OCR MEI Further Mechanics B AS Specimen — Question 4 8 marks

Exam BoardOCR MEI
ModuleFurther Mechanics B AS (Further Mechanics B AS)
SessionSpecimen
Marks8
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Mark schemeDownload PDF ↗
TopicOblique and successive collisions
TypeOblique collision, find velocities/angles
DifficultyStandard +0.8 This oblique collision problem requires understanding that perpendicular components are unchanged in smooth collisions, applying both momentum conservation and coefficient of restitution in the parallel direction, then using Pythagoras to find resultant speed. It's a multi-step problem requiring conceptual understanding beyond routine mechanics, typical of Further Mechanics content which is inherently more challenging than standard A-level.
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
4 Two uniform circular discs with the same radius, A of mass 1 kg and B of mass 5.25 kg , slide on a smooth horizontal surface and collide obliquely with smooth contact. Fig. 4 gives information about the velocities of the discs just before and just after the collision.
  • The line XY passes through the centres of the discs at the moment of collision
  • The components parallel and perpendicular to XY of the velocities of A are shown
  • Before the collision, B is at rest and after it is moving at \(2 \mathrm {~ms} ^ { - 1 }\) in the direction XY
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a01b2e46-e213-4f20-bc2e-5852061d8b91-4_582_1716_721_155} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure} The coefficient of restitution between the two discs is \(\frac { 2 } { 3 }\).
  1. Find the values of \(U\) and \(u\).
  2. What information in the question tells you that \(v = V\) ? The speed of disc A before the collision is \(8.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
  3. Find the speed of disc A after the collision. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{a01b2e46-e213-4f20-bc2e-5852061d8b91-5_398_396_397_475} \captionsetup{labelformat=empty} \caption{Fig. 5.1}
    \end{figure} \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{a01b2e46-e213-4f20-bc2e-5852061d8b91-5_399_332_399_945} \captionsetup{labelformat=empty} \caption{Fig. 5.2}
    \end{figure} \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{a01b2e46-e213-4f20-bc2e-5852061d8b91-5_305_326_493_1354} \captionsetup{labelformat=empty} \caption{Fig. 5.3}
    \end{figure} Fig. 5.1 shows a vertical light elastic spring. It is fixed to a horizontal table at one end. Fig 5.2 shows the spring with a particle of mass \(m \mathrm {~kg}\) attached to it at the other end. The system is in equilibrium when the spring is compressed by a distance \(h \mathrm {~m}\).
Question 4:
AnswerMarks Guidance
4(i) AB → PCLM U + 0 = –u + 2×5.25
so U = –u + 10.5
2(cid:16)((cid:16)u) 2
NEL (cid:32)(cid:16)
0(cid:16)U 3
so 6 + 3u = 2U
Solving
AnswerMarks
u = 3, U = 7.5M1
A1
M1
A1
A1
AnswerMarks
[5]3.4
1.1
3.4
1.1
AnswerMarks
1.1Any form
Must be right way up. Accept sign errors.
N
Any form
BC
AnswerMarks Guidance
4(ii) Contact is smooth
[1]3.3 E
4(iii) For disc A,
initial speed of 8.5 m s -1 gives
V (cid:32) 8.52 (cid:16)7.52 (cid:32)4 m s -1
AnswerMarks
final speed is ((cid:16)3)2 (cid:14)42 (cid:32)5 m s -1M1
A1
AnswerMarks
[2]3.1b
I
AnswerMarks
1.1M
FT their U and u
Question 4:
4 | (i) | AB → PCLM U + 0 = –u + 2×5.25
so U = –u + 10.5
2(cid:16)((cid:16)u) 2
NEL (cid:32)(cid:16)
0(cid:16)U 3
so 6 + 3u = 2U
Solving
u = 3, U = 7.5 | M1
A1
M1
A1
A1
[5] | 3.4
1.1
3.4
1.1
1.1 | Any form
Must be right way up. Accept sign errors.
N
Any form
BC
4 | (ii) | Contact is smooth | B1
[1] | 3.3 | E
4 | (iii) | For disc A,
initial speed of 8.5 m s -1 gives
V (cid:32) 8.52 (cid:16)7.52 (cid:32)4 m s -1
final speed is ((cid:16)3)2 (cid:14)42 (cid:32)5 m s -1 | M1
A1
[2] | 3.1b
I
1.1 | M
FT their U and u
4 Two uniform circular discs with the same radius, A of mass 1 kg and B of mass 5.25 kg , slide on a smooth horizontal surface and collide obliquely with smooth contact.

Fig. 4 gives information about the velocities of the discs just before and just after the collision.

\begin{itemize}
  \item The line XY passes through the centres of the discs at the moment of collision
  \item The components parallel and perpendicular to XY of the velocities of A are shown
  \item Before the collision, B is at rest and after it is moving at $2 \mathrm {~ms} ^ { - 1 }$ in the direction XY
\end{itemize}

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{a01b2e46-e213-4f20-bc2e-5852061d8b91-4_582_1716_721_155}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}

The coefficient of restitution between the two discs is $\frac { 2 } { 3 }$.\\
(i) Find the values of $U$ and $u$.\\
(ii) What information in the question tells you that $v = V$ ?

The speed of disc A before the collision is $8.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(iii) Find the speed of disc A after the collision.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{a01b2e46-e213-4f20-bc2e-5852061d8b91-5_398_396_397_475}
\captionsetup{labelformat=empty}
\caption{Fig. 5.1}
\end{center}
\end{figure}

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{a01b2e46-e213-4f20-bc2e-5852061d8b91-5_399_332_399_945}
\captionsetup{labelformat=empty}
\caption{Fig. 5.2}
\end{center}
\end{figure}

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{a01b2e46-e213-4f20-bc2e-5852061d8b91-5_305_326_493_1354}
\captionsetup{labelformat=empty}
\caption{Fig. 5.3}
\end{center}
\end{figure}

Fig. 5.1 shows a vertical light elastic spring. It is fixed to a horizontal table at one end. Fig 5.2 shows the spring with a particle of mass $m \mathrm {~kg}$ attached to it at the other end. The system is in equilibrium when the spring is compressed by a distance $h \mathrm {~m}$.\\

\hfill \mbox{\textit{OCR MEI Further Mechanics B AS  Q4 [8]}}
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