OCR Further Mechanics 2019 June — Question 3 13 marks

Exam BoardOCR
ModuleFurther Mechanics (Further Mechanics)
Year2019
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAdvanced work-energy problems
TypeWork done by vector force displacement
DifficultyStandard +0.3 This is a straightforward vector mechanics question requiring systematic application of standard kinematic equations and the work-energy relationship. While it involves vectors and multiple parts, each step follows directly from standard formulas (v²=u²+2as, P=F·v, KE=½mv²) with no novel problem-solving insight required. The calculations are routine for Further Maths students, making it slightly easier than average.
Spec1.10d Vector operations: addition and scalar multiplication1.10h Vectors in kinematics: uniform acceleration in vector form6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product
3 A particle \(Q\) of mass \(m \mathrm {~kg}\) is acted on by a single force so that it moves with constant acceleration \(\mathbf { a } = \binom { 1 } { 2 } \mathrm {~ms} ^ { - 2 }\). Initially \(Q\) is at the point \(O\) and is moving with velocity \(\mathbf { u } = \binom { 2 } { - 5 } \mathrm {~ms} ^ { - 1 }\). After \(Q\) has been moving for 5 seconds it reaches the point \(A\).
  1. Use the equation \(\mathbf { v . v } = \mathbf { u . u } + 2 \mathbf { a x }\) to show that at \(A\) the kinetic energy of \(Q\) is 37 m J .
    1. Show that the power initially generated by the force is - 8 mW .
    2. The power in part (b)(i) is negative. Explain what this means about the initial motion of \(Q\).
    1. Find the time at which the power generated by the force is instantaneously zero.
    2. Find the minimum kinetic energy of \(Q\) in terms of \(m\).
Question 3:
AnswerMarks Guidance
3(a)  2  1 1
x(5)5  52
5 2 2
22.5
x(5)
0
 2   2  1 22.5
v.v . 2 .
5 5 2 0
v.v4254574
KE 1m7437m
AnswerMarks
2M1
A1
M1
M1
AnswerMarks
A13.1b
1.1
3.1b
1.1
AnswerMarks
2.2asoi
AG wwwAward SC2 for using v = u +
at to find v and then KE
Must be complete and
correct if not SC0
Alternate method
Substitute xut1at2into v.vu.u2a.x
AnswerMarks Guidance
2M1 or v.vwith vuat
to get v.vu.u2a.uta.at2A1
 2   2  1  2  1 1
v.v . 25 . 25 .
AnswerMarks
5 5 2 5 2 2M1
v.v42510(210)25(14)74M1
KE 1m7437m
AnswerMarks Guidance
2A1 AG
[5]
AnswerMarks Guidance
3(b) (i)
𝑃 =𝑚( ).( ) or equiv vector form
2 −5
AnswerMarks
𝑃 =𝑚(2−10)= −8𝑚M1
A1
AnswerMarks Guidance
[2]2.2a
1.1AG www Must see 2-10 or 2m – 10m
3(b) (ii)
[1]2.2a or Kinetic Energy is decreasing
3(c) (i)
v  tor v = ( )
5 2 −5+2𝑡
1  2t 
ma.vm .  02t2(52t)0
2 52t
AnswerMarks
Time taken is 1.6 sB1
M1
AnswerMarks
A11.1
2.2a
AnswerMarks
1.1v = u + at soi
Allow missing m; must have = 0
AnswerMarks
wwwMay be derived from
integrating a
Must be equation in terms of
t. Condone one slip
Alternate method
 2  1 2+𝑡
v  tor v = ( )
AnswerMarks Guidance
5 2 −5+2𝑡B1 v = u + at soi
1mv.v 1m(5t2 16t29)10t160
AnswerMarks Guidance
2 21mv.v 1m(5t2 16t29)10t160
2 2M1 Find minimum
2
AnswerMarks Guidance
Time taken is 1.6 sA1 www
[3]
AnswerMarks Guidance
3(c) (ii)
1mv.v 1m .
2 2 521.6 521.6
81
m Jor 8.1m J
AnswerMarks
10M1
A1
AnswerMarks
[2]2.2a
1.1With their 1.6 
Question 3:
3 | (a) |  2  1 1
x(5)5  52
5 2 2
22.5
x(5)
0
 2   2  1 22.5
v.v . 2 .
5 5 2 0
v.v4254574
KE 1m7437m
2 | M1
A1
M1
M1
A1 | 3.1b
1.1
3.1b
1.1
2.2a | soi
AG www | Award SC2 for using v = u +
at to find v and then KE
Must be complete and
correct if not SC0
Alternate method
Substitute xut1at2into v.vu.u2a.x
2 | M1 | or v.vwith vuat
to get v.vu.u2a.uta.at2 | A1
 2   2  1  2  1 1
v.v . 25 . 25 .
5 5 2 5 2 2 | M1
v.v42510(210)25(14)74 | M1
KE 1m7437m
2 | A1 | AG
[5]
3 | (b) | (i) | 1 2
𝑃 =𝑚( ).( ) or equiv vector form
2 −5
𝑃 =𝑚(2−10)= −8𝑚 | M1
A1
[2] | 2.2a
1.1 | AG www | Must see 2-10 or 2m – 10m
3 | (b) | (ii) | E.g. Q is slowing down | B1
[1] | 2.2a | or Kinetic Energy is decreasing
3 | (c) | (i) |  2  1 2+𝑡
v  tor v = ( )
5 2 −5+2𝑡
1  2t 
ma.vm .  02t2(52t)0
2 52t
Time taken is 1.6 s | B1
M1
A1 | 1.1
2.2a
1.1 | v = u + at soi
Allow missing m; must have = 0
www | May be derived from
integrating a
Must be equation in terms of
t. Condone one slip
Alternate method
 2  1 2+𝑡
v  tor v = ( )
5 2 −5+2𝑡 | B1 | v = u + at soi
1mv.v 1m(5t2 16t29)10t160
2 2 | 1mv.v 1m(5t2 16t29)10t160
2 2 | M1 | Find minimum | Allow missing 1m
2
Time taken is 1.6 s | A1 | www
[3]
3 | (c) | (ii) |  21.6   21.6 
1mv.v 1m .
2 2 521.6 521.6
81
m Jor 8.1m J
10 | M1
A1
[2] | 2.2a
1.1 | With their 1.6
3 A particle $Q$ of mass $m \mathrm {~kg}$ is acted on by a single force so that it moves with constant acceleration $\mathbf { a } = \binom { 1 } { 2 } \mathrm {~ms} ^ { - 2 }$. Initially $Q$ is at the point $O$ and is moving with velocity $\mathbf { u } = \binom { 2 } { - 5 } \mathrm {~ms} ^ { - 1 }$.

After $Q$ has been moving for 5 seconds it reaches the point $A$.
\begin{enumerate}[label=(\alph*)]
\item Use the equation $\mathbf { v . v } = \mathbf { u . u } + 2 \mathbf { a x }$ to show that at $A$ the kinetic energy of $Q$ is 37 m J .
\item \begin{enumerate}[label=(\roman*)]
\item Show that the power initially generated by the force is - 8 mW .
\item The power in part (b)(i) is negative. Explain what this means about the initial motion of $Q$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find the time at which the power generated by the force is instantaneously zero.
\item Find the minimum kinetic energy of $Q$ in terms of $m$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR Further Mechanics 2019 Q3 [13]}}
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