| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Position from velocity and initial conditions |
| Difficulty | Moderate -0.3 This is a standard M1 kinematics question involving differentiation, integration with initial conditions, and vector interpretation. Parts (i)-(iv) require routine calculus and algebraic manipulation (differentiating a quadratic, integrating a quadratic with boundary condition, solving simultaneous equations). Part (v) involves plotting, which is straightforward. While multi-part with 19 marks total, each component uses well-practiced techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02b Kinematic graphs: displacement-time and velocity-time3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | Mark | Comment |
| (i) | v =8−4t |
| Answer | Marks |
|---|---|
| x | M1 |
| Answer | Marks |
|---|---|
| F1 | either Differentiating |
| Answer | Marks |
|---|---|
| x | 3 |
| (ii) | y=∫( 3t2−8t+4 ) dt |
| Answer | Marks |
|---|---|
| so c = 3 – 1 = 2 and y=t3−4t2+4t+2 | M1 |
| Answer | Marks |
|---|---|
| E1 | Integrating v with at least one correct integrated |
| Answer | Marks |
|---|---|
| Clearly shown and stated | 4 |
| (iii) | We need x = 0 so 8t−2t2 =0 |
| Answer | Marks |
|---|---|
| m | M1 |
| Answer | Marks |
|---|---|
| A1 | May be implied. |
| Answer | Marks |
|---|---|
| Condone 18j | 4 |
| (iv) | We need v =v =0 |
| Answer | Marks |
|---|---|
| Distance is 82+22 = 68 m ( 8.25 3 s.f.) | M1 |
| Answer | Marks |
|---|---|
| B1 | either Recognises v =0 when t = 2 |
| Answer | Marks | Guidance |
|---|---|---|
| followed through correctly. | 5 | |
| (v) | t = 0, 1 give (0, 2) and (6, 3) | B1 |
| Answer | Marks |
|---|---|
| B1 | At least one value 0≤t<2 correctly calc. This |
| Answer | Marks |
|---|---|
| use of line segments. | 3 |
Question 6:
6 | Mark | Comment | Sub
(i) | v =8−4t
x
v =0⇔t=2 so at t = 2
x | M1
A1
F1 | either Differentiating
or Finding ‘u’ and ‘a’ from x and use of v = u + at
FT their v =0
x | 3
(ii) | y=∫( 3t2−8t+4 ) dt
=t3−4t2+4t+c
y = 3 when t = 1 so 3=1−4+4+c
so c = 3 – 1 = 2 and y=t3−4t2+4t+2 | M1
A1
MM11
E1 | Integrating v with at least one correct integrated
y
term.
All correct. Accept no arbitrary constant.
Clea evidence
Clearly shown and stated | 4
(iii) | We need x = 0 so 8t−2t2 =0
so t = 0 or t = 4
t = 0 gives y = 2 so 2 m
t = 4 gives y = 43−43+16+2=18 so 18
m | M1
A1
A1
A1 | May be implied.
Must have both
Condone 2j
Condone 18j | 4
(iv) | We need v =v =0
x y
From above, v =0 only when t = 2 so
x
evaluatev (2)
y
v (2)=0 [(t – 2) is a factor] so yes only
y
at t = 2
At t = 2, the position is (8, 2)
Distance is 82+22 = 68 m ( 8.25 3 s.f.) | M1
M1
A1
B1
B1 | either Recognises v =0 when t = 2
x
or Finds time(s) whenv =0
y
or States or implies v =v =0
x y
Considers v =0 and v = 0with their time(s)
x y
t = 2 recognised as only value (accept as evidence
only
t = 2 used below).
For the last 2 marks, no credit lost for reference
to t= 2 .
3
May be implied
FT from their position. Accept one position
followed through correctly. | 5
(v) | t = 0, 1 give (0, 2) and (6, 3) | B1
B1
B1 | At least one value 0≤t<2 correctly calc. This
need not be plotted
Must be x-y curve. Accept sketch. Ignore curve
outside interval for t.
Accept unlabelled axes. Condone use of line
segments.
At least three correct points used in x-y graph or
sketch. General shape correct. Do not condone
use of line segments. | 3
19A toy boat moves in a horizontal plane with position vector $\mathbf{r} = x\mathbf{i} + y\mathbf{j}$, where $\mathbf{i}$ and $\mathbf{j}$ are the standard unit vectors east and north respectively. The origin of the position vectors is at O. The displacements $x$ and $y$ are in metres.
First consider only the motion of the boat parallel to the $x$-axis. For this motion
$$x = 8t - 2t^2.$$
The velocity of the boat in the $x$-direction is $v_x$ ms$^{-1}$.
\begin{enumerate}[label=(\roman*)]
\item Find an expression in terms of $t$ for $v_x$ and determine when the boat instantaneously has zero speed in the $x$-direction. [3]
\end{enumerate}
Now consider only the motion of the boat parallel to the $y$-axis. For this motion
$$v_y = (t - 2)(3t - 2),$$
where $v_y$ ms$^{-1}$ is the velocity of the boat in the $y$-direction at time $t$ seconds.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Given that $y = 3$ when $t = 1$, use integration to show that $y = t^3 - 4t^2 + 4t + 2$. [4]
\end{enumerate}
The position vector of the boat is given in terms of $t$ by $\mathbf{r} = (8t - 2t^2)\mathbf{i} + (t^3 - 4t^2 + 4t + 2)\mathbf{j}$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the time(s) when the boat is due north of O and also the distance of the boat from O at any such times. [4]
\item Find the time(s) when the boat is instantaneously at rest. Find the distance of the boat from O at any such times. [5]
\item Plot a graph of the path of the boat for $0 \leq t \leq 2$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M1 Q6 [19]}}