OCR Further Mechanics 2019 June — Question 6 9 marks

Exam BoardOCR
ModuleFurther Mechanics (Further Mechanics)
Year2019
SessionJune
Marks9
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TopicImpulse and momentum (advanced)
TypeEnergy change from impulse
DifficultyChallenging +1.8 This is a challenging Further Mechanics question requiring geometric insight (recognizing the 60° angle when the string becomes taut), vector resolution of impulse-momentum in two perpendicular directions, conservation principles, and algebraic manipulation. Part (c) requires physical interpretation of limiting behavior. Significantly harder than standard A-level mechanics but typical of Further Mechanics material.
Spec6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form
6 Two particles \(A\) and \(B\), of masses \(m \mathrm {~kg}\) and 1 kg respectively, are connected by a light inextensible string of length \(d \mathrm {~m}\) and placed at rest on a smooth horizontal plane a distance of \(\frac { 1 } { 2 } d \mathrm {~m}\) apart. \(B\) is then projected horizontally with speed \(v \mathrm {~ms} ^ { - 1 }\) in a direction perpendicular to \(A B\).
  1. Show that, at the instant that the string becomes taut, the magnitude of the instantaneous impulse in the string, \(I \mathrm { Ns }\), is given by \(\mathrm { I } = \frac { \sqrt { 3 } \mathrm { mv } } { 2 ( 1 + \mathrm { m } ) }\).
  2. Find, in terms of \(m\) and \(v\), the kinetic energy of \(B\) at the instant after the string becomes taut. Give your answer as a single algebraic fraction.
  3. In the case where \(m\) is very large, describe, with justification, the approximate motion of \(B\) after the string becomes taut.
Question 6:
AnswerMarks Guidance
6(a) √3
cos 1 or sin𝜃 =
2 2
vsin(1m)V
√3𝑣
𝑉 =
2(1+𝑚)
I mV or I (V vsin)
v1 3 3mv
V  2 I 
AnswerMarks
1m 2(1m)B1
M1
A1
A1
AnswerMarks
[4]3.1b
3.1b
3.1b
AnswerMarks
2.2aMust be clear where it comes from
ie. Shown in diagram or clear use of
distances. May be other way round
Conservation of momentum in
direction of string
Working must be seen
AG Justification of using
AnswerMarks
Impulse = change in momentumθ is the angle between the
string and the direction of
AB when the string initially
becomes taut
AnswerMarks Guidance
6(b) KE 1 V2(vcos)2 soi
2
 2 2
1  3v   1
KE    v  
2 2(1m)  2 
 
v2(42mm2)
KE
AnswerMarks
8(1m)2M1
M1
A1
AnswerMarks
[3]3.1b
1.1
AnswerMarks
1.1Transverse component unchanged
and using their longitudinal
component
Substituting in for V and cosθ
(could just be in speed equation)
Or any equivalent single algebraic
AnswerMarks
fractionCondone consideration of
speed squared or inclusion of
m in KE
AnswerMarks Guidance
6(c) 3v
V  0 asm
21m
AnswerMarks
B moves (approximately) in a circle around AB1
B1
AnswerMarks
[2]3.2a
2.2bIf m is very large then A is
approximately stationary or B has
only its transverse velocity of 1v
2
after the string becomes taut
Question 6:
6 | (a) | √3
cos 1 or sin𝜃 =
2 2
vsin(1m)V
√3𝑣
𝑉 =
2(1+𝑚)
I mV or I (V vsin)
v1 3 3mv
V  2 I 
1m 2(1m) | B1
M1
A1
A1
[4] | 3.1b
3.1b
3.1b
2.2a | Must be clear where it comes from
ie. Shown in diagram or clear use of
distances. May be other way round
Conservation of momentum in
direction of string
Working must be seen
AG Justification of using
Impulse = change in momentum | θ is the angle between the
string and the direction of
AB when the string initially
becomes taut
6 | (b) | KE 1 V2(vcos)2 soi
2
 2 2
1  3v   1
KE    v  
2 2(1m)  2 
 
v2(42mm2)
KE
8(1m)2 | M1
M1
A1
[3] | 3.1b
1.1
1.1 | Transverse component unchanged
and using their longitudinal
component
Substituting in for V and cosθ
(could just be in speed equation)
Or any equivalent single algebraic
fraction | Condone consideration of
speed squared or inclusion of
m in KE
6 | (c) | 3v
V  0 asm
21m
B moves (approximately) in a circle around A | B1
B1
[2] | 3.2a
2.2b | If m is very large then A is
approximately stationary or B has
only its transverse velocity of 1v
2
after the string becomes taut
6 Two particles $A$ and $B$, of masses $m \mathrm {~kg}$ and 1 kg respectively, are connected by a light inextensible string of length $d \mathrm {~m}$ and placed at rest on a smooth horizontal plane a distance of $\frac { 1 } { 2 } d \mathrm {~m}$ apart. $B$ is then projected horizontally with speed $v \mathrm {~ms} ^ { - 1 }$ in a direction perpendicular to $A B$.
\begin{enumerate}[label=(\alph*)]
\item Show that, at the instant that the string becomes taut, the magnitude of the instantaneous impulse in the string, $I \mathrm { Ns }$, is given by $\mathrm { I } = \frac { \sqrt { 3 } \mathrm { mv } } { 2 ( 1 + \mathrm { m } ) }$.
\item Find, in terms of $m$ and $v$, the kinetic energy of $B$ at the instant after the string becomes taut. Give your answer as a single algebraic fraction.
\item In the case where $m$ is very large, describe, with justification, the approximate motion of $B$ after the string becomes taut.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Mechanics 2019 Q6 [9]}}
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