Question 6 - A-Level Maths

AQA Further AS Paper 2 Mechanics 2021 June — Question 6 5 marks

Exam Board	AQA
Module	Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics)
Year	2021
Session	June
Marks	5
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Momentum and Collisions
Type	Calculate impulse from force-time data
Difficulty	Moderate -0.3 This is a straightforward two-part impulse-momentum question. Part (a) requires direct application of the impulse formula with careful attention to sign (2 marks of routine calculation). Part (b) involves integrating a quadratic force function and equating to the impulse from part (a) - standard mechanics technique requiring no novel insight, though the integration and algebra add modest computational demand for 3 marks total.
Spec	6.03f Impulse-momentum: relation 6.03h Variable force impulse: using integration

A ball of mass 0.15 kg is hit directly by a vertical cricket bat. Immediately before the impact, the ball is travelling horizontally with speed 28 m s\(^{-1}\) Immediately after the impact, the ball is travelling horizontally with speed 14 m s\(^{-1}\) in the opposite direction.

Find the magnitude of the impulse exerted by the bat on the ball. [2 marks]
In a simple model the force, \(F\) newtons, exerted by the bat on the ball, \(t\) seconds after the initial impact, is given by \(F = 10kt (0.05 - t)\) where \(k\) is a constant. Given the ball is in contact with the bat for 0.05 seconds, find the value of \(k\) [3 marks]

Show mark scheme Show mark scheme source

Question 6:

Answer	Marks	Guidance
6(a)	Uses correct formula for impulse	1.1a

I = 0.15(14) – (0.15)(–28)

I = 6.3 Ns

Obtains correct answer of 6.3

Condone missing or incorrect

Answer	Marks	Guidance
units	1.1b	A1
Total	2
Q	Marking instructions	AO

Answer	Marks
6(b)	Forms an equation involving

appropriate integral using ‘their’

Answer	Marks	Guidance
value from part (a)	3.4	M1

0.05 𝑘𝑘∫0 10𝑡𝑡(0.05−𝑡𝑡)𝑑𝑑𝑡𝑡

0.05

1 � 10𝑡𝑡(0.05− 𝑡𝑡)𝑑𝑑𝑡𝑡 =

0 k = 4800 × 6.3 = 30244800 0

Evaluates definite integral

Answer	Marks	Guidance
correctly	1.1b	B1

Solves equation to find ‘their’

value of k

FT their impulse from part (a) or

their incorrect value for the

Answer	Marks	Guidance
definite integral	1.1b	A1F
Total	3
Question total	5
Q	Marking instructions	AO

Question 6:
--- 6(a) ---
6(a) | Uses correct formula for impulse | 1.1a | M1 | I = mv – mu
I = 0.15(14) – (0.15)(–28)
I = 6.3 Ns
Obtains correct answer of 6.3
Condone missing or incorrect
units | 1.1b | A1
Total | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 6(b) ---
6(b) | Forms an equation involving
appropriate integral using ‘their’
value from part (a) | 3.4 | M1 | 6.3 =
0.05
𝑘𝑘∫0 10𝑡𝑡(0.05−𝑡𝑡)𝑑𝑑𝑡𝑡
0.05
1
� 10𝑡𝑡(0.05− 𝑡𝑡)𝑑𝑑𝑡𝑡 =
0 k = 4800 × 6.3 = 30244800 0
Evaluates definite integral
correctly | 1.1b | B1
Solves equation to find ‘their’
value of k
FT their impulse from part (a) or
their incorrect value for the
definite integral | 1.1b | A1F
Total | 3
Question total | 5
Q | Marking instructions | AO | Marks | Typical solution

Show LaTeX source

A ball of mass 0.15 kg is hit directly by a vertical cricket bat.

Immediately before the impact, the ball is travelling horizontally with speed 28 m s$^{-1}$

Immediately after the impact, the ball is travelling horizontally with speed 14 m s$^{-1}$ in the opposite direction.

\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the impulse exerted by the bat on the ball. [2 marks]
\item In a simple model the force, $F$ newtons, exerted by the bat on the ball, $t$ seconds after the initial impact, is given by

$F = 10kt (0.05 - t)$

where $k$ is a constant.

Given the ball is in contact with the bat for 0.05 seconds, find the value of $k$ [3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2021 Q6 [5]}}

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