AQA Further Paper 3 Mechanics 2024 June — Question 7 10 marks

Exam BoardAQA
ModuleFurther Paper 3 Mechanics (Further Paper 3 Mechanics)
Year2024
SessionJune
Marks10
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Mark schemeDownload PDF ↗
TopicImpulse and momentum (advanced)
TypeCollision with fixed wall
DifficultyStandard +0.3 This is a standard oblique collision mechanics problem requiring resolution of velocities parallel and perpendicular to the wall, application of Newton's experimental law (e=0.7), and impulse calculation. While it involves multiple steps and careful component work, it follows a well-established procedure taught in Further Maths mechanics with no novel insight required. The final part tests conceptual understanding of friction effects.
Spec6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts
A sphere, of mass 0.2 kg, moving on a smooth horizontal surface, collides with a fixed wall. Before the collision the sphere moves with speed 5 m s\(^{-1}\) at an angle of 60° to the wall. After the collision the sphere moves with speed \(v\) m s\(^{-1}\) at an angle of \(\theta\)° to the wall. The velocities are shown in the diagram below. \includegraphics{figure_7} The coefficient of restitution between the wall and the sphere is 0.7
  1. Assume that the wall is smooth.
    1. Find the value of \(v\) Give your answer to two significant figures. [4 marks]
    2. Find the value of \(\theta\) Give your answer to the nearest whole number. [2 marks]
    3. Find the magnitude of the impulse exerted on the sphere by the wall. Give your answer to two significant figures. [2 marks]
  2. In reality the wall is not smooth. Explain how this would cause a change in the magnitude of the impulse calculated in part (a)(iii). [2 marks]
Question 7:
AnswerMarks
7(a)(i)Forms two expressions for motion
by resolving parallel and
AnswerMarks Guidance
perpendicular to the wall.3.3 M1
0.7 × 5sin 60° = vsin θ
v = ( 5 c o s 6 0  ) 2 + ( 3 . 5 s in 6 0  ) 2
= 3 . 9
AnswerMarks Guidance
Obtains two correct expressions.1.1b A1
Eliminates θ to find a value for v1.1a M1
Obtains AWRT 3.91.1b A1
Subtotal4
QMarking Instructions AO
AnswerMarks
7(a)(ii)Uses their value for v or eliminates
v to find an expression for tan θ,
AnswerMarks Guidance
sin θ or cos θ3.4 M1
tan θ =
5cos60
θ =50
AnswerMarks Guidance
Obtains AWRT 501.1b A1
Subtotal2
QMarking Instructions AO
AnswerMarks
7(a)(iii)Uses the impulse equation with
components of velocities
perpendicular to the wall with
AnswerMarks Guidance
opposite signs.3.4 M1
– (–0.2 × 5sin 60°)
= 1.5 Ns
Obtains AWRT 1.5.
Must be positive.
AnswerMarks Guidance
Condone missing/incorrect units.1.1b A1
Subtotal2
QMarking Instructions AO
AnswerMarks Guidance
7(b)Explains what happens to one
component of the impulse.2.4 M1
change in the velocities
perpendicular to the wall and so
the magnitude of the component of
the impulse perpendicular to the
wall would be unchanged.
There will be a component of the
impulse parallel to the wall and so
the magnitude of the total impulse
would increase.
Explains what will happen to the
other component of the impulse
and that the magnitude will
AnswerMarks Guidance
increase.2.4 R1
Subtotal2
Question total10
QMarking Instructions AO 
Question 7:
--- 7(a)(i) ---
7(a)(i) | Forms two expressions for motion
by resolving parallel and
perpendicular to the wall. | 3.3 | M1 | 5cos 60° = vcos θ
0.7 × 5sin 60° = vsin θ
v = ( 5 c o s 6 0  ) 2 + ( 3 . 5 s in 6 0  ) 2
= 3 . 9
Obtains two correct expressions. | 1.1b | A1
Eliminates θ to find a value for v | 1.1a | M1
Obtains AWRT 3.9 | 1.1b | A1
Subtotal | 4
Q | Marking Instructions | AO | Marks | Typical Solution
--- 7(a)(ii) ---
7(a)(ii) | Uses their value for v or eliminates
v to find an expression for tan θ,
sin θ or cos θ | 3.4 | M1 | 3.5sin 60
tan θ =
5cos60
θ =50
Obtains AWRT 50 | 1.1b | A1
Subtotal | 2
Q | Marking Instructions | AO | Marks | Typical Solution
--- 7(a)(iii) ---
7(a)(iii) | Uses the impulse equation with
components of velocities
perpendicular to the wall with
opposite signs. | 3.4 | M1 | I = 0.2 × 3.929sin 50.48°
– (–0.2 × 5sin 60°)
= 1.5 Ns
Obtains AWRT 1.5.
Must be positive.
Condone missing/incorrect units. | 1.1b | A1
Subtotal | 2
Q | Marking Instructions | AO | Marks | Typical Solution
--- 7(b) ---
7(b) | Explains what happens to one
component of the impulse. | 2.4 | M1 | Any friction would not cause a
change in the velocities
perpendicular to the wall and so
the magnitude of the component of
the impulse perpendicular to the
wall would be unchanged.
There will be a component of the
impulse parallel to the wall and so
the magnitude of the total impulse
would increase.
Explains what will happen to the
other component of the impulse
and that the magnitude will
increase. | 2.4 | R1
Subtotal | 2
Question total | 10
Q | Marking Instructions | AO | Marks | Typical Solution
A sphere, of mass 0.2 kg, moving on a smooth horizontal surface, collides with a fixed wall.

Before the collision the sphere moves with speed 5 m s$^{-1}$ at an angle of 60° to the wall.

After the collision the sphere moves with speed $v$ m s$^{-1}$ at an angle of $\theta$° to the wall.

The velocities are shown in the diagram below.

\includegraphics{figure_7}

The coefficient of restitution between the wall and the sphere is 0.7

\begin{enumerate}[label=(\alph*)]
\item Assume that the wall is smooth.

\begin{enumerate}[label=(\roman*)]
\item Find the value of $v$

Give your answer to two significant figures.
[4 marks]

\item Find the value of $\theta$

Give your answer to the nearest whole number.
[2 marks]

\item Find the magnitude of the impulse exerted on the sphere by the wall.

Give your answer to two significant figures.
[2 marks]
\end{enumerate}

\item In reality the wall is not smooth.

Explain how this would cause a change in the magnitude of the impulse calculated in part (a)(iii).
[2 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2024 Q7 [10]}}
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