OCR MEI Further Mechanics Major 2022 June — Question 8 13 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
Year2022
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeDeriving trajectory equation
DifficultyStandard +0.8 This is a multi-part Further Maths projectile motion question requiring vector integration, elimination of parameters to find a Cartesian equation, completing the square to find maximum height, and solving a constraint equation. While the techniques are standard (integration, parameter elimination, optimization), it requires careful algebraic manipulation across multiple connected parts and is from a Further Maths module, placing it above average difficulty.
Spec3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model
A particle P is projected from a fixed point O with initial velocity \(u\mathbf{i} + ku\mathbf{j}\), where \(k\) is a positive constant. The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are in the horizontal and vertically upward directions respectively. P moves with constant gravitational acceleration of magnitude \(g\). At time \(t \geq 0\), particle P has position vector \(\mathbf{r}\) relative to O.
  1. Starting from an expression for \(\ddot{\mathbf{r}}\), use integration to derive the formula $$\mathbf{r} = ut\mathbf{i} + \left(kut - \frac{1}{2}gt^2\right)\mathbf{j}.$$ [4]
The position vector \(\mathbf{r}\) of P at time \(t \geq 0\) can be expressed as \(\mathbf{r} = x\mathbf{i} + y\mathbf{j}\), where the axes Ox and Oy are horizontally and vertically upwards through O respectively. The axis Ox lies on horizontal ground.
  1. Show that the path of P has cartesian equation $$gy^2 - 2ku^2x + 2u^2y = 0.$$ [3]
  2. Hence find, in terms of \(g\), \(k\) and \(u\), the maximum height of P above the ground during its motion. [3]
The maximum height P reaches above the ground is equal to the distance OA, where A is the point where P first hits the ground.
  1. Determine the value of \(k\). [3]
Question 8:
AnswerMarks Guidance
8(a) r=−gj
r =−gtj+c
1
t =0,r =ui+kuj⇒c =...
1
r =ui+(ku−gt) j⇒r=...
 1 
r=uti+ kut− gt2 j+c and showing c =0leading to
2 2
 2 
 1 
r=uti+ kut− gt2 j
AnswerMarks
 2 B1
M1
M1
A1
AnswerMarks
[4]1.2
3.4
1.1
AnswerMarks
2.2aCorrectly integrates their rand attempt to find c
1
using correct initial conditions – correct expression
for rcan imply this mark
Integrates their expression for r(all powers increased
by one)
AG – must show (or explicitly) state that second
constant of integration is 0
Deriving the given result from constant
acceleration formulae scores no marks
AnswerMarks
(b)1
x=ut,y=kut− gt2
2
2
x 1 x
y=ku − g 
u 2 u
gx2
y=kx− ⇒gx2 −2ku2x+2u2y=0
AnswerMarks
2u2B1
M1
A1
AnswerMarks
[3]1.1
1.1
AnswerMarks
1.1Correctly remove vectors and states the correct
equations for x and y
Eliminates t
AG - sufficient working must be shown
AnswerMarks
(c)( )2 ( )
∆= −2ku2 −4g 2u2y
4k2u4 −8gu2y=0⇒ y=...
4k2u4 k2u2
y = =
AnswerMarks
max 8gu2 2gM1*
M1dep*
A1
AnswerMarks
[3]3.1a
2.1
AnswerMarks
2.2aConsiders equation of the path as a quadratic in x and
calculates its discriminant
Sets discriminant equal to zero and solves for y
Alternative method
dy dy
2gx−2ku2 +2u2 =0and =0
AnswerMarks Guidance
dx dxM1* M1*
derivative equal to zero
2
ku2 ku2  ku2 
x= ⇒g  −2ku2  +2u2y=0
   
AnswerMarks Guidance
g  g   g M1dep* Solves for x and substitutes into original equation
1 2k2u4 k2u4  k2u2
y =  − =
AnswerMarks
max 2u2   g g   2 gA1
[3]
AnswerMarks
(d)2ku2
y=0⇒OA=
AnswerMarks Guidance
gB1 3.4
Correct expression for OA or implied e.g. kx=
2u2
2ku
(where x = OA) or finding the time of flight T as
g
k2u2 2ku2
= ⇒k =...
AnswerMarks Guidance
2g gM1 3.4
and solving for k or for equating the time of flight
from y = 0 with the time of flight from OA = uT and
solving for k
AnswerMarks Guidance
k2 −4k =0⇒k =4A1 1.1
[3]
Question 8:
8 | (a) | r=−gj
r =−gtj+c
1
t =0,r =ui+kuj⇒c =...
1
r =ui+(ku−gt) j⇒r=...
 1 
r=uti+ kut− gt2 j+c and showing c =0leading to
2 2
 2 
 1 
r=uti+ kut− gt2 j
 2  | B1
M1
M1
A1
[4] | 1.2
3.4
1.1
2.2a | Correctly integrates their rand attempt to find c
1
using correct initial conditions – correct expression
for rcan imply this mark
Integrates their expression for r(all powers increased
by one)
AG – must show (or explicitly) state that second
constant of integration is 0
Deriving the given result from constant
acceleration formulae scores no marks
(b) | 1
x=ut,y=kut− gt2
2
2
x 1 x
y=ku − g 
u 2 u
gx2
y=kx− ⇒gx2 −2ku2x+2u2y=0
2u2 | B1
M1
A1
[3] | 1.1
1.1
1.1 | Correctly remove vectors and states the correct
equations for x and y
Eliminates t
AG - sufficient working must be shown
(c) | ( )2 ( )
∆= −2ku2 −4g 2u2y
4k2u4 −8gu2y=0⇒ y=...
4k2u4 k2u2
y = =
max 8gu2 2g | M1*
M1dep*
A1
[3] | 3.1a
2.1
2.2a | Considers equation of the path as a quadratic in x and
calculates its discriminant
Sets discriminant equal to zero and solves for y
Alternative method
dy dy
2gx−2ku2 +2u2 =0and =0
dx dx | M1* | M1* | Differentiates (powers decreased by one) and sets the
derivative equal to zero
2
ku2 ku2  ku2 
x= ⇒g  −2ku2  +2u2y=0
   
g  g   g  | M1dep* | Solves for x and substitutes into original equation | Solves for x and substitutes into original equation
1 2k2u4 k2u4  k2u2
y =  − =
max 2u2   g g   2 g | A1
[3]
(d) | 2ku2
y=0⇒OA=
g | B1 | 3.4 | gx2
Correct expression for OA or implied e.g. kx=
2u2
2ku
(where x = OA) or finding the time of flight T as
g
k2u2 2ku2
= ⇒k =...
2g g | M1 | 3.4 | Sets answer to (c) equal to their expression for OA
and solving for k or for equating the time of flight
from y = 0 with the time of flight from OA = uT and
solving for k
k2 −4k =0⇒k =4 | A1 | 1.1
[3]
A particle P is projected from a fixed point O with initial velocity $u\mathbf{i} + ku\mathbf{j}$, where $k$ is a positive constant. The unit vectors $\mathbf{i}$ and $\mathbf{j}$ are in the horizontal and vertically upward directions respectively. P moves with constant gravitational acceleration of magnitude $g$.
At time $t \geq 0$, particle P has position vector $\mathbf{r}$ relative to O.

\begin{enumerate}[label=(\alph*)]
\item Starting from an expression for $\ddot{\mathbf{r}}$, use integration to derive the formula
$$\mathbf{r} = ut\mathbf{i} + \left(kut - \frac{1}{2}gt^2\right)\mathbf{j}.$$ [4]
\end{enumerate}

The position vector $\mathbf{r}$ of P at time $t \geq 0$ can be expressed as $\mathbf{r} = x\mathbf{i} + y\mathbf{j}$, where the axes Ox and Oy are horizontally and vertically upwards through O respectively. The axis Ox lies on horizontal ground.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that the path of P has cartesian equation
$$gy^2 - 2ku^2x + 2u^2y = 0.$$ [3]

\item Hence find, in terms of $g$, $k$ and $u$, the maximum height of P above the ground during its motion. [3]
\end{enumerate}

The maximum height P reaches above the ground is equal to the distance OA, where A is the point where P first hits the ground.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Determine the value of $k$. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2022 Q8 [13]}}
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