OCR MEI M1 — Question 4 8 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeVelocity from acceleration and initial conditions
DifficultyModerate -0.3 This is a straightforward M1 kinematics question requiring standard differentiation of position vectors and basic vector interpretation. Part (i) involves simple substitution and bearing calculation, part (ii) requires differentiation and showing a quadratic has no real roots, and part (iii) involves second differentiation and solving a linear equation. All techniques are routine for M1 with no novel problem-solving required, making it slightly easier than average.
Spec1.10a Vectors in 2D: i,j notation and column vectors3.02a Kinematics language: position, displacement, velocity, acceleration3.02b Kinematic graphs: displacement-time and velocity-time
At time \(t\) seconds, a particle has position with respect to an origin O given by the vector $$\mathbf{r} = \begin{pmatrix} 8t \\ 10t^2 - 2t^3 \end{pmatrix},$$ where \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\) and \(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\) are perpendicular unit vectors east and north respectively and distances are in metres.
  1. When \(t = 1\), the particle is at P. Find the bearing of P from O. [2]
  2. Find the velocity of the particle at time \(t\) and show that it is never zero. [3]
  3. Determine the time(s), if any, when the acceleration of the particle is zero. [3]
Question 4:
AnswerMarks Guidance
4mark notes
(i) 8  8
When t = 1, r =   
102 8
[8i +(10 – 2)j = 8i + 8j]
AnswerMarks
Bearing OP is 045°B1
F1
AnswerMarks
2Accept column or ai + bj notation
May be implied
Accept 45°. Accept NE and northeast. Condone r given as well.
AnswerMarks
(ii) 8 
v =   [8i + (20t – 6t²)j]
20t6t2
AnswerMarks
The i cpt is always 8 so v0for any tM1
A1
E1
AnswerMarks
3Differentiating both components. Condone 1 error if clearly attempting differentiation.
Must be a vector answer.
Accept any correct argument e.g. based on i cpt never 0.
AnswerMarks
(iii) 0 
a =  [ (20 – 12t)j ]
2012t
20 5
a = 0 when t 
12 3
5
so s (1.67 s (3 s. f.))
AnswerMarks
3M1
F1
B1
AnswerMarks
3Differentiating as a vector. Condone 1 error if clearly attempting differentiation of their v.
FT their v.
cao. Condone obtained from scalar equation.
8
5
5 (i)
((iiii))
AnswerMarks
(iii)12  2 
= +4a
  9     −3 
2.5
so a =
 
 3 
eit
 −1  2  1
r= + ×4+ a×42
  2     −3   2
27 27
r= so m
   
14 14
or
Using N2L
12.5 12.5
F = 5a = so N
   
AnswerMarks
 15   15 MM11
A1
M1
A1
A1
M1
A1
A1
M1
AnswerMarks
F1Us of v=u+at
If vector a seen, isw.
For use of s=ut+ 1at2with their a. Initial
2
position may be omitted.
FT their a. Initial position may be omitted.
cao. Do not condone magnitude as final answer.
Use of s=0.5t ( u+v ) Initial position may be
omitted.
Correct substitution. Initial position may be
omitted.
cao Do not condone mag as final answer.
28
SC2 for
 
12
Use of F = ma or F = mga.
FT their a only. Do not accept magnitude as final
AnswerMarks
ans.2
3
2
7
Question 4:
4 | mark | notes
(i) |  8  8
When t = 1, r =   
102 8
[8i +(10 – 2)j = 8i + 8j]
Bearing OP is 045° | B1
F1
2 | Accept column or ai + bj notation
May be implied
Accept 45°. Accept NE and northeast. Condone r given as well.
(ii) |  8 
v =   [8i + (20t – 6t²)j]
20t6t2

The i cpt is always 8 so v0for any t | M1
A1
E1
3 | Differentiating both components. Condone 1 error if clearly attempting differentiation.
Must be a vector answer.
Accept any correct argument e.g. based on i cpt never 0.
(iii) |  0 
a =  [ (20 – 12t)j ]
2012t
20 5
a = 0 when t 
12 3
5
so s (1.67 s (3 s. f.))
3 | M1
F1
B1
3 | Differentiating as a vector. Condone 1 error if clearly attempting differentiation of their v.
FT their v.
cao. Condone obtained from scalar equation.
8
5
5 (i)
((iiii))
(iii) | 12  2 
= +4a
  9     −3 
2.5
so a =
 
 3 
eit
 −1  2  1
r= + ×4+ a×42
  2     −3   2
27 27
r= so m
   
14 14
or
Using N2L
12.5 12.5
F = 5a = so N
   
 15   15  | MM11
A1
M1
A1
A1
M1
A1
A1
M1
F1 | Us of v=u+at
If vector a seen, isw.
For use of s=ut+ 1at2with their a. Initial
2
position may be omitted.
FT their a. Initial position may be omitted.
cao. Do not condone magnitude as final answer.
Use of s=0.5t ( u+v ) Initial position may be
omitted.
Correct substitution. Initial position may be
omitted.
cao Do not condone mag as final answer.
28
SC2 for
 
12
Use of F = ma or F = mga.
FT their a only. Do not accept magnitude as final
ans. | 2
3
2
7
At time $t$ seconds, a particle has position with respect to an origin O given by the vector
$$\mathbf{r} = \begin{pmatrix} 8t \\ 10t^2 - 2t^3 \end{pmatrix},$$

where $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ are perpendicular unit vectors east and north respectively and distances are in metres.

\begin{enumerate}[label=(\roman*)]
\item When $t = 1$, the particle is at P. Find the bearing of P from O. [2]
\item Find the velocity of the particle at time $t$ and show that it is never zero. [3]
\item Determine the time(s), if any, when the acceleration of the particle is zero. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI M1  Q4 [8]}}
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