AQA AS Paper 1 2021 June — Question 15 10 marks

Exam BoardAQA
ModuleAS Paper 1 (AS Paper 1)
Year2021
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPulley systems
TypeString breaks during motion
DifficultyModerate -0.3 This is a standard AS-level mechanics problem involving connected particles with friction. Part (b) requires setting up two F=ma equations and solving simultaneously (routine but multi-step), while part (c) applies SUVAT equations. The 'show that' format in (b) reduces difficulty, and all techniques are standard textbook exercises requiring no novel insight, making it slightly easier than average.
Spec3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium
In this question, use \(g = 10\) m s⁻² A box, B, of mass 4 kg lies at rest on a fixed rough horizontal shelf. One end of a light string is connected to B. The string passes over a smooth peg, attached to the end of the shelf. The other end of the string is connected to particle, P, of mass 1 kg, which hangs freely below the shelf as shown in the diagram below. \includegraphics{figure_15} B is initially held at rest with the string taut. B is then released. B and P both move with constant acceleration \(a\) m s⁻² As B moves across the shelf it experiences a total resistance force of 5 N
  1. State one type of force that would be included in the total resistance force. [1 mark]
  2. Show that \(a = 1\) [4 marks]
  3. When B has moved forward exactly 20 cm the string breaks. Find how much further B travels before coming to rest. [4 marks]
  4. State one assumption you have made when finding your solutions in parts (b) or (c). [1 mark]
Question 15:
AnswerMarks
15(a)States either friction or air
resistance or both but no
AnswerMarks Guidance
incorrect forces.3.3 B1
Subtotal1
QMarking instructions AO
AnswerMarks
15(b)Uses to form a three
term equation modelling the
AnswerMarks Guidance
motion𝐹𝐹 o=f Pπ‘šπ‘š π‘Žπ‘Ž3.3 M1
π‘‡π‘‡βˆ’5 = 4Γ— π‘Žπ‘Ž
5 = 5π‘Žπ‘Ž
π‘Žπ‘Ž = 1
Uses to form a three
term equation modelling the
AnswerMarks Guidance
motion𝐹𝐹 o=f Bπ‘šπ‘š π‘Žπ‘Ž3.3 M1
Obtains two correct equations1.1b A1
Completes a rigorous argument
AnswerMarks Guidance
to show that2.1 R1
π‘Žπ‘Ž =1
AnswerMarks Guidance
Subtotal4
QMarking instructions AO
AnswerMarks Guidance
15(c)Uses appropriate suvat equation
with given values to find v or v23.3 M1
𝑣𝑣 = 0+2Γ— 1Γ—0.2 = 0.4
–5 = 4
π‘Žπ‘Ž
π‘Žπ‘Ž = βˆ’ 1.25
2
0 = 0.4+( 2)(βˆ’1.25)𝑠𝑠
m
𝑠𝑠 = 0m.1,6 to 1 sf
Uses to find
acceleration of B after string
AnswerMarks Guidance
breaks𝐹𝐹 = π‘šπ‘šπ‘Žπ‘Ž3.4 M1
Obtains1.1b A1
π‘Žπ‘Ž = βˆ’1.25
Uses appropriate suvat equation
to obtain 0.16m.
AnswerMarks Guidance
Must include units.1.1b A1
Subtotal4 𝑠𝑠 = 0.2
QMarking instructions AO
AnswerMarks
15(d)States a valid assumption
Accept:
β€’ No air resistance to motion of
P
β€’ string remains parallel to table
β€’ B does not reach end of table
before it stops
β€’ P does not hit floor before
AnswerMarks Guidance
string breaks3.5b E1
Subtotal1
Question Total10 
Question 15:
--- 15(a) ---
15(a) | States either friction or air
resistance or both but no
incorrect forces. | 3.3 | B1 | Friction
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 15(b) ---
15(b) | Uses to form a three
term equation modelling the
motion𝐹𝐹 o=f Pπ‘šπ‘š π‘Žπ‘Ž | 3.3 | M1 | 10βˆ’π‘‡π‘‡ = 1 Γ— π‘Žπ‘Ž
π‘‡π‘‡βˆ’5 = 4Γ— π‘Žπ‘Ž
5 = 5π‘Žπ‘Ž
π‘Žπ‘Ž = 1
Uses to form a three
term equation modelling the
motion𝐹𝐹 o=f Bπ‘šπ‘š π‘Žπ‘Ž | 3.3 | M1
Obtains two correct equations | 1.1b | A1
Completes a rigorous argument
to show that | 2.1 | R1
π‘Žπ‘Ž =1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 15(c) ---
15(c) | Uses appropriate suvat equation
with given values to find v or v2 | 3.3 | M1 | 2
𝑣𝑣 = 0+2Γ— 1Γ—0.2 = 0.4
–5 = 4
π‘Žπ‘Ž
π‘Žπ‘Ž = βˆ’ 1.25
2
0 = 0.4+( 2)(βˆ’1.25)𝑠𝑠
m
𝑠𝑠 = 0m.1,6 to 1 sf
Uses to find
acceleration of B after string
breaks𝐹𝐹 = π‘šπ‘šπ‘Žπ‘Ž | 3.4 | M1
Obtains | 1.1b | A1
π‘Žπ‘Ž = βˆ’1.25
Uses appropriate suvat equation
to obtain 0.16m.
Must include units. | 1.1b | A1
Subtotal | 4 | 𝑠𝑠 = 0.2
Q | Marking instructions | AO | Marks | Typical solution
--- 15(d) ---
15(d) | States a valid assumption
Accept:
β€’ No air resistance to motion of
P
β€’ string remains parallel to table
β€’ B does not reach end of table
before it stops
β€’ P does not hit floor before
string breaks | 3.5b | E1 | String is inextensible
Subtotal | 1
Question Total | 10
In this question, use $g = 10$ m s⁻²

A box, B, of mass 4 kg lies at rest on a fixed rough horizontal shelf.

One end of a light string is connected to B.

The string passes over a smooth peg, attached to the end of the shelf.

The other end of the string is connected to particle, P, of mass 1 kg, which hangs freely below the shelf as shown in the diagram below.

\includegraphics{figure_15}

B is initially held at rest with the string taut.

B is then released.

B and P both move with constant acceleration $a$ m s⁻²

As B moves across the shelf it experiences a total resistance force of 5 N

\begin{enumerate}[label=(\alph*)]
\item State one type of force that would be included in the total resistance force.
[1 mark]

\item Show that $a = 1$
[4 marks]

\item When B has moved forward exactly 20 cm the string breaks.

Find how much further B travels before coming to rest.
[4 marks]

\item State one assumption you have made when finding your solutions in parts (b) or (c).
[1 mark]
\end{enumerate}

\hfill \mbox{\textit{AQA AS Paper 1 2021 Q15 [10]}}
This paper (15 questions)
View full paper
Q1 1 Q2 1 Q3 3 Q4 9 Q5 6 Q6 7 Q7 12 Q8 7 Q9 7 Q10 1 Q11 1 Q12 4 Q13 5 Q14 6 Q15 10