OCR Further Mechanics 2020 November — Question 6 12 marks

Exam BoardOCR
ModuleFurther Mechanics (Further Mechanics)
Year2020
SessionNovember
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicOblique and successive collisions
TypeOblique collision, direction deflected given angle
DifficultyChallenging +1.8 This is an oblique collision problem requiring resolution of velocities, conservation of momentum in the line of centres, Newton's experimental law, and the geometric constraint that final velocity is perpendicular to initial velocity. It demands careful vector decomposition, simultaneous equations, and multi-step reasoning across several mechanics principles, making it significantly harder than standard collision questions but still within established Further Mechanics techniques.
Spec6.03a Linear momentum: p = mv6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts
6 Two smooth circular discs \(A\) and \(B\) are moving on a horizontal plane. The masses of \(A\) and \(B\) are 3 kg and 4 kg respectively. At the instant before they collide
  • the velocity of \(A\) is \(2 \mathrm {~ms} ^ { - 1 }\) at an angle of \(60 ^ { \circ }\) to the line joining their centres,
  • the velocity of \(B\) is \(5 \mathrm {~ms} ^ { - 1 }\) towards \(A\) along the line joining their centres (see Fig. 6).
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{831ba5da-df19-43bb-b163-02bbddb4e2b8-5_490_1047_470_244} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure} Given that the velocity of \(A\) after the collision is perpendicular to the velocity of \(A\) before the collision, find
  1. the coefficient of restitution between \(A\) and \(B\),
  2. the total loss of kinetic energy as a result of the collision.
Question 6:
AnswerMarks Guidance
6(a) u Ay = 3 or awrt 1.73
v = u (= 3)
Ay Ay
 1  v 
 . Ax =0⇒v =−3
    Ax
 3  3
3×2cos60° + 4×–5 = 3×–3 + 4v
Bx
v = –2
Bx
−2−−3
e=
2cos60°−−5
1
e= or awrt 0.17
AnswerMarks
6B1
B1
M1
M1
A1
M1
AnswerMarks
A13.1b
3.1b
2.2a
3.1b
1.1
3.1b
AnswerMarks
1.1Correct perpendicular
component of velocity of A
before collision
This component of A’s
velocity is unchanged by
collision
3
Or tan30°= . Using
v
Ax
perpendicularity of A’s
velocities to derive a value for
v
Ax
Conservation of momentum
with 4 terms
AnswerMarks
Restitutionor seen
𝑣𝑣𝐴𝐴 =2√3
Could be positive if shown in
diagram
Allow one sign slip
Could be positive
Allow one sign slip
Alternative method for last 5 marks
3×2cos60° + 4×–5 = 3v + 4v
AnswerMarks Guidance
Ax BxM1 Conservation of momentum
Ax Bx
v −v
e= Bx Ax
AnswerMarks Guidance
2cos60°−−5M1 M1
Bx Axv – v = 6e
Bx Ax
−17−24e
v =
AnswerMarks Guidance
Ax 7A1 −17+18e
v =
Bx 7
−17−24e
 1 
 
 . 7 =0
   
 3  
AnswerMarks
 3 −17−24e
 1 
 
 . 7 =0
   
 3  
AnswerMarks Guidance
 3 M1 3
Or tan30°= .
17+24e
 
 7 
Using perpendicularity of A’s
velocities.
1
e= or awrt 0.17
AnswerMarks
6A1
[7]
AnswerMarks Guidance
6(b) ( )2
v2 = 3 +(−3)2 soi
A
1 1
×4×52 + ×3×22
2 2
1 1
×4×22 + ×3×12
2 2
56 or 26
AnswerMarks
56 – 26 = 30 JM1
M1
M1
A1
AnswerMarks
A13.1b
1.1
1.1
1.1
AnswerMarks
1.1Attempt to find final speed
(squared) of A vectorially
Attempt to find total initial
KE
Attempt to find total final KE
Either correctly calculated
AnswerMarks
Not –30v 2 = 12
A
Can be implied by correct
expressions and 30 seen.
AnswerMarks Guidance
Alternative method for last 4 marksM1 Attempt to find KE change
for B
1 1
×4×52 − ×4×22
2 2
1 1
×3×22 − ×3×12
AnswerMarks Guidance
2 2M1 Attempt to find KE change
for A
AnswerMarks Guidance
42 or 12 so loss = 42 + (–12)42 or 12 so loss = 42 + (–12) M1
expressions and 30 seen.
and correctly used to find

total energy change

AnswerMarks Guidance
= 30 JA1 Not –30
[5]
M1
Attempt to find KE change
for B
Attempt to find KE change
for A
Can be implied by correct
expressions and 30 seen.
Question 6:
6 | (a) | u Ay = 3 or awrt 1.73
v = u (= 3)
Ay Ay
 1  v 
 . Ax =0⇒v =−3
    Ax
 3  3
3×2cos60° + 4×–5 = 3×–3 + 4v
Bx
v = –2
Bx
−2−−3
e=
2cos60°−−5
1
e= or awrt 0.17
6 | B1
B1
M1
M1
A1
M1
A1 | 3.1b
3.1b
2.2a
3.1b
1.1
3.1b
1.1 | Correct perpendicular
component of velocity of A
before collision
This component of A’s
velocity is unchanged by
collision
3
Or tan30°= . Using
v
Ax
perpendicularity of A’s
velocities to derive a value for
v
Ax
Conservation of momentum
with 4 terms
Restitution | or seen
𝑣𝑣𝐴𝐴 =2√3
Could be positive if shown in
diagram
Allow one sign slip
Could be positive
Allow one sign slip
Alternative method for last 5 marks
3×2cos60° + 4×–5 = 3v + 4v
Ax Bx | M1 | Conservation of momentum | 3v + 4v = –17
Ax Bx
v −v
e= Bx Ax
2cos60°−−5 | M1 | M1 | Restitution | Restitution | v – v = 6e
Bx Ax | v – v = 6e
Bx Ax
−17−24e
v =
Ax 7 | A1 | −17+18e
v =
Bx 7
−17−24e
 1 
 
 . 7 =0
   
 3  
 3  | −17−24e
 1 
 
 . 7 =0
   
 3  
 3  | M1 | 3
Or tan30°= .
17+24e
 
 7 
Using perpendicularity of A’s
velocities.
1
e= or awrt 0.17
6 | A1
[7]
6 | (b) | ( )2
v2 = 3 +(−3)2 soi
A
1 1
×4×52 + ×3×22
2 2
1 1
×4×22 + ×3×12
2 2
56 or 26
56 – 26 = 30 J | M1
M1
M1
A1
A1 | 3.1b
1.1
1.1
1.1
1.1 | Attempt to find final speed
(squared) of A vectorially
Attempt to find total initial
KE
Attempt to find total final KE
Either correctly calculated
Not –30 | v 2 = 12
A
Can be implied by correct
expressions and 30 seen.
Alternative method for last 4 marks | M1 | Attempt to find KE change
for B
1 1
×4×52 − ×4×22
2 2
1 1
×3×22 − ×3×12
2 2 | M1 | Attempt to find KE change
for A
42 or 12 so loss = 42 + (–12) | 42 or 12 so loss = 42 + (–12) | M1 | Either correctly calculated | Can be implied by correct
expressions and 30 seen.
and correctly used to find
total energy change
= 30 J | A1 | Not –30
[5]
M1
Attempt to find KE change
for B
Attempt to find KE change
for A
Can be implied by correct
expressions and 30 seen.
6 Two smooth circular discs $A$ and $B$ are moving on a horizontal plane. The masses of $A$ and $B$ are 3 kg and 4 kg respectively. At the instant before they collide

\begin{itemize}
  \item the velocity of $A$ is $2 \mathrm {~ms} ^ { - 1 }$ at an angle of $60 ^ { \circ }$ to the line joining their centres,
  \item the velocity of $B$ is $5 \mathrm {~ms} ^ { - 1 }$ towards $A$ along the line joining their centres (see Fig. 6).
\end{itemize}

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{831ba5da-df19-43bb-b163-02bbddb4e2b8-5_490_1047_470_244}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}

Given that the velocity of $A$ after the collision is perpendicular to the velocity of $A$ before the collision, find
\begin{enumerate}[label=(\alph*)]
\item the coefficient of restitution between $A$ and $B$,
\item the total loss of kinetic energy as a result of the collision.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Mechanics 2020 Q6 [12]}}
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