OCR MEI Further Mechanics B AS 2019 June — Question 2 7 marks

Exam BoardOCR MEI
ModuleFurther Mechanics B AS (Further Mechanics B AS)
Year2019
SessionJune
Marks7
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TopicVariable acceleration (1D)
TypeVariable acceleration with initial conditions
DifficultyModerate -0.8 This is a straightforward mechanics question requiring basic application of Newton's second law (F=ma gives 3mt=m(dΒ²x/dtΒ²)), verification by differentiation (routine calculus), and solving simultaneous equations using initial conditions. All steps are standard textbook procedures with no problem-solving insight required.
Spec4.10a General/particular solutions: of differential equations6.06a Variable force: dv/dt or v*dv/dx methods
2 A particle P of mass \(m\) travels in a straight line on a smooth horizontal surface.
At time \(t , \mathrm { P }\) is a distance \(x\) from a fixed point O and is moving with speed \(v\) away from O . A horizontal force of magnitude \(3 m t\) acts on P , in a direction away from O .
  1. Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = 3 t\).
  2. Verify that the general solution of this differential equation is \(x = \frac { 1 } { 2 } t ^ { 3 } + A t + k\), where \(A\) and \(k\) are constants.
  3. Given that \(x = 6\) and \(v = 12\) when \(t = 1\), find the values of \(A\) and \(k\).
Question 2:
AnswerMarks Guidance
2(a) d2x
Must show 3π‘šπ‘‘ = π‘šπ‘Ž or 3mt ο€½m
AnswerMarks
dt2Using Newton’s second law
d2π‘₯
= 3𝑑
AnswerMarks Guidance
d𝑑2B1 1.1a
[1]
AnswerMarks
(b)3
[π‘₯β€²] = 𝑑2 +𝐴
AnswerMarks Guidance
2M1 1.1
integration (any constant)
d2π‘₯
= 3𝑑
AnswerMarks Guidance
d𝑑2A1 1.1
(consts need not be A/k)
AnswerMarks
[2]Candidate needs to be
consistent in whether they
are integrating the
acceleration or
differentiating the given
solution
AnswerMarks
(c)1
6 = +𝐴+π‘˜
AnswerMarks Guidance
2B1 3.4
Or 6 = +10.5+π‘˜
AnswerMarks
2Might have the value of A
already substituted
AnswerMarks Guidance
12 = 1.5+𝐴A1 3.1b
1
𝐴 = 10
AnswerMarks Guidance
2A1 1.1
k = βˆ’ 5A1 1.1
the correct value of k
[4]
Question 2:
2 | (a) | d2x
Must show 3π‘šπ‘‘ = π‘šπ‘Ž or 3mt ο€½m
dt2 | Using Newton’s second law
d2π‘₯
= 3𝑑
d𝑑2 | B1 | 1.1a | AG
[1]
(b) | 3
[π‘₯β€²] = 𝑑2 +𝐴
2 | M1 | 1.1 | Diff wrt t | Or x’ = … found from
integration (any constant)
d2π‘₯
= 3𝑑
d𝑑2 | A1 | 1.1 | CWO | Or second integration here
(consts need not be A/k)
[2] | Candidate needs to be
consistent in whether they
are integrating the
acceleration or
differentiating the given
solution
(c) | 1
6 = +𝐴+π‘˜
2 | B1 | 3.4 | 1
Or 6 = +10.5+π‘˜
2 | Might have the value of A
already substituted
12 = 1.5+𝐴 | A1 | 3.1b
1
𝐴 = 10
2 | A1 | 1.1
k = βˆ’ 5 | A1 | 1.1 | B1 can be assumed from
the correct value of k
[4]
2 A particle P of mass $m$ travels in a straight line on a smooth horizontal surface.\\
At time $t , \mathrm { P }$ is a distance $x$ from a fixed point O and is moving with speed $v$ away from O . A horizontal force of magnitude $3 m t$ acts on P , in a direction away from O .
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = 3 t$.
\item Verify that the general solution of this differential equation is $x = \frac { 1 } { 2 } t ^ { 3 } + A t + k$, where $A$ and $k$ are constants.
\item Given that $x = 6$ and $v = 12$ when $t = 1$, find the values of $A$ and $k$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics B AS 2019 Q2 [7]}}
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