AS October 2020 Q4

EdexcelAS paperCurrent spec11 marksCollisions: 1 Sphere Direct Work-Energy Principle

4. A small ball, of mass \(m\), is thrown vertically upwards with speed \(\sqrt{8gH}\) from a point \(O\) on a smooth horizontal floor. The ball moves towards a smooth horizontal ceiling that is a vertical distance \(H\) above \(O\). The coefficient of restitution between the ball and the ceiling is \(\dfrac{1}{2}\)

In a model of the motion of the ball, it is assumed that the ball, as it moves up or down, is subject to air resistance of constant magnitude \(\dfrac{1}{2}mg\).

Using this model,

(a) use the work-energy principle to find, in terms of \(g\) and \(H\), the speed of the ball immediately before it strikes the ceiling, (5)

(b) find, in terms of \(g\) and \(H\), the speed of the ball immediately before it strikes the floor at \(O\) for the first time. (5)

In a simplified model of the motion of the ball, it is assumed that the ball, as it moves up or down, is subject to no air resistance.

Using this simplified model,

(c) explain, without any detailed calculation, why the speed of the ball, immediately before it strikes the floor at \(O\) for the first time, would still be less than \(\sqrt{8gH}\) (1)

Mark scheme (a)

Scheme	Marks	AO
\(\dfrac{1}{2}mgH\)	B1	1.1b
\(\dfrac{1}{2}m(8gH - v^2)\)	B1	1.1b
Apply the work-energy principle	M1	3.3
\(\dfrac{1}{2}mgH = \dfrac{1}{2}m(8gH - v^2) - mgH\)	A1	1.1b
\(v = \sqrt{5gH}\)	A1	1.1b
	(5)

Notes

B1: Work done against resistance (allow -ve) Can be implied by use of \(\dfrac{3}{2}mgH\) (work done against resistance + work done against weight)

B1: KE loss (allow -ve)

M1: Correct no. of terms, dimensionally correct. Condone sign errors.

A1: Correct unsimplified equation

A1: Correct answer (any equivalent but must be in terms of \(g\) and \(H\))
Accept \(2.2\sqrt{gH}\) or better

Mark scheme (b)

Scheme	Marks	AO
Use NLR to find rebound speed: \(\dfrac{1}{2}\sqrt{5gH}\)	M1	3.4
Apply the work-energy principle or suvat with \(a = \dfrac{1}{2}g\)	M1	3.3
\(\dfrac{1}{2}mgH = mgH - \dfrac{1}{2}m\left({v_1}^2 - \dfrac{1}{4} \times 5gH\right)\) or \((v_1)^2 = \dfrac{5gH}{4} + 2 \times \dfrac{g}{2} \times H\)	A1ft A1	1.1b 1.1b
\(v_1 = \dfrac{3}{2}\sqrt{gH}\)	A1	2.2a
	(5)

Notes

M1: Use of NLR

M1: Correct no. of terms, dimensionally correct

A1ft: Correct equation with at most one error ft on their answer to (a)

M1A1ft is available to a candidate who has not scored the first M1

A1: Correct equation (no ft)

A1: Correct answer (any equivalent but must be in terms of \(g\) and \(H\))

Mark scheme (c)

Scheme	Marks	AO
Since \(e \lt 1\), ball loses energy in its collision with the ceiling.	B1	2.4
	(1)
	(11 marks)

Notes

B1: Clear explanation

Need to identify that the loss of KE occurs in the impact with the ceiling. Do not insist on seeing \(e \lt 1\) or equivalent.
If they include incorrect additional statements then B0