AQA Further AS Paper 2 Mechanics 2021 June — Question 8 11 marks

Exam BoardAQA
ModuleFurther AS Paper 2 Mechanics (Further AS Paper 2 Mechanics)
Year2021
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions 1
TypeCollision with unchanged direction
DifficultyStandard +0.3 This is a standard Further Maths mechanics collision problem requiring conservation of momentum and Newton's restitution law. Parts (a) and (b) are routine algebraic manipulation (showing a given result and finding velocity), part (b) requires understanding that 'same direction' means both velocities have the same sign, and part (c) tests understanding that impulses are equal and opposite by Newton's third law. All techniques are standard textbook exercises with no novel insight required, making it slightly easier than average even for Further Maths.
Spec6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation6.03k Newton's experimental law: direct impact
Two spheres A and B are free to move on a smooth horizontal surface. The masses of A and B are 2 kg and 3 kg respectively. Both A and B are initially at rest. Sphere A is set in motion directly towards sphere B with speed 4 m s\(^{-1}\) and subsequently collides with sphere B The coefficient of restitution between the spheres is \(e\)
    1. Show that the speed of B immediately after the collision is $$\frac{8(1 + e)}{5}$$ [4 marks]
    2. Find an expression, in terms of \(e\), for the velocity of A immediately after the collision. [2 marks]
  2. It is given that the spheres both move in the same direction after the collision. Find the range of possible values of \(e\) [2 marks]
    1. The impulse of sphere A on sphere B is \(I\) The impulse of sphere B on sphere A is \(J\) Given that the collision is perfectly inelastic, find the value of \(I + J\) [1 mark]
    2. State, giving a reason for your answer, whether the value found in part (c)(i) would change if the collision was not perfectly inelastic. [2 marks]
Question 8:
AnswerMarks Guidance
8(a)(i)Forms an equation using
conservation of momentum1.1a M1
Velocity of B = w
C of M
NLR 2(4) = 2𝑣𝑣+3𝑀𝑀
8 = 2𝑣𝑣+3𝑀𝑀
π‘€π‘€βˆ’π‘£π‘£ = 4𝑒𝑒
8 = 2(π‘€π‘€βˆ’4𝑒𝑒)+ 3𝑀𝑀
8+8𝑒𝑒 = 5𝑀𝑀
8( 1+𝑒𝑒)
𝑀𝑀 =
5
Obtains a correct momentum
AnswerMarks Guidance
equation – can be unsimplified1.1b A1
Forms a correct equation using
AnswerMarks Guidance
Newton’s law of restitution1.1b B1
Completes a reasoned
argument using both
conservation of momentum and
the coefficient of restitution to
AnswerMarks Guidance
verify the correct speed of B2.1 R1
Total4
QMarking instructions AO
AnswerMarks
8(a)(ii)Substitutes into either of
8(1+𝑒𝑒)
their equations
5
or
Subtracts original equations and
AnswerMarks Guidance
eliminates the velocity of B1.1a M1
𝑣𝑣 = βˆ’4𝑒𝑒
5
8+8π‘’π‘’βˆ’20𝑒𝑒
𝑣𝑣 =
5
4(2 βˆ’3𝑒𝑒)
𝑣𝑣 =
Obtains the correct velocity for A
AnswerMarks Guidance
ACF1.1b A1
Total2 5
QMarking instructions AO
AnswerMarks
8(b)Solves v > 0 or v = 0 and
deduces upper bound for e
provided that their answer is
AnswerMarks Guidance
between 0 and 12.2a M1
> 0
5
2
𝑒𝑒 <
0 ≀ 3
2
𝑒𝑒 < 3
Correctly states full range of
AnswerMarks Guidance
values for e1.1b A1
Total2
QMarking instructions AO
AnswerMarks Guidance
8(c)(i)Deduces that I + J = 0 2.2a
Total1
QMarking instructions AO
AnswerMarks
8(c)(ii)Explains that corresponding
impulses for any collision are
equal in magnitude and opposite
AnswerMarks Guidance
in direction2.4 M1
and opposite in direction so will
always have a sum of 0
Deduces that the value of I + J
AnswerMarks Guidance
will not change2.2a R1
Total2
Question total11
Paper total40 
Question 8:
--- 8(a)(i) ---
8(a)(i) | Forms an equation using
conservation of momentum | 1.1a | M1 | Velocity of A = v
Velocity of B = w
C of M
NLR 2(4) = 2𝑣𝑣+3𝑀𝑀
8 = 2𝑣𝑣+3𝑀𝑀
π‘€π‘€βˆ’π‘£π‘£ = 4𝑒𝑒
8 = 2(π‘€π‘€βˆ’4𝑒𝑒)+ 3𝑀𝑀
8+8𝑒𝑒 = 5𝑀𝑀
8( 1+𝑒𝑒)
𝑀𝑀 =
5
Obtains a correct momentum
equation – can be unsimplified | 1.1b | A1
Forms a correct equation using
Newton’s law of restitution | 1.1b | B1
Completes a reasoned
argument using both
conservation of momentum and
the coefficient of restitution to
verify the correct speed of B | 2.1 | R1
Total | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 8(a)(ii) ---
8(a)(ii) | Substitutes into either of
8(1+𝑒𝑒)
their equations
5
or
Subtracts original equations and
eliminates the velocity of B | 1.1a | M1 | 8(1+𝑒𝑒)
𝑣𝑣 = βˆ’4𝑒𝑒
5
8+8π‘’π‘’βˆ’20𝑒𝑒
𝑣𝑣 =
5
4(2 βˆ’3𝑒𝑒)
𝑣𝑣 =
Obtains the correct velocity for A
ACF | 1.1b | A1
Total | 2 | 5
Q | Marking instructions | AO | Marks | Typical solution
--- 8(b) ---
8(b) | Solves v > 0 or v = 0 and
deduces upper bound for e
provided that their answer is
between 0 and 1 | 2.2a | M1 | 4(2βˆ’3𝑒𝑒)
> 0
5
2
𝑒𝑒 <
0 ≀ 3
2
𝑒𝑒 < 3
Correctly states full range of
values for e | 1.1b | A1
Total | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 8(c)(i) ---
8(c)(i) | Deduces that I + J = 0 | 2.2a | R1 | I + J = 0
Total | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 8(c)(ii) ---
8(c)(ii) | Explains that corresponding
impulses for any collision are
equal in magnitude and opposite
in direction | 2.4 | M1 | Impulses are equal in magnitude
and opposite in direction so will
always have a sum of 0
Deduces that the value of I + J
will not change | 2.2a | R1
Total | 2
Question total | 11
Paper total | 40
Two spheres A and B are free to move on a smooth horizontal surface.

The masses of A and B are 2 kg and 3 kg respectively.

Both A and B are initially at rest.

Sphere A is set in motion directly towards sphere B with speed 4 m s$^{-1}$ and subsequently collides with sphere B

The coefficient of restitution between the spheres is $e$

\begin{enumerate}[label=(\alph*)]
\item 
\begin{enumerate}[label=(\roman*)]
\item Show that the speed of B immediately after the collision is

$$\frac{8(1 + e)}{5}$$
[4 marks]

\item Find an expression, in terms of $e$, for the velocity of A immediately after the collision. [2 marks]
\end{enumerate}

\item It is given that the spheres both move in the same direction after the collision.

Find the range of possible values of $e$ [2 marks]

\item 
\begin{enumerate}[label=(\roman*)]
\item The impulse of sphere A on sphere B is $I$
The impulse of sphere B on sphere A is $J$

Given that the collision is perfectly inelastic, find the value of $I + J$ [1 mark]

\item State, giving a reason for your answer, whether the value found in part (c)(i) would change if the collision was not perfectly inelastic. [2 marks]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2021 Q8 [11]}}
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