OCR MEI Further Mechanics Major 2021 November — Question 11 16 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
Year2021
SessionNovember
Marks16
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeRotating disc with friction
DifficultyChallenging +1.2 This is a multi-part collision problem requiring conservation of momentum, Newton's law of restitution, and circular motion kinematics. Part (a) is standard A-level mechanics, part (b) requires careful tracking of relative motion around a circle to reach a given result, and part (c) extends the analysis. While it involves several steps and careful algebra, the techniques are all standard Further Mechanics content with no novel conceptual leaps required.
Spec6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03j Perfectly elastic/inelastic: collisions6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r
11 Two small uniform smooth spheres A and B , of equal radius, have masses 4 kg and 3 kg respectively. The spheres are placed in a smooth horizontal circular groove. The coefficient of restitution between the spheres is \(e\), where \(e > \frac { 2 } { 5 }\). At a given instant B is at rest and A is set moving along the groove with speed \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\). It may be assumed that in the subsequent motion the two spheres do not leave the groove.
  1. Determine, in terms of \(e\) and \(V\), the speeds of A and B immediately after the first collision.
  2. Show that the arc through which A moves between the first and second collisions subtends an angle at the centre of the circular groove of $$\frac { 2 \pi ( 4 - 3 e ) } { 7 e } \text { radians. }$$
    1. Determine, in terms of \(e\) and \(V\), the speed of B immediately after the second collision.
    2. What can be said about the motion of A and B if the collisions between A and B are perfectly elastic?
Question 11:
AnswerMarks Guidance
11(a) 4V = 4v + 3v
A B
v − v = −eV
A B
V (4 − 3e) 4V (1+ e)
v = and v =
AnswerMarks
A 7 B 7M1*
A1
M1*
A1
M1dep*
A1
AnswerMarks
[6]3.3
1.1
3.3
1.1
1.1
AnswerMarks
1.1Conservation of linear momentum with
correct number of terms
cao
Newton’s experimental law with correct
number of terms
Must be consistent with CLM
Solve the simultaneous equations to find
AnswerMarks
both speedsWhere v is the speed
A
of A after 1st impact
and similarly for v
B
AnswerMarks Guidance
11(b) Let θ be the angle subtended by A in time t
rθ
For A, t =
V (4 − 3e)
7
2π r + rθ
For B, t =
4V (1+ e)
7
2π + θ θ
=
4V ( 1 + e ) V ( 4 − 3e )
2π (4 − 3e)
θ =
AnswerMarks
7eM1
M1
M1
A1
AnswerMarks
[4]3.1b
1.1
3.4
AnswerMarks
2.2aUse of s = ut with their v and s = rθ
A
Use of s = ut with their v and
B
s = 2π r + rθ
Equate expressions for t to form an
equation in terms of θ ,V and e
AnswerMarks
AGWhere r is the radius
of the circular groove
Alternative method
4V (1+ e) V (4 − 3e)
AnswerMarks Guidance
ALT: v − v = − = eVM1* Difference in speeds calculated
B A
7 7
2π r
AnswerMarks Guidance
Time for B to catch up to A isM1dep* Using their eV
of the circular groove
eV
AnswerMarks Guidance
2π r  V (4 − 3e)  2π rM1 Where d is the distance travelled by A
A
d A =   = (4 − 3e)
eV  7  7e
2π r (4 − 3e) 2π (4 − 3e)
AnswerMarks Guidance
θ = =A1 AG
7er 7e
AnswerMarks
11(c)
(i)12 4
3w + 4w = V (1 + e) + V (4 − 3e)
B A
7 7
 4 1 
w − w = −e V (1+ e) − V (4 − 3e)
 
B A  7 7 
3w + 4w = 4V and w − w = −e2V
B A B A
4 ( )
w = V 1 − e2
AnswerMarks
B 7M1*
M1*
A1
M1dep*
A1
AnswerMarks
[5]3.3
3.3
1.1
1.1
AnswerMarks
1.1CLM correct number of terms using their
expressions from (a)
NEL correct number of terms
oe
Solve simultaneously for w
B
AnswerMarks
caoWhere w is the
A
speed of A after the
second collision
For reference:
1 ( )
w = V 4 + 3e2
A 7
AnswerMarks
11(c)
(ii)If the collision is perfectly elastic (e = 1) B is brought to
rest by the second collision and A is moving with speed
AnswerMarks Guidance
V (which is the situation before the first collision)B1
[1]3.5a oe correct statement 
Question 11:
11 | (a) | 4V = 4v + 3v
A B
v − v = −eV
A B
V (4 − 3e) 4V (1+ e)
v = and v =
A 7 B 7 | M1*
A1
M1*
A1
M1dep*
A1
[6] | 3.3
1.1
3.3
1.1
1.1
1.1 | Conservation of linear momentum with
correct number of terms
cao
Newton’s experimental law with correct
number of terms
Must be consistent with CLM
Solve the simultaneous equations to find
both speeds | Where v is the speed
A
of A after 1st impact
and similarly for v
B
11 | (b) | Let θ be the angle subtended by A in time t
rθ
For A, t =
V (4 − 3e)
7
2π r + rθ
For B, t =
4V (1+ e)
7
2π + θ θ
=
4V ( 1 + e ) V ( 4 − 3e )
2π (4 − 3e)
θ =
7e | M1
M1
M1
A1
[4] | 3.1b
1.1
3.4
2.2a | Use of s = ut with their v and s = rθ
A
Use of s = ut with their v and
B
s = 2π r + rθ
Equate expressions for t to form an
equation in terms of θ ,V and e
AG | Where r is the radius
of the circular groove
Alternative method
4V (1+ e) V (4 − 3e)
ALT: v − v = − = eV | M1* | Difference in speeds calculated
B A
7 7
2π r
Time for B to catch up to A is | M1dep* | Using their eV | Where r is the radius
of the circular groove
eV
2π r  V (4 − 3e)  2π r | M1 | Where d is the distance travelled by A
A
d A =   = (4 − 3e)
eV  7  7e
2π r (4 − 3e) 2π (4 − 3e)
θ = = | A1 | AG
7er 7e
11 | (c)
(i) | 12 4
3w + 4w = V (1 + e) + V (4 − 3e)
B A
7 7
 4 1 
w − w = −e V (1+ e) − V (4 − 3e)
 
B A  7 7 
3w + 4w = 4V and w − w = −e2V
B A B A
4 ( )
w = V 1 − e2
B 7 | M1*
M1*
A1
M1dep*
A1
[5] | 3.3
3.3
1.1
1.1
1.1 | CLM correct number of terms using their
expressions from (a)
NEL correct number of terms
oe
Solve simultaneously for w
B
cao | Where w is the
A
speed of A after the
second collision
For reference:
1 ( )
w = V 4 + 3e2
A 7
11 | (c)
(ii) | If the collision is perfectly elastic (e = 1) B is brought to
rest by the second collision and A is moving with speed
V (which is the situation before the first collision) | B1
[1] | 3.5a | oe correct statement
11 Two small uniform smooth spheres A and B , of equal radius, have masses 4 kg and 3 kg respectively. The spheres are placed in a smooth horizontal circular groove. The coefficient of restitution between the spheres is $e$, where $e > \frac { 2 } { 5 }$.

At a given instant B is at rest and A is set moving along the groove with speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$. It may be assumed that in the subsequent motion the two spheres do not leave the groove.
\begin{enumerate}[label=(\alph*)]
\item Determine, in terms of $e$ and $V$, the speeds of A and B immediately after the first collision.
\item Show that the arc through which A moves between the first and second collisions subtends an angle at the centre of the circular groove of

$$\frac { 2 \pi ( 4 - 3 e ) } { 7 e } \text { radians. }$$
\item \begin{enumerate}[label=(\roman*)]
\item Determine, in terms of $e$ and $V$, the speed of B immediately after the second collision.
\item What can be said about the motion of A and B if the collisions between A and B are perfectly elastic?
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2021 Q11 [16]}}
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