Question 2 - A-Level Maths

OCR MEI Further Mechanics B AS 2021 November — Question 2 8 marks

Exam Board	OCR MEI
Module	Further Mechanics B AS (Further Mechanics B AS)
Year	2021
Session	November
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Variable acceleration (vectors)
Type	3D vector motion problems
Difficulty	Standard +0.8 This question requires differentiating a vector velocity to find acceleration, then solving conditions where acceleration components satisfy specific directional constraints (parallel to i, and at 45° to i). The multi-step nature, involving vector calculus, geometric interpretation of angles, and coordinate geometry to find distance, places it moderately above average difficulty for A-level, though it's systematic rather than requiring deep insight.
Spec	1.10a Vectors in 2D: i,j notation and column vectors 1.10b Vectors in 3D: i,j,k notation 1.10c Magnitude and direction: of vectors 1.10f Distance between points: using position vectors 3.02a Kinematics language: position, displacement, velocity, acceleration 3.02b Kinematic graphs: displacement-time and velocity-time

2 A particle, Q , moves so that its velocity, \(\mathbf { v }\), at time \(t\) is given by \(\mathbf { v } = ( 6 t - 6 ) \mathbf { i } + \left( 3 - 2 t + t ^ { 2 } \right) \mathbf { j } + 4 \mathbf { k }\), where \(0 \leqslant t \leqslant 6\).

Explain how you know that Q is never stationary. When Q is at a point A the direction of the acceleration of Q is parallel to the \(\mathbf { i }\) direction. When Q is at a point B the direction of the acceleration of Q makes an angle of \(45 ^ { \circ }\) with the \(\mathbf { i }\) direction.
Determine the straight-line distance AB .

Show mark scheme Show mark scheme source

Question 2:

Answer	Marks	Guidance
2	(a)	Speed in k direction is always 4

component is always 4”

[1]

Answer	Marks
(b)	6

𝐚 = (−2+2𝑡)

Answer	Marks
0	M1
A1	3.1a
1.1	Attempt to differentiate; allow if one

component correct

Answer	Marks
All correct	Or component form

At A when t = 1

Answer	Marks	Guidance
At B when t = 4	A1	2.1

3𝑡2−6𝑡

𝑐

1 𝐫 = (3𝑡−𝑡2+ 𝑡3)+(𝑐 2)

3 𝑐

3

Answer	Marks
4𝑡	M1
A1	1.1
1.1	Attempt to integrate

Correct, including constants in some

Answer	Marks
form	Allow constants to be

implied by use of limits

(( ) ( ))2

342 −64 − 312 −61

(( ) ( ))2

d = + 34−42 +143 − 31−12 +113

3 3

+(44−41)2

Answer	Marks	Guidance
= 272 +152 +122 = 1098	M1	1.1
Dist is 33.1 (cm)	A1	1.1

[7]

Question 2:
2 | (a) | Speed in k direction is always 4 | B1 | 2.4 | Oe involving i and j | Allow “Because the k
component is always 4”
[1]
(b) | 6
𝐚 = (−2+2𝑡)
0 | M1
A1 | 3.1a
1.1 | Attempt to differentiate; allow if one
component correct
All correct | Or component form
At A when t = 1
At B when t = 4 | A1 | 2.1 | Both correct
3𝑡2−6𝑡
𝑐
1
1
𝐫 = (3𝑡−𝑡2+ 𝑡3)+(𝑐 2)
3 𝑐
3
4𝑡 | M1
A1 | 1.1
1.1 | Attempt to integrate
Correct, including constants in some
form | Allow constants to be
implied by use of limits
(( ) ( ))2
342 −64 − 312 −61
(( ) ( ))2
d = + 34−42 +143 − 31−12 +113
3 3
+(44−41)2
= 272 +152 +122 = 1098 | M1 | 1.1 | For using dist when t = 4 and t = 1
Dist is 33.1 (cm) | A1 | 1.1 | 33.136…
[7]

Show LaTeX source

2 A particle, Q , moves so that its velocity, $\mathbf { v }$, at time $t$ is given by $\mathbf { v } = ( 6 t - 6 ) \mathbf { i } + \left( 3 - 2 t + t ^ { 2 } \right) \mathbf { j } + 4 \mathbf { k }$, where $0 \leqslant t \leqslant 6$.
\begin{enumerate}[label=(\alph*)]
\item Explain how you know that Q is never stationary.

When Q is at a point A the direction of the acceleration of Q is parallel to the $\mathbf { i }$ direction.

When Q is at a point B the direction of the acceleration of Q makes an angle of $45 ^ { \circ }$ with the $\mathbf { i }$ direction.
\item Determine the straight-line distance AB .
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics B AS 2021 Q2 [8]}}

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