OCR MEI Further Mechanics Major 2019 June — Question 5 7 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
Year2019
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeGiven velocity function find force
DifficultyStandard +0.3 This is a straightforward Further Mechanics question requiring standard differentiation of the position vector to find velocity and acceleration, followed by routine calculations. Part (a) involves differentiating to get velocity, finding speed, and applying KE = ½mv². Part (b) requires finding acceleration by differentiating again, calculating its magnitude, and solving a simple exponential equation. While it's Further Maths content, the techniques are mechanical and well-practiced, making it slightly easier than average overall.
Spec3.02f Non-uniform acceleration: using differentiation and integration6.02d Mechanical energy: KE and PE concepts
A particle P of mass 4 kilograms moves in such a way that its position vector at time \(t\) seconds is \(\mathbf{r}\) metres, where $$\mathbf{r} = 3t\mathbf{i} + 2e^{-3t}\mathbf{j}.$$
  1. Find the initial kinetic energy of P. [4]
  2. Find the time when the acceleration of P is 2 metres per second squared. [3]
Question 5:
AnswerMarks Guidance
5(a) r(t)3i6e 3t j
r(0)3i6j
KE 1  4  32 62 
2
AnswerMarks
90 (J)M1*
M1dep*
M1dep*
A1
AnswerMarks
[4]3.1b
3.4
3.3
AnswerMarks
1.1Attempt at differentiation – must be of
the form 3ike 3t j where k 0
Substitute t = 0 into their r(t)
Correct method for finding KE with
AnswerMarks
their r(0)Dependent on first two
M marks
AnswerMarks Guidance
5(b) 18e3t 2
1
e 3t 
9
1
3tln t...
9
AnswerMarks
t0.732M1*
M1dep*
A1
AnswerMarks
[3]3.4
1.1
AnswerMarks
1.1Differentiating their r(t)and equating
to 2
Correct method (i.e. taking logs
correctly) to find t
1
oe e.g. ln9 - if not given exact then
3
AnswerMarks
must be given to at least 2 sfTheir acceleration must
be of the form ke 3t j
1
0.7324081…
Question 5:
5 | (a) | r(t)3i6e 3t j
r(0)3i6j
KE 1  4  32 62 
2
90 (J) | M1*
M1dep*
M1dep*
A1
[4] | 3.1b
3.4
3.3
1.1 | Attempt at differentiation – must be of
the form 3ike 3t j where k 0
Substitute t = 0 into their r(t)
Correct method for finding KE with
their r(0) | Dependent on first two
M marks
5 | (b) | 18e3t 2
1
e 3t 
9
1
3tln t...
9
t0.732 | M1*
M1dep*
A1
[3] | 3.4
1.1
1.1 | Differentiating their r(t)and equating
to 2
Correct method (i.e. taking logs
correctly) to find t
1
oe e.g. ln9 - if not given exact then
3
must be given to at least 2 sf | Their acceleration must
be of the form ke 3t j
1
0.7324081…
A particle P of mass 4 kilograms moves in such a way that its position vector at time $t$ seconds is $\mathbf{r}$ metres, where

$$\mathbf{r} = 3t\mathbf{i} + 2e^{-3t}\mathbf{j}.$$

\begin{enumerate}[label=(\alph*)]
\item Find the initial kinetic energy of P. [4]
\item Find the time when the acceleration of P is 2 metres per second squared. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2019 Q5 [7]}}
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Q1 5 Q2 4 Q3 5 Q4 6 Q5 7 Q6 7 Q7 8 Q8 11 Q9 12 Q10 8 Q11 14 Q12 16 Q13 17