OCR MEI Further Mechanics Minor Specimen — Question 6 14 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Minor (Further Mechanics Minor)
SessionSpecimen
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMomentum and Collisions
TypeCollision with coefficient of restitution
DifficultyModerate -0.8 This is a straightforward multi-part mechanics question testing standard conservation of momentum and coefficient of restitution formulas. Parts (i)-(iv) involve direct application of well-rehearsed techniques with no conceptual surprises—part (i) is a 'show that' with the answer given, part (ii) uses the standard e-formula, part (iii) requires only a qualitative statement about momentum conservation, and part (iv) is another routine momentum calculation. The numbers are simple and the question structure is typical of textbook exercises, making it easier than average despite being Further Maths content.
Spec6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03k Newton's experimental law: direct impact

6 My cat Jeoffry has a mass of 4 kg and is sitting on rough ground near a sledge of mass 8 kg . The sledge is on a large area of smooth horizontal ice. Initially, the sledge is at rest and Jeoffry jumps and lands on it with a horizontal velocity of \(2.25 \mathrm {~ms} ^ { - 1 }\) parallel to the runners of the sledge. On landing, Jeoffry grips the sledge with his claws so that he does not move relative to the sledge in the subsequent motion.
  1. Show that the sledge with Jeoffry on it moves off with a speed of \(0.75 \mathrm {~ms} ^ { - 1 }\). With the sledge and Jeoffry moving at \(0.75 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the sledge collides directly with a stationary stone of mass 3 kg . The stone may move freely over the ice. The coefficient of restitution in the collision is \(\frac { 4 } { 15 }\).
  2. Calculate the velocity of the sledge and Jeoffry immediately after the collision. In a new situation, Jeoffry is initially sitting at rest on the sledge when it is stationary on the ice. He then walks from the back to the front of the sledge.
  3. Giving a brief reason for your answer, describe what happens to the sledge during his walk. Jeoffry is again sitting on the sledge when it is stationary on the ice. He jumps off and, after he has lost contact with the sledge, has a horizontal speed relative to the sledge of \(3 \mathrm {~ms} ^ { - 1 }\).
  4. Determine the speed of the sledge after Jeoffry loses contact with it. Fig. 7 shows a container for flowers which is a vertical cylindrical shell with a closed horizontal base. Its radius and its height are both \(\frac { 1 } { 2 } \mathrm {~m}\). Both the curved surface and the base are made of the same thin uniform material. The mass of the container is \(M \mathrm {~kg}\). \begin{figure}[h]
    \includegraphics[width=0.8\textwidth]{54711a46-83ce-4fb9-b6d3-53b264725c74-6_323_709_447_767} \caption{Fig. 7}
    \end{figure}
  5. Find, as a fraction, the height above the base of the centre of mass of the container. The container would hold \(\frac { 3 } { 2 } M \mathrm {~kg}\) of soil when full to the top. Some soil is put into the empty container and levelled with its top surface \(y \mathrm {~m}\) above the base. The centre of mass of the container with this much soil is zm above the base.
  6. Show that \(z = \frac { 1 + 9 y ^ { 2 } } { 6 ( 1 + 3 y ) }\).
  7. It is given that \(\frac { \mathrm { d } z } { \mathrm {~d} y } = 0\) when \(y = 0.14\) (to 2 significant figures) and that \(\frac { \mathrm { d } ^ { 2 } z } { \mathrm {~d} y ^ { 2 } } > 0\) at this value of \(y\). When putting in the soil, how might you use this information if the container is to be placed on slopes without it tipping over? \section*{END OF QUESTION PAPER} }{www.ocr.org.uk}) after the live examination series. If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.
    OCR is part of the

Question 6:
AnswerMarks Guidance
6(i) PCLM in direction of motion: 4 × 2.25 + 0 = 12v
so v = 0.75 so 0.75ms -1.M1
E1
AnswerMarks Guidance
[2]3.3
1.1Use of PCLM PCLM is Principle of
conservation of
linear momentum
AnswerMarks Guidance
6(ii) 0.75 m s -1 0
before
12 kg 3 kg
after
v m s -1 Vms -1
PCLM in initial direction of the sledge
12×0.75 = 12v + 3V
so 4v + V = 3
NEL C
V (cid:16)v 4
(cid:32)(cid:16)
0(cid:16)0.75 15
so V – v = 0.2
AnswerMarks
Solving, v = 0.56B1
M1
IA1
M1
A1
A1
AnswerMarks
[6]3.3
M
3.4
1.1
3.4
2.1
AnswerMarks
1.1Noe Award also if final answer correct
E
PCLM
Any form
NEL Must be the right way up.
Any form
AnswerMarks Guidance
6(iii) E
P
The sledge moves (in the opposite direction to Jeoffry)
No horizontal external force acts on the cat + sledge system,
AnswerMarks
so linear momentum is conserved.E1
E1
AnswerMarks
[2]2.2a
2.4
12kg
3kg
AnswerMarks Guidance
6(iv) Let the absolute speeds of J and sledge be u and U
respectively
0 0
before
4 kg 8 kg
after
u m s -1 Ums -1
PCLM
4u = 8U
relative motion
u + U = 3
AnswerMarks
so U = 1 so 1ms -1M1
B1
B1
I
A1
AnswerMarks
[4]M3.3
1.1
3.1b
AnswerMarks
1.1N
E
Correct analysis
PCLM
relative motion. May use, say, v and
v + 3
4kg
8kg
Question 6:
6 | (i) | PCLM in direction of motion: 4 × 2.25 + 0 = 12v
so v = 0.75 so 0.75ms -1. | M1
E1
[2] | 3.3
1.1 | Use of PCLM | PCLM is Principle of
conservation of
linear momentum
6 | (ii) | 0.75 m s -1 0
before
12 kg 3 kg
after
v m s -1 Vms -1
PCLM in initial direction of the sledge
12×0.75 = 12v + 3V
so 4v + V = 3
NEL C
V (cid:16)v 4
(cid:32)(cid:16)
0(cid:16)0.75 15
so V – v = 0.2
Solving, v = 0.56 | B1
M1
IA1
M1
A1
A1
[6] | 3.3
M
3.4
1.1
3.4
2.1
1.1 | Noe Award also if final answer correct
E
PCLM
Any form
NEL Must be the right way up.
Any form
6 | (iii) | E
P
The sledge moves (in the opposite direction to Jeoffry)
No horizontal external force acts on the cat + sledge system,
so linear momentum is conserved. | E1
E1
[2] | 2.2a
2.4
12kg
3kg
6 | (iv) | Let the absolute speeds of J and sledge be u and U
respectively
0 0
before
4 kg 8 kg
after
u m s -1 Ums -1
PCLM
4u = 8U
relative motion
u + U = 3
so U = 1 so 1ms -1 | M1
B1
B1
I
A1
[4] | M3.3
1.1
3.1b
1.1 | N
E
Correct analysis
PCLM
relative motion. May use, say, v and
v + 3
4kg
8kg
6 My cat Jeoffry has a mass of 4 kg and is sitting on rough ground near a sledge of mass 8 kg . The sledge is on a large area of smooth horizontal ice.

Initially, the sledge is at rest and Jeoffry jumps and lands on it with a horizontal velocity of $2.25 \mathrm {~ms} ^ { - 1 }$ parallel to the runners of the sledge. On landing, Jeoffry grips the sledge with his claws so that he does not move relative to the sledge in the subsequent motion.\\
(i) Show that the sledge with Jeoffry on it moves off with a speed of $0.75 \mathrm {~ms} ^ { - 1 }$.

With the sledge and Jeoffry moving at $0.75 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the sledge collides directly with a stationary stone of mass 3 kg . The stone may move freely over the ice. The coefficient of restitution in the collision is $\frac { 4 } { 15 }$.\\
(ii) Calculate the velocity of the sledge and Jeoffry immediately after the collision.

In a new situation, Jeoffry is initially sitting at rest on the sledge when it is stationary on the ice. He then walks from the back to the front of the sledge.\\
(iii) Giving a brief reason for your answer, describe what happens to the sledge during his walk.

Jeoffry is again sitting on the sledge when it is stationary on the ice. He jumps off and, after he has lost contact with the sledge, has a horizontal speed relative to the sledge of $3 \mathrm {~ms} ^ { - 1 }$.\\
(iv) Determine the speed of the sledge after Jeoffry loses contact with it.

Fig. 7 shows a container for flowers which is a vertical cylindrical shell with a closed horizontal base. Its radius and its height are both $\frac { 1 } { 2 } \mathrm {~m}$. Both the curved surface and the base are made of the same thin uniform material. The mass of the container is $M \mathrm {~kg}$.

\begin{figure}[h]
\begin{center}
  \includegraphics[width=0.8\textwidth]{54711a46-83ce-4fb9-b6d3-53b264725c74-6_323_709_447_767}
\caption{Fig. 7}
\end{center}
\end{figure}

(i) Find, as a fraction, the height above the base of the centre of mass of the container.

The container would hold $\frac { 3 } { 2 } M \mathrm {~kg}$ of soil when full to the top.

Some soil is put into the empty container and levelled with its top surface $y \mathrm {~m}$ above the base. The centre of mass of the container with this much soil is zm above the base.\\
(ii) Show that $z = \frac { 1 + 9 y ^ { 2 } } { 6 ( 1 + 3 y ) }$.\\
(iii) It is given that $\frac { \mathrm { d } z } { \mathrm {~d} y } = 0$ when $y = 0.14$ (to 2 significant figures) and that $\frac { \mathrm { d } ^ { 2 } z } { \mathrm {~d} y ^ { 2 } } > 0$ at this value of $y$.

When putting in the soil, how might you use this information if the container is to be placed on slopes without it tipping over?

\section*{END OF QUESTION PAPER}

}{www.ocr.org.uk}) after the live examination series.

If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity.

For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.\\
OCR is part of the

\hfill \mbox{\textit{OCR MEI Further Mechanics Minor  Q6 [14]}}