OCR Further Mechanics 2022 June — Question 3 6 marks

Exam BoardOCR
ModuleFurther Mechanics (Further Mechanics)
Year2022
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeMotion with exponential force
DifficultyChallenging +1.2 This is a Further Mechanics question requiring integration of an exponential force expression to find impulse, then applying impulse-momentum and work-energy theorems. While the exponential form looks intimidating, the actual calculus is straightforward (substitution u=t²), and parts (b) and (c) follow directly from standard mechanics principles once (a) is solved. Slightly above average due to the Further Maths context and multi-step nature, but the techniques are standard.
Spec6.02a Work done: concept and definition6.02c Work by variable force: using integration6.03f Impulse-momentum: relation
3 A particle \(P\) of mass 6 kg moves in a straight line under the action of a single force of magnitude \(F N\) which acts in the direction of motion of \(P\).
At time \(t\) seconds, where \(t \geqslant 0 , F\) is given by \(\mathrm { F } = \frac { 1 } { 5 - 4 \mathrm { e } ^ { - \mathrm { t } ^ { 2 } } }\).
When \(t = 0\), the speed of \(P\) is \(1.9 \mathrm {~ms} ^ { - 1 }\).
  1. Find the impulse of the force over the period \(0 \leqslant t \leqslant 2\).
  2. Find the speed of \(P\) at the instant when \(t = 2\).
  3. Find the work done by the force on \(P\) over the period \(0 \leqslant t \leqslant 2\).
Question 3:
AnswerMarks Guidance
3(a) 2
1
I = dt
−t2
5−4e
0
AnswerMarks
= 0.81365... so impulse is awrt 0.814 NsM1
A1
AnswerMarks
[2]1.1
1.1Correct form of integral and correct
limits.
AnswerMarks
BCAccept awrt 0.81 or 0.82 if correct
integral shown
AnswerMarks
(b)Impulse = change in momentum
so “0.81365...” = ±(6v – 61.9)
v = 1.9 + 0.81365.../6 = 2.0356... so P’s speed is
AnswerMarks
awrt 2.04 ms–1M1
A1
AnswerMarks Guidance
[2]1.1
1.1Use of impulse-momentum principle Accept awrt 2.03 or 2.04
(c)WD = change in KE = ½6(“2.0356...”2 – 1.92) J
= awrt 1.60 JM1
A1
AnswerMarks
[2]1.1
1.1Use of work-energy principle with
values for v, m and u substituted inAccept anything awrt 1.60 – 1.62 
Question 3:
3 | (a) | 2
1
I = dt
−t2
5−4e
0
= 0.81365... so impulse is awrt 0.814 Ns | M1
A1
[2] | 1.1
1.1 | Correct form of integral and correct
limits.
BC | Accept awrt 0.81 or 0.82 if correct
integral shown
(b) | Impulse = change in momentum
so “0.81365...” = ±(6v – 61.9)
v = 1.9 + 0.81365.../6 = 2.0356... so P’s speed is
awrt 2.04 ms–1 | M1
A1
[2] | 1.1
1.1 | Use of impulse-momentum principle | Accept awrt 2.03 or 2.04
(c) | WD = change in KE = ½6(“2.0356...”2 – 1.92) J
= awrt 1.60 J | M1
A1
[2] | 1.1
1.1 | Use of work-energy principle with
values for v, m and u substituted in | Accept anything awrt 1.60 – 1.62
3 A particle $P$ of mass 6 kg moves in a straight line under the action of a single force of magnitude $F N$ which acts in the direction of motion of $P$.\\
At time $t$ seconds, where $t \geqslant 0 , F$ is given by $\mathrm { F } = \frac { 1 } { 5 - 4 \mathrm { e } ^ { - \mathrm { t } ^ { 2 } } }$.\\
When $t = 0$, the speed of $P$ is $1.9 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the impulse of the force over the period $0 \leqslant t \leqslant 2$.
\item Find the speed of $P$ at the instant when $t = 2$.
\item Find the work done by the force on $P$ over the period $0 \leqslant t \leqslant 2$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Mechanics 2022 Q3 [6]}}
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