OCR MEI Further Mechanics Major 2020 November — Question 5 8 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
Year2020
SessionNovember
Marks8
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Mark schemeDownload PDF ↗
TopicPower and driving force
TypeFind acceleration given power
DifficultyStandard +0.3 This is a standard Further Maths mechanics problem involving power, resistance, and differential equations. Part (a) requires setting up P=Fv and F=ma to derive the given differential equation (routine manipulation). Part (b) is a 'verify' question where students integrate by separation of variables and check the given answer—straightforward but requires careful algebraic manipulation. The techniques are well-practiced in FM1/FM2 courses, making this slightly easier than average overall.
Spec6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product6.06a Variable force: dv/dt or v*dv/dx methods
A car of mass \(900\) kg moves along a straight level road. The power developed by the car is constant and equal to \(60\) kW. The resistance to the motion of the car is constant and equal to \(1500\) N. At time \(t\) seconds the velocity of the car is denoted by \(v\) m s\(^{-1}\). Initially the car is at rest.
  1. Show that \(\frac{3v\,dv}{5\,dt} = 40 - v\). [3]
  2. Verify that \(t = 24\ln\left(\frac{40}{40-v}\right) - \frac{3}{5}v\). [5]
Question 5:
AnswerMarks Guidance
5(a) 60000
Driving force =
AnswerMarks Guidance
vB1 1.2
60000 dv
−1500=900
AnswerMarks Guidance
v dtM1 3.3
terms – allow a for dv/dt
dv 3v dv
200−5v=3v ⇒ =40−v
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dt 5 dtA1 2.2a
as answer given
[3]
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5(b) 40 3
v=0,t =24ln − (0)=24ln1=0
AnswerMarks Guidance
40 5B1 1.1
 40  3 dt
t =24ln  − v⇒ =...
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40−v 5 dvM1* 1.1
1 (40−v)−2
terms and
 40 
 
40−v
 
 
dt =24 1 (−1)(−40)(40−v)−2 − 3
dv  40  5
 
AnswerMarks Guidance
40−vA1 1.1
dt −24 3
= (−1)− from
dv 40−v 5
t =24ln40−24ln(40−v)−3v
5
dt 24 3 3v dv
= − = ⇒ =...
AnswerMarks Guidance
dv 40−v 5 5(40−v) dtM1dep* 1.1
reciprocal
5(40−v)
dv 3v dv
= ⇒ =40−v
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dt 3v 5 dtA1 1.1
as answer given
[5]
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5(b) ALT
3v dv =40−v⇒ 5∫dt =∫ v dv
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5 dt 3 40−vB1 B1
5 t =−∫1− 40 dv
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3 40−v5 t =−∫1− 40 dv
3 40−vM1* Rewrite in an integratable form and
attempt to integrate (must contain a log
term of 40 – v )
5 t =−( v+40ln(40−v)) (+c)
AnswerMarks Guidance
3A1 +c not required for
this mark
AnswerMarks Guidance
t =0,v=0⇒c=40ln40M1dep* Use given initial conditions to find c
5 t =40ln40−40ln(40−v)−v
3
( 40 ) 3
⇒t =24ln − v
AnswerMarks Guidance
40−v 5A1 A1
[5]
Question 5:
5 | (a) | 60000
Driving force =
v | B1 | 1.2
60000 dv
−1500=900
v dt | M1 | 3.3 | Applying N,II with correct number of
terms – allow a for dv/dt
dv 3v dv
200−5v=3v ⇒ =40−v
dt 5 dt | A1 | 2.2a | AG – sufficient working must be shown
as answer given
[3]
5 | (b) | 40 3
v=0,t =24ln − (0)=24ln1=0

40 5 | B1 | 1.1 | Must explicitly verify initial conditions
 40  3 dt
t =24ln  − v⇒ =...
40−v 5 dv | M1* | 1.1 | Attempt at differentiation – must see
1 (40−v)−2
terms and
 40 
 
40−v
 
 
dt =24 1 (−1)(−40)(40−v)−2 − 3
dv  40  5
 
40−v | A1 | 1.1 | Correct derivative (allow unsimplified) or
dt −24 3
= (−1)− from
dv 40−v 5
t =24ln40−24ln(40−v)−3v
5
dt 24 3 3v dv
= − = ⇒ =...
dv 40−v 5 5(40−v) dt | M1dep* | 1.1 | Simplify to single fraction before taking
reciprocal
5(40−v)
dv 3v dv
= ⇒ =40−v
dt 3v 5 dt | A1 | 1.1 | AG – sufficient working must be shown
as answer given
[5]
5 | (b) | ALT
3v dv =40−v⇒ 5∫dt =∫ v dv
5 dt 3 40−v | B1 | B1 | Separate variables correctly | Separate variables correctly
5 t =−∫1− 40 dv
3 40−v | 5 t =−∫1− 40 dv
3 40−v | M1* | Rewrite in an integratable form and
attempt to integrate (must contain a log
term of 40 – v )
5 t =−( v+40ln(40−v)) (+c)
3 | A1 | +c not required for
this mark
t =0,v=0⇒c=40ln40 | M1dep* | Use given initial conditions to find c
5 t =40ln40−40ln(40−v)−v
3
( 40 ) 3
⇒t =24ln − v
40−v 5 | A1 | A1 | AG | AG
[5]
A car of mass $900$ kg moves along a straight level road. The power developed by the car is constant and equal to $60$ kW. The resistance to the motion of the car is constant and equal to $1500$ N.

At time $t$ seconds the velocity of the car is denoted by $v$ m s$^{-1}$. Initially the car is at rest.

\begin{enumerate}[label=(\alph*)]
\item Show that $\frac{3v\,dv}{5\,dt} = 40 - v$. [3]
\item Verify that $t = 24\ln\left(\frac{40}{40-v}\right) - \frac{3}{5}v$. [5]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2020 Q5 [8]}}
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