Question 2 - A-Level Maths

OCR MEI Further Mechanics A AS 2023 June — Question 2 10 marks

Exam Board	OCR MEI
Module	Further Mechanics A AS (Further Mechanics A AS)
Year	2023
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Momentum and Collisions 1
Type	Vertical drop and bounce
Difficulty	Standard +0.8 This is a multi-part mechanics question requiring understanding of coefficient of restitution, energy methods with air resistance, and comparing two physical models. Part (a) is routine, but parts (b) and (c) require careful energy accounting across multiple bounces and showing equivalence between models—demanding more sophisticated problem-solving than typical A-level mechanics questions.
Spec	6.02e Calculate KE and PE: using formulae 6.02i Conservation of energy: mechanical energy principle 6.02j Conservation with elastics: springs and strings 6.03i Coefficient of restitution: e 6.03j Perfectly elastic/inelastic: collisions

2 A ball P of mass \(m \mathrm {~kg}\) is held at a height of 12.8 m above a horizontal floor. P is released from rest and rebounds from the floor. After the first bounce, P reaches a maximum height of 5 m above the floor. Two models, A and B , are suggested for the motion of P .
Model A assumes that air resistance may be neglected.

Determine, according to model A , the coefficient of restitution between P and the floor. Model B assumes that the collision between P and the floor is perfectly elastic, but that work is done against air resistance at a constant rate of \(E\) joules per metre.
Show that, according to model \(\mathrm { B } , \mathrm { E } = \frac { 39 } { 89 } \mathrm { mg }\).
Show that both models predict that P will attain the same maximum height after the second bounce.

Show mark scheme Show mark scheme source

Question 2:

Answer	Marks	Guidance
2	(a)	Let the speed of the ball just before and just after hitting the

ground be u and v m s-1. Let the coefficient of restitution be e.

𝑢2 = 02+2𝑔×12.8 or 12 m u 2 = m g  1 2 .8

𝑢2 = 128 𝑔 = 250.88, 𝑢 = 15.84

Answer	Marks	Guidance
5	M1	3.3

02 = 𝑣2−2𝑔×5 or 12 m v 2 = m g  5

Answer	Marks	Guidance
𝑣2 = 10𝑔 = 98, 𝑣 = 9.90	M1	3.3
OR	5 = 12.8𝑒2	M2

𝑣 5

𝑒 = = (= 0.625)

Answer	Marks	Guidance
𝑢 8	A1	1.1

(e.g. 9.9/15.8 = 0.627)

A0 for −5/8 (as final answer)

[3]

Answer	Marks	Guidance
(b)	‘Perfectly elastic’ means no energy loss	B1

Just 𝑒 = 1 is not sufficient

Answer	Marks	Guidance
mg12.8−E ( 12.8+5 )=mg5	M1	3.3
OR	1 𝑚𝑣2 = 𝑚𝑔×12.8−12.8𝐸 , 1 𝑚𝑣2 = 𝑚𝑔×5+5𝐸

1 2

2 2

𝑣 = 𝑣

Answer	Marks
2 1	M1
B1	WEP in two stages

39 ⇒ 17.8𝐸 = 7.8𝑚𝑔 ⇒ 𝐸 = 𝑚𝑔

Answer	Marks	Guidance
89	A1	2.2a

Note that (39/89)g = 1911/445

[3]

Answer	Marks
(c)	Model A. Speed just after second bounce is

𝑤 = 𝑒𝑣 = 5 ×√98 = 6.187, 𝑤2 = 125 𝑔 = 38.28125

Answer	Marks	Guidance
8 32	M1	3.1b
OR	2

Maximum height is 5𝑒2 = 5×( 5 ) or 5× 5

Answer	Marks
8 12.8	M1

125 Maximum height is (= 1.953125) m

Answer	Marks	Guidance
64	A1	1.1

Model B. Let maximum height be h m after second bounce.

mg5−39mg ( 5+h )=mgh

Answer	Marks	Guidance
89	M1	1.1

128h= 250 h=125(=1.953125 )

(which is the same)

Answer	Marks	Guidance
89 89 64	A1	2.2a

at least 3 sf

[4]

1 𝑚𝑣2 = 𝑚𝑔×12.8−12.8𝐸 , 1 𝑚𝑣2 = 𝑚𝑔×5+5𝐸

1 2

2 2

𝑣 = 𝑣

2 1

M1

B1

Question 2:
2 | (a) | Let the speed of the ball just before and just after hitting the
ground be u and v m s-1. Let the coefficient of restitution be e.
𝑢2 = 02+2𝑔×12.8 or 12 m u 2 = m g  1 2 .8
𝑢2 = 128 𝑔 = 250.88, 𝑢 = 15.84
5 | M1 | 3.3 | Attempt at equation for u
02 = 𝑣2−2𝑔×5 or 12 m v 2 = m g  5
𝑣2 = 10𝑔 = 98, 𝑣 = 9.90 | M1 | 3.3 | Attempt at equation for v
OR | 5 = 12.8𝑒2 | M2
𝑣 5
𝑒 = = (= 0.625)
𝑢 8 | A1 | 1.1 | Accept 0.62 to 0.63 from correct method
(e.g. 9.9/15.8 = 0.627)
A0 for −5/8 (as final answer)
[3]
(b) | ‘Perfectly elastic’ means no energy loss | B1 | 1.2 | Or Speed is unchanged by the impact
Just 𝑒 = 1 is not sufficient
mg12.8−E ( 12.8+5 )=mg5 | M1 | 3.3 | WEP Allow sign errors M1 normally implies B1
OR | 1 𝑚𝑣2 = 𝑚𝑔×12.8−12.8𝐸 , 1 𝑚𝑣2 = 𝑚𝑔×5+5𝐸
1 2
2 2
𝑣 = 𝑣
2 1 | M1
B1 | WEP in two stages
39
⇒ 17.8𝐸 = 7.8𝑚𝑔 ⇒ 𝐸 = 𝑚𝑔
89 | A1 | 2.2a | AG Must be convincingly shown.
Note that (39/89)g = 1911/445
[3]
(c) | Model A. Speed just after second bounce is
𝑤 = 𝑒𝑣 = 5 ×√98 = 6.187, 𝑤2 = 125 𝑔 = 38.28125
8 32 | M1 | 3.1b | Attempt to find w or w2
OR | 2
Maximum height is 5𝑒2 = 5×( 5 ) or 5× 5
8 12.8 | M1
125
Maximum height is (= 1.953125) m
64 | A1 | 1.1 | Accept 1.9 to 2.0 from correct method
Model B. Let maximum height be h m after second bounce.
mg5−39mg ( 5+h )=mgh
89 | M1 | 1.1 | WEP
128h= 250 h=125(=1.953125 )
(which is the same)
89 89 64 | A1 | 2.2a | Both A1’s can only be awarded if values agree to
at least 3 sf
[4]
1 𝑚𝑣2 = 𝑚𝑔×12.8−12.8𝐸 , 1 𝑚𝑣2 = 𝑚𝑔×5+5𝐸
1 2
2 2
𝑣 = 𝑣
2 1
M1
B1

Show LaTeX source

2 A ball P of mass $m \mathrm {~kg}$ is held at a height of 12.8 m above a horizontal floor. P is released from rest and rebounds from the floor. After the first bounce, P reaches a maximum height of 5 m above the floor.

Two models, A and B , are suggested for the motion of P .\\
Model A assumes that air resistance may be neglected.
\begin{enumerate}[label=(\alph*)]
\item Determine, according to model A , the coefficient of restitution between P and the floor.

Model B assumes that the collision between P and the floor is perfectly elastic, but that work is done against air resistance at a constant rate of $E$ joules per metre.
\item Show that, according to model $\mathrm { B } , \mathrm { E } = \frac { 39 } { 89 } \mathrm { mg }$.
\item Show that both models predict that P will attain the same maximum height after the second bounce.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2023 Q2 [10]}}

This paper (6 questions)

View full paper

Q1 7 Q2 10 Q3 9 Q4 10 Q5 13 Q6 11