OCR MEI Further Mechanics B AS 2022 June — Question 5 15 marks

Exam BoardOCR MEI
ModuleFurther Mechanics B AS (Further Mechanics B AS)
Year2022
SessionJune
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicOblique and successive collisions
TypeOblique collision, find velocities/angles
DifficultyChallenging +1.8 This is a challenging Further Mechanics oblique collision problem requiring multiple conservation principles, Newton's experimental law with coefficient of restitution, and optimization to find maximum/minimum speeds. The multi-part structure with conceptual explanation, extremal cases, and energy loss calculation demands sophisticated understanding beyond standard A-level mechanics, though the mathematical techniques themselves are accessible to Further Maths students.
Spec6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts
5 Two small uniform discs, A of mass \(2 m \mathrm {~kg}\) and B of mass \(3 m \mathrm {~kg}\), slide on a smooth horizontal surface and collide obliquely with smooth contact. Immediately before the collision, A is moving towards B along the line of centres with speed \(2 \mathrm {~ms} ^ { - 1 }\) and B is moving towards A with speed \(\sqrt { 3 } \mathrm {~ms} ^ { - 1 }\) in a direction making an angle of \(30 ^ { \circ }\) with the line of centres, as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{feb9a438-26b0-41d3-b044-6acd6efccde0-5_366_976_539_244}
  1. Explain how you know that the motion of A will be along the line of centres after the collision.
  2. - Determine the maximum possible speed of A after the collision.
    When the speed of B after the collision is a minimum, the loss of kinetic energy in the collision is 1.4625 J .
  3. Determine the value of \(m\).
Question 5:
AnswerMarks Guidance
5(a) Because the impact is smooth, so there is no
change in mom of A perp to loc.B1 2.4
equivalent
[1]
AnswerMarks Guidance
(b)2π‘šΓ—2βˆ’3π‘šΓ—βˆš3cos30Β° = 2π‘šπ‘Ž+3π‘šπ‘ M1
correct number of termsa, b are vels to right after
impact
AnswerMarks Guidance
βˆ’0.5 = 2π‘Ž+3𝑏A1 1.1
π‘βˆ’π‘Ž = βˆ’π‘’(βˆ’βˆš3cos30Β°βˆ’2)M1 3.4
π‘βˆ’π‘Ž = 3.5𝑒A1 1.1
5π‘Ž = βˆ’0.5βˆ’10.5𝑒M1 1.1
Max speed for A is 2.2A1 1.1
expression for a
AnswerMarks Guidance
When e = 1A1FT 1.1
speed for A is achieved when e = 1Must follow attempt to find
a in terms of e
[7]
AnswerMarks Guidance
(c)Solve for b M1
Allow a slipNote: first 4 marks for (b)
may be gained in (c)
1
[Min speed for B along loc (0) is] when e =
AnswerMarks Guidance
14A1 2.2a
1
Min speed for B is √3
AnswerMarks Guidance
2A1 1.1
1
for min speed is √3 o.e.
2
[3]
AnswerMarks
(d)2
Init KE = 1 Γ—2π‘šΓ—22+ 1 Γ—3π‘šΓ—( 3 ) (= 59 π‘š)
AnswerMarks Guidance
2 2 2 8M1 1.1
2 2
2 17
(√3) (= π‘š) [Total KE
2
method]
Fin KE = 1 Γ—2π‘šΓ— 12 [+ 1 Γ—3π‘šΓ—02] (= 1 π‘š)
AnswerMarks Guidance
2 4 2 16M1 3.1b
OR Γ—2π‘šΓ— + Γ—3π‘šΓ—
2 4 2
2
1 19
( √3) (= π‘š)
AnswerMarks
2 16Note – the 2 method marks
can only be awarded for
CONSISTENT KE, so either
both for LOC KE, or both
for total KE.
Allow use of β€˜their’ ΒΌ for
speed of A from use of 5π‘Ž =
βˆ’0.5βˆ’10.5𝑒
= 1 Γ—2π‘šΓ—22+ 1 Γ—3π‘šΓ—( 3 ) 2 βˆ’ 1 Γ—2π‘šΓ— 12 =
2 2 2 2 4
AnswerMarks Guidance
1.4625M1 1.1
2 2
2 1 12 1
(√3) βˆ’ Γ—2π‘šΓ— + Γ—3π‘šΓ—
2 4 2
2
1
( √3) = 1.4625
AnswerMarks
2Can be awarded for a mix of
KE methods, so M1 M0 M1
possible.
AnswerMarks Guidance
[m =] 0.2A1 1.1
can still result in an answer
very close to correct one.
[4]
Question 5:
5 | (a) | Because the impact is smooth, so there is no
change in mom of A perp to loc. | B1 | 2.4 | Allow β€œNo impulse perp to LOC” or
equivalent
[1]
(b) | 2π‘šΓ—2βˆ’3π‘šΓ—βˆš3cos30Β° = 2π‘šπ‘Ž+3π‘šπ‘ | M1 | 3.4 | Conservation of Momentum –
correct number of terms | a, b are vels to right after
impact
βˆ’0.5 = 2π‘Ž+3𝑏 | A1 | 1.1 | oe
π‘βˆ’π‘Ž = βˆ’π‘’(βˆ’βˆš3cos30Β°βˆ’2) | M1 | 3.4 | NEL (correct number of terms)
π‘βˆ’π‘Ž = 3.5𝑒 | A1 | 1.1 | Oe (must be consistent with CLM)
5π‘Ž = βˆ’0.5βˆ’10.5𝑒 | M1 | 1.1 | Solve for a in terms of e | Allow a slip
Max speed for A is 2.2 | A1 | 1.1 | Dep correct working to find
expression for a
When e = 1 | A1FT | 1.1 | Follow through as long as maximum
speed for A is achieved when e = 1 | Must follow attempt to find
a in terms of e
[7]
(c) | Solve for b | M1 | 1.1 | 5𝑏 = 7π‘’βˆ’0.5 oe
Allow a slip | Note: first 4 marks for (b)
may be gained in (c)
1
[Min speed for B along loc (0) is] when e =
14 | A1 | 2.2a
1
Min speed for B is √3
2 | A1 | 1.1 | Accept 0.866… or 0.87 | SC If M0 then can award B1
1
for min speed is √3 o.e.
2
[3]
(d) | 2
Init KE = 1 Γ—2π‘šΓ—22+ 1 Γ—3π‘šΓ—( 3 ) (= 59 π‘š)
2 2 2 8 | M1 | 1.1 | OR 1 Γ—2π‘šΓ—22+ 1 Γ—3π‘šΓ—
2 2
2 17
(√3) (= π‘š) [Total KE
2
method]
Fin KE = 1 Γ—2π‘šΓ— 12 [+ 1 Γ—3π‘šΓ—02] (= 1 π‘š)
2 4 2 16 | M1 | 3.1b | 1 12 1
OR Γ—2π‘šΓ— + Γ—3π‘šΓ—
2 4 2
2
1 19
( √3) (= π‘š)
2 16 | Note – the 2 method marks
can only be awarded for
CONSISTENT KE, so either
both for LOC KE, or both
for total KE.
Allow use of β€˜their’ ΒΌ for
speed of A from use of 5π‘Ž =
βˆ’0.5βˆ’10.5𝑒
= 1 Γ—2π‘šΓ—22+ 1 Γ—3π‘šΓ—( 3 ) 2 βˆ’ 1 Γ—2π‘šΓ— 12 =
2 2 2 2 4
1.4625 | M1 | 1.1 | OR 1 Γ—2π‘šΓ—22+ 1 Γ—3π‘šΓ—
2 2
2 1 12 1
(√3) βˆ’ Γ—2π‘šΓ— + Γ—3π‘šΓ—
2 4 2
2
1
( √3) = 1.4625
2 | Can be awarded for a mix of
KE methods, so M1 M0 M1
possible.
[m =] 0.2 | A1 | 1.1 | Correct working only! | Note a mistake in speed of A
can still result in an answer
very close to correct one.
[4]
5 Two small uniform discs, A of mass $2 m \mathrm {~kg}$ and B of mass $3 m \mathrm {~kg}$, slide on a smooth horizontal surface and collide obliquely with smooth contact.

Immediately before the collision, A is moving towards B along the line of centres with speed $2 \mathrm {~ms} ^ { - 1 }$ and B is moving towards A with speed $\sqrt { 3 } \mathrm {~ms} ^ { - 1 }$ in a direction making an angle of $30 ^ { \circ }$ with the line of centres, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{feb9a438-26b0-41d3-b044-6acd6efccde0-5_366_976_539_244}
\begin{enumerate}[label=(\alph*)]
\item Explain how you know that the motion of A will be along the line of centres after the collision.
\item - Determine the maximum possible speed of A after the collision.

\begin{itemize}
  \item Find the value of the coefficient of restitution in this case.
\item - Determine the minimum possible speed of B after the collision.
  \item Find the value of the coefficient of restitution in this case.
\end{itemize}

When the speed of B after the collision is a minimum, the loss of kinetic energy in the collision is 1.4625 J .
\item Determine the value of $m$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics B AS 2022 Q5 [15]}}
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