impulse on \(B\) in the collision between \(B\) and \(C\) is \(3mu\)

(b) Find the value of \(k\). (4)

Diagram: particles A, B (4m) and C (km); B moving towards C with speed u; after the collision B and C move in the same direction with speeds v_B and v_C — Mark scheme (a)

Scheme	Marks	AO

Use of CLM	M1	3.1a
\(4mu = 4mv_B + kmv_C\)	A1	1.1b
Use of NLR	M1	3.1a
\(\dfrac{1}{4}u = -v_B + v_C\)	A1	1.1b
Solve for \(v_B\)	M1	1.1b
\(v_B = \dfrac{u(16-k)}{4(k+4)} \qquad \left(v_C = \dfrac{5u}{k+4}\right)\)	A1	1.1b
Use of \(v_B \geqslant 0\) and solve for \(k\)	M1	3.4
\((0 \lt)\ k \leqslant 16\)	A1	1.1b
	(8)

Scheme	Marks	AO
Solve for \(v_B\) in terms of \(v_C\) only	M1
\(v_B = \dfrac{(16-k)v_C}{20}\)	A1
Use of \(v_B \geqslant 0\) and \(v_C \gt 0\) to solve for \(k\)	M1
\((0 \lt)\ k \leqslant 16\)	A1

Mark scheme (b)

Scheme	Marks	AO
Impulse-momentum equation	M1	3.1a
\(-3mu = 4m(v_B - u) \quad \left(v_B = \dfrac{u}{4}\right)\) or \(3mu = kmv_C\)	A1	1.1b
Complete method to solve for \(k\)	M1	1.1b
\(k = 6\)	A1	2.2a
	(4)
	(12 marks)

Notes

M1: Correct no. of terms, condone sign errors, but must be subtracting momentum terms

A1: Correct equation

M1: Eliminate and solve for \(k\)

A1: \(k = 6\)

Notes

Alternative for last 4 marks

Notes