OCR MEI Further Mechanics Major 2021 November — Question 7 12 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
Year2021
SessionNovember
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeLifting objects vertically
DifficultyChallenging +1.2 This is a multi-part Further Mechanics question requiring application of power equations, differential equations, and verification of a given solution. While it involves several steps and Further Maths content (making it harder than typical A-level), the parts are scaffolded: (a) is a standard derivation from P=Fv, (b) is verification (not derivation), and (c) applies work-energy principle in a guided way. The techniques are bookwork-level for FM students.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)6.02a Work done: concept and definition6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product
7 A box B of mass \(m \mathrm {~kg}\) is raised vertically by an engine working at a constant rate of \(k m g \mathrm {~W}\). Initially B is at rest. The speed of B when it has been raised a distance \(x \mathrm {~m}\) is denoted by \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
  1. Show that \(v ^ { 2 } \frac { d v } { d x } = ( k - v ) g\).
  2. Verify that \(\mathrm { gx } = \mathrm { k } ^ { 2 } \ln \left( \frac { \mathrm { k } } { \mathrm { k } - \mathrm { v } } \right) - \mathrm { kv } - \frac { 1 } { 2 } \mathrm { v } ^ { 2 }\).
  3. By using the work-energy principle, show that the time taken for B to reach a speed \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from rest is given by \(\frac { \mathrm { k } } { \mathrm { g } } \ln \left( \frac { \mathrm { k } } { \mathrm { k } - \mathrm { V } } \right) - \frac { \mathrm { V } } { \mathrm { g } }\).
Question 7:
AnswerMarks Guidance
7(a) kmg
Driving force of engine is
v
kmg dv
− mg = mv
v dx
dv dv
kg − gv = v2 ⇒ v2 = (k − v) g
AnswerMarks
dx dxB1
M1
A1
AnswerMarks
[3]1.1
3.3
AnswerMarks
2.2aUse of N2L, correct number of terms,
kmg
allow D (oe) for and a (oe) for the
v
acceleration
AG – sufficient working must be shown
as answer given
AnswerMarks Guidance
7(b)  k  1
gx = k 2 ln   − kv − v2
k − v 2
 
x = 0,v = 0 ⇒ g (0) = k 2 ln  k  − k (0) − 1 (0)2 so
 
k − 0 2
 
initial conditions are consistent with given equation
 
 
g dx = k 2 1 k ( )−2 − k − v
 k − v 
dv  k  
  
 k − v
  
dx −kv + v2 − k 2 + kv + k 2
g =
dv (k − v)
dx dv
v2 = g (k − v) ⇒ v2 = (k − v) g
AnswerMarks
dv dxB1
M1*
A1
M1dep*
A1
AnswerMarks
[5]1.1
2.1
1.1
1.1
AnswerMarks
2.2aAttempt to differentiate using chain rule
cao oe e.g.
  dv  
 −k  −  
 k − v  dx dv dv
2  
g = k   k    (k − v)2   − k dx − v dx
 
 
Correct method to obtain an expression
dx
for as a single fraction or as a single
dv
dv
fraction with
dx
 k 2 − k 2 + kv − kv + v2  dv
e.g. g =
 
k − v dx
 
AG – sufficient working required as
AnswerMarks
answer givenOr equivalent (e.g.
solving using
separation of
variables)
AnswerMarks Guidance
7(c) Work done by engine is kmgt
1
kgmt = mV 2 + mgx
2
1  k  1
kgt = V 2 + k 2 ln   − kV − V 2
2 k − V 2
 
 k  k  k  V
kgt = k 2 ln − kV ⇒ t = ln −
   
k − V g k − V g
AnswerMarks
   B1
M1*
M1dep*
A1
AnswerMarks
[4]1.1
3.3
3.4
AnswerMarks
2.2aUse work-energy principle – correct
number of terms
Use given result from (b) in work-energy
equation to eliminate x
AG – sufficient working required as
answer given
SC if correctly found by solving
kmg dv
− mg = m this can score 3/4 max.
v dt
Question 7:
7 | (a) | kmg
Driving force of engine is
v
kmg dv
− mg = mv
v dx
dv dv
kg − gv = v2 ⇒ v2 = (k − v) g
dx dx | B1
M1
A1
[3] | 1.1
3.3
2.2a | Use of N2L, correct number of terms,
kmg
allow D (oe) for and a (oe) for the
v
acceleration
AG – sufficient working must be shown
as answer given
7 | (b) |  k  1
gx = k 2 ln   − kv − v2
k − v 2
 
x = 0,v = 0 ⇒ g (0) = k 2 ln  k  − k (0) − 1 (0)2 so
 
k − 0 2
 
initial conditions are consistent with given equation
 
 
g dx = k 2 1 k ( )−2 − k − v
 k − v 
dv  k  
  
 k − v
  
dx −kv + v2 − k 2 + kv + k 2
g =
dv (k − v)
dx dv
v2 = g (k − v) ⇒ v2 = (k − v) g
dv dx | B1
M1*
A1
M1dep*
A1
[5] | 1.1
2.1
1.1
1.1
2.2a | Attempt to differentiate using chain rule
cao oe e.g.
  dv  
 −k  −  
 k − v  dx dv dv
2  
g = k   k    (k − v)2   − k dx − v dx
 
 
Correct method to obtain an expression
dx
for as a single fraction or as a single
dv
dv
fraction with
dx
 k 2 − k 2 + kv − kv + v2  dv
e.g. g =
 
k − v dx
 
AG – sufficient working required as
answer given | Or equivalent (e.g.
solving using
separation of
variables)
7 | (c) | Work done by engine is kmgt
1
kgmt = mV 2 + mgx
2
1  k  1
kgt = V 2 + k 2 ln   − kV − V 2
2 k − V 2
 
 k  k  k  V
kgt = k 2 ln − kV ⇒ t = ln −
   
k − V g k − V g
    | B1
M1*
M1dep*
A1
[4] | 1.1
3.3
3.4
2.2a | Use work-energy principle – correct
number of terms
Use given result from (b) in work-energy
equation to eliminate x
AG – sufficient working required as
answer given
SC if correctly found by solving
kmg dv
− mg = m this can score 3/4 max.
v dt
7 A box B of mass $m \mathrm {~kg}$ is raised vertically by an engine working at a constant rate of $k m g \mathrm {~W}$. Initially B is at rest. The speed of B when it has been raised a distance $x \mathrm {~m}$ is denoted by $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $v ^ { 2 } \frac { d v } { d x } = ( k - v ) g$.
\item Verify that $\mathrm { gx } = \mathrm { k } ^ { 2 } \ln \left( \frac { \mathrm { k } } { \mathrm { k } - \mathrm { v } } \right) - \mathrm { kv } - \frac { 1 } { 2 } \mathrm { v } ^ { 2 }$.
\item By using the work-energy principle, show that the time taken for B to reach a speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from rest is given by\\
$\frac { \mathrm { k } } { \mathrm { g } } \ln \left( \frac { \mathrm { k } } { \mathrm { k } - \mathrm { V } } \right) - \frac { \mathrm { V } } { \mathrm { g } }$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2021 Q7 [12]}}
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