Question 9 - A-Level Maths

OCR MEI Further Mechanics Major 2021 November — Question 9 15 marks

Exam Board	OCR MEI
Module	Further Mechanics Major (Further Mechanics Major)
Year	2021
Session	November
Marks	15
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Projectiles
Type	Projectile on inclined plane
Difficulty	Challenging +1.2 This is a standard Further Maths projectile-on-inclined-plane question requiring coordinate transformation and trigonometric manipulation. Part (a) involves deriving a given result through standard projectile equations, part (b) applies a provided identity to find maximum range, and part (c) is straightforward substitution. While it requires multiple steps and careful algebra, the techniques are well-established for Further Maths students and the identity is given.
Spec	1.05b Sine and cosine rules: including ambiguous case 1.05c Area of triangle: using 1/2 ab sin(C)1.05l Double angle formulae: and compound angle formulae 3.02i Projectile motion: constant acceleration model

9 A small ball P is projected with speed $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of elevation of $( \alpha + \theta )$ from a point O at the bottom of a plane inclined at $\alpha$ to the horizontal. P subsequently hits the plane at a point R , where OR is a line of greatest slope, as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{17e92314-d7df-49b8-a441-8d18c91dbbb0-07_456_862_406_242}

By deriving an expression, in terms of $\theta$, $\alpha$ and $g$, for the time of flight of P , show that the distance OR, in metres, is $$\frac { 50 \sin \theta \cos ( \theta + \alpha ) } { g \cos ^ { 2 } \alpha }$$
By using the identity $2 \sin \mathrm {~A} \cos \mathrm {~B} \equiv \sin ( \mathrm {~A} + \mathrm { B } ) - \sin ( \mathrm { B } - \mathrm { A } )$, determine, in terms of $g$ and $\sin \alpha$, an expression for the maximum range of P up the plane, as $\theta$ varies.
Given that OR is the maximum range of P up the plane and is equal to 1.8 m , determine the value of $\theta$. \includegraphics[max width=\textwidth, alt={}, center]{17e92314-d7df-49b8-a441-8d18c91dbbb0-08_625_1180_255_239} A rigid wire ABC is fixed in a vertical plane. The section AB of the wire, of length $b$, is straight and horizontal. The section BC of the wire is smooth and in the form of a circular arc of radius $a$ and length $\frac { 1 } { 2 } a \pi$. The centre of the arc is O , which is vertically above B . A bead P of mass $m$ is threaded on the wire and projected from B with speed $u$ towards C . The angle BOP when P is between B and C is denoted by $\theta$, as shown in the diagram.

Show mark scheme Show mark scheme source

Question 9:

Answer	Marks	Guidance
9	(a)	x = −g sinα , y = −g cosα

x = 5cosθ − gt sinα , y = 5sinθ − gt cosα

x = 5t cosθ − 0.5gt2 sinα

y = 5t sinθ − 0.5gt2 cosα

y = 0 ⇒ t = ...

10sinθ

t =

g cosα

2  10sinθ   10sinθ 

x = 5 cosθ − 0.5g   sinα

 g cosα   g cosα 

50sinθ

x = (cosθ cosα − sinθ sinα )

g cos2 α

50sinθ cos(θ + α )

⇒ OR =

Answer	Marks
g cos2 α	B1

M1*

A1

M1dep*

A1

M1

A1

Answer	Marks
[8]	2.1

3.4

1.1

3.3

1.1

3.4

Answer	Marks
2.2a	Attempt to integrate (twice) and use of

initial conditions

1 Or M1 for use of s = ut + at 2 parallel

2 to line of greatest slope and then A1 for

correct expression for x

Sets y = 0 and solve for t

Substitute expression for t into equation

for x

Answer	Marks
AG	Similarly M1 A1 for

correct expression for

y (following SUVAT

perpendicular to

slope)

Dependent on both

previous M marks

Answer	Marks	Guidance
9	(b)	sinθ cos(θ + α ) = 1 ( sin (2θ + α ) − sinα )

2

25 OR = ( sin(2θ + α ) − sinα )

g cos2 α

25 R = ( ) ( 1 − sinα )

g 1 − sin α

Answer	Marks
max 2	M1

A1

Answer	Marks
[3]	1.1

1.1

Answer	Marks
3.1a	Use of given identity to re-write

numerator from (a) as a difference of two

sines

Use of correct trig. identity and setting

sin(2θ + α ) equal to 1 – oe e.g.

25 R =

Answer	Marks
max g ( 1 + sinα )	R occurs when

max

sin(2θ + α ) = 1

Answer	Marks	Guidance
9	(c)	25 25(1− sinα )

= 1.8 or = 1.8

g ( 1 + sinα ) g ( 1 − sin2 α )

25 = 1.8 ⇒ sinα = ...

g ( 1 + sinα )

θ = 45 − 0.5α

Answer	Marks
θ = 32.7	M1*

M1dep*

M1

A1

Answer	Marks
[4]	3.4

1.1 3.1a

Answer	Marks
1.1	Setting their expression equal to 1.8

Attempting to solve for sinα or α - for

reference sinα = 184 or α = 24.660053...

441 (or 0.430399… in radians)

Answer	Marks
Follow through their α	Expression must only

contain sinα terms

If solving a 3TQ in

sine then must solve

using a correct method

32.6699733… or

0.5701986… (in

radians)

Question 9:
9 | (a) | x = −g sinα , y = −g cosα
x = 5cosθ − gt sinα , y = 5sinθ − gt cosα
x = 5t cosθ − 0.5gt2 sinα
y = 5t sinθ − 0.5gt2 cosα
y = 0 ⇒ t = ...
10sinθ
t =
g cosα
2
 10sinθ   10sinθ 
x = 5 cosθ − 0.5g   sinα
 g cosα   g cosα 
50sinθ
x = (cosθ cosα − sinθ sinα )
g cos2 α
50sinθ cos(θ + α )
⇒ OR =
g cos2 α | B1
M1*
A1
A1
M1dep*
A1
M1
A1
[8] | 2.1
3.4
1.1
1.1
3.3
1.1
3.4
2.2a | Attempt to integrate (twice) and use of
initial conditions
1
Or M1 for use of s = ut + at 2 parallel
2
to line of greatest slope and then A1 for
correct expression for x
Sets y = 0 and solve for t
Substitute expression for t into equation
for x
AG | Similarly M1 A1 for
correct expression for
y (following SUVAT
perpendicular to
slope)
Dependent on both
previous M marks
9 | (b) | sinθ cos(θ + α ) = 1 ( sin (2θ + α ) − sinα )
2
25
OR = ( sin(2θ + α ) − sinα )
g cos2 α
25
R = ( ) ( 1 − sinα )
g 1 − sin α
max 2 | M1
A1
A1
[3] | 1.1
1.1
3.1a | Use of given identity to re-write
numerator from (a) as a difference of two
sines
Use of correct trig. identity and setting
sin(2θ + α ) equal to 1 – oe e.g.
25
R =
max g ( 1 + sinα ) | R occurs when
max
sin(2θ + α ) = 1
9 | (c) | 25 25(1− sinα )
= 1.8 or = 1.8
g ( 1 + sinα ) g ( 1 − sin2 α )
25
= 1.8 ⇒ sinα = ...
g ( 1 + sinα )
θ = 45 − 0.5α
θ = 32.7 | M1*
M1dep*
M1
A1
[4] | 3.4
1.1
3.1a
1.1 | Setting their expression equal to 1.8
Attempting to solve for sinα or α - for
reference sinα = 184 or α = 24.660053...
441
(or 0.430399… in radians)
Follow through their α | Expression must only
contain sinα terms
If solving a 3TQ in
sine then must solve
using a correct method
32.6699733… or
0.5701986… (in
radians)

Show LaTeX source

9 A small ball P is projected with speed $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of elevation of $( \alpha + \theta )$ from a point O at the bottom of a plane inclined at $\alpha$ to the horizontal. P subsequently hits the plane at a point R , where OR is a line of greatest slope, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{17e92314-d7df-49b8-a441-8d18c91dbbb0-07_456_862_406_242}
\begin{enumerate}[label=(\alph*)]
\item By deriving an expression, in terms of $\theta$, $\alpha$ and $g$, for the time of flight of P , show that the distance OR, in metres, is

$$\frac { 50 \sin \theta \cos ( \theta + \alpha ) } { g \cos ^ { 2 } \alpha }$$
\item By using the identity $2 \sin \mathrm {~A} \cos \mathrm {~B} \equiv \sin ( \mathrm {~A} + \mathrm { B } ) - \sin ( \mathrm { B } - \mathrm { A } )$, determine, in terms of $g$ and $\sin \alpha$, an expression for the maximum range of P up the plane, as $\theta$ varies.
\item Given that OR is the maximum range of P up the plane and is equal to 1.8 m , determine the value of $\theta$.\\
\includegraphics[max width=\textwidth, alt={}, center]{17e92314-d7df-49b8-a441-8d18c91dbbb0-08_625_1180_255_239}

A rigid wire ABC is fixed in a vertical plane. The section AB of the wire, of length $b$, is straight and horizontal. The section BC of the wire is smooth and in the form of a circular arc of radius $a$ and length $\frac { 1 } { 2 } a \pi$. The centre of the arc is O , which is vertically above B .

A bead P of mass $m$ is threaded on the wire and projected from B with speed $u$ towards C . The angle BOP when P is between B and C is denoted by $\theta$, as shown in the diagram.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2021 Q9 [15]}}

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