OCR MEI Further Mechanics B AS 2022 June — Question 4 10 marks

Exam BoardOCR MEI
ModuleFurther Mechanics B AS (Further Mechanics B AS)
Year2022
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeProjectile on inclined plane
DifficultyStandard +0.8 This is a Further Maths mechanics question requiring resolution perpendicular to an inclined plane (non-standard coordinate system), derivation of a time-of-flight formula, numerical calculation, and qualitative reasoning about parameter effects. The perpendicular resolution and proof in part (a) requires careful geometric thinking beyond standard projectile questions, placing it moderately above average difficulty.
Spec3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model
4 A plane is inclined at an angle \(\theta ^ { \circ }\) to the horizontal. A particle is projected from a point A on the plane with speed \(V \mathrm {~ms} ^ { - 1 }\) in a direction making an angle of \(\phi ^ { \circ }\) with a line of greatest slope of the plane. The particle lands at a point B on the plane, as shown in the diagram, and the time of flight is \(T\) seconds. \includegraphics[max width=\textwidth, alt={}, center]{feb9a438-26b0-41d3-b044-6acd6efccde0-4_332_872_461_246} \begin{enumerate}[label=(\alph*)] \item By considering the motion of the particle perpendicular to the plane, show that \(\mathrm { T } = \frac { 2 \mathrm {~V} \sin \phi } { \mathrm {~g} \cos \theta }\). Consider the case when \(\theta = 30 , \phi = 25\) and \(V = 20\). \item
  1. Calculate the distance AB .
  2. State, with reasons but without any detailed calculations, what effect each of the following actions would have on the distance AB .
Question 4:
AnswerMarks Guidance
4(a) 1
𝑠 = 𝑉sinβˆ…π‘‘βˆ’ π‘Žπ‘‘2
AnswerMarks Guidance
2M1 3.3
Accel = 𝑔cosπœƒM1 1.1
1
0 = 𝑉sinβˆ…π‘‡βˆ’ 𝑔cosπœƒπ‘‡2
AnswerMarks Guidance
2A1 3.4
1
Eg 𝑔cosπœƒπ‘‡ = 𝑉sinβˆ…
AnswerMarks
2Correct intermediate step
2𝑉sinβˆ…
𝑇 =
AnswerMarks Guidance
𝑔cosπœƒA1 2.2a
ALT𝑉_π‘π‘’π‘Ÿπ‘ = 𝑉sinβˆ…βˆ’π‘Žπ‘‘ M1
Accel = 𝑔cosπœƒM1 Attempt to resolve g
𝑉sinβˆ…βˆ’π‘”cosπœƒπ‘‡ = 0
AnswerMarks Guidance
π‘‘π‘œπ‘π‘‰sinβˆ…βˆ’π‘”cosπœƒπ‘‡ = 0
π‘‘π‘œπ‘A1 A1
comp in attempt to find time to top
AnswerMarks Guidance
of motionUsing velocity 0 and correct accel Or could solve
comp in attempt to find time to top𝑉sinβˆ…βˆ’π‘”cosπœƒπ‘‡
of motion= βˆ’π‘‰sinβˆ…
𝑉sinβˆ…
𝑇 =
AnswerMarks
π‘‘π‘œπ‘ 𝑔cosπœƒCorrect intermediate step
2𝑉sinβˆ…
𝑇 = 2𝑇 =
AnswerMarks Guidance
π‘‘π‘œπ‘ 𝑔cosπœƒA1 AG
[4]
AnswerMarks
(b)2Γ—20sin25Β°
𝑇 = = 1.99
𝑔cos30Β°
(
i
AnswerMarks Guidance
)B1 3.4
1
𝑋 = 20cos25Β°π‘‡βˆ’ 𝑔sin30°𝑇2
AnswerMarks Guidance
2M1 3.4
Range = 26.4A1 1.1
[3]
AnswerMarks Guidance
(ii)AB is bigger because greater V means particle goes
higher and furtherB1 1.1
detailed calculations
AB gets bigger to start with as time of flight is
increased, but for even bigger  velocity parallel
to plane is reduced and AB decreases.
AnswerMarks
(Eventually AB becomes negative.)B1
B12.2b
3.5aDo not accept answers coming from
detailed calculationsSC If no reason given, then
allow B1 for statement that
AB will increase at first and
then decrease.
[3]
Question 4:
4 | (a) | 1
𝑠 = 𝑉sinβˆ…π‘‘βˆ’ π‘Žπ‘‘2
2 | M1 | 3.3
Accel = 𝑔cosπœƒ | M1 | 1.1 | Attempt to resolve g
1
0 = 𝑉sinβˆ…π‘‡βˆ’ 𝑔cosπœƒπ‘‡2
2 | A1 | 3.4 | Using dist 0 and correct accel comp
1
Eg 𝑔cosπœƒπ‘‡ = 𝑉sinβˆ…
2 | Correct intermediate step
2𝑉sinβˆ…
𝑇 =
𝑔cosπœƒ | A1 | 2.2a | AG
ALT | 𝑉_π‘π‘’π‘Ÿπ‘ = 𝑉sinβˆ…βˆ’π‘Žπ‘‘ | M1
Accel = 𝑔cosπœƒ | M1 | Attempt to resolve g
𝑉sinβˆ…βˆ’π‘”cosπœƒπ‘‡ = 0
π‘‘π‘œπ‘ | 𝑉sinβˆ…βˆ’π‘”cosπœƒπ‘‡ = 0
π‘‘π‘œπ‘ | A1 | A1 | Using velocity 0 and correct accel
comp in attempt to find time to top
of motion | Using velocity 0 and correct accel | Or could solve
comp in attempt to find time to top | 𝑉sinβˆ…βˆ’π‘”cosπœƒπ‘‡
of motion | = βˆ’π‘‰sinβˆ…
𝑉sinβˆ…
𝑇 =
π‘‘π‘œπ‘ 𝑔cosπœƒ | Correct intermediate step
2𝑉sinβˆ…
𝑇 = 2𝑇 =
π‘‘π‘œπ‘ 𝑔cosπœƒ | A1 | AG
[4]
(b) | 2Γ—20sin25Β°
𝑇 = = 1.99
𝑔cos30Β°
(
i
) | B1 | 3.4 | 1.991…
1
𝑋 = 20cos25Β°π‘‡βˆ’ 𝑔sin30°𝑇2
2 | M1 | 3.4
Range = 26.4 | A1 | 1.1 | 26.377… Accept 26
[3]
(ii) | AB is bigger because greater V means particle goes
higher and further | B1 | 1.1 | Do not accept answers coming from
detailed calculations
AB gets bigger to start with as time of flight is
increased, but for even bigger  velocity parallel
to plane is reduced and AB decreases.
(Eventually AB becomes negative.) | B1
B1 | 2.2b
3.5a | Do not accept answers coming from
detailed calculations | SC If no reason given, then
allow B1 for statement that
AB will increase at first and
then decrease.
[3]
4 A plane is inclined at an angle $\theta ^ { \circ }$ to the horizontal. A particle is projected from a point A on the plane with speed $V \mathrm {~ms} ^ { - 1 }$ in a direction making an angle of $\phi ^ { \circ }$ with a line of greatest slope of the plane. The particle lands at a point B on the plane, as shown in the diagram, and the time of flight is $T$ seconds.\\
\includegraphics[max width=\textwidth, alt={}, center]{feb9a438-26b0-41d3-b044-6acd6efccde0-4_332_872_461_246}
\begin{enumerate}[label=(\alph*)]
\item By considering the motion of the particle perpendicular to the plane, show that $\mathrm { T } = \frac { 2 \mathrm {~V} \sin \phi } { \mathrm {~g} \cos \theta }$.

Consider the case when $\theta = 30 , \phi = 25$ and $V = 20$.
\item \begin{enumerate}[label=(\roman*)]
\item Calculate the distance AB .
\item State, with reasons but without any detailed calculations, what effect each of the following actions would have on the distance AB .

\begin{itemize}
\end{enumerate}\item Increasing $V$ while leaving $\theta$ and $\phi$ unchanged.
  \item Increasing $\phi$ while leaving $\theta$ and $V$ unchanged.
\end{itemize}
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics B AS 2022 Q4 [10]}}
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