Question 5 - A-Level Maths

Edexcel M2 2024 October — Question 5

Exam Board	Edexcel
Module	M2 (Mechanics 2)
Year	2024
Session	October
Paper	Download PDF ↗
Topic	Projectiles
Type	Projectile clearing obstacle
Difficulty	Standard +0.3 Part (a) is a standard bookwork derivation of the range formula that appears in every M2 textbook. Parts (b) and (c) require applying projectile equations with given values and solving for unknowns, but involve straightforward substitution and calculation with no novel problem-solving insight. This is slightly easier than a typical M2 question due to the guided structure and standard techniques.
Spec	3.02h Motion under gravity: vector form 3.02i Projectile motion: constant acceleration model

The fixed points \(X\) and \(Y\) lie on horizontal ground.

At time \(t = 0\), a particle \(P\) is projected from \(X\) with speed \(u \mathrm {~ms} ^ { - 1 }\) at angle \(\theta\) to the ground. Particle \(P\) moves freely under gravity and first hits the ground at \(Y\).

Show that \(X Y = \frac { u ^ { 2 } \sin 2 \theta } { g }\) The points \(A\) and \(B\) lie on horizontal ground. A vertical pole \(C D\) has length 5 m .
The end \(C\) is fixed to the ground between \(A\) and \(B\), where \(A C = 12 \mathrm {~m}\).
At time \(t = 0\), a particle \(Q\) is projected from \(A\) with speed \(20 \mathrm {~ms} ^ { - 1 }\) at \(60 ^ { \circ }\) to the ground.
Particle \(Q\) moves freely under gravity, passes over the pole and first hits the ground at \(B\), as shown in Figure 4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{3e78f951-041d-4227-aa4b-e67a6ab5b4cd-14_335_1179_1032_443} \captionsetup{labelformat=empty} \caption{Figure 4}
\end{figure}
Find the distance \(C B\).
Find the height of \(Q\) above \(D\) at the instant when \(Q\) passes over the pole.

Show mark scheme Show mark scheme source

Question 5:

Answer	Marks	Guidance
5(a)	Horizontal motion to find expression for the distance	M1
( X Y ) u c o s t  =	A1

Method using vertical motion to find relevant equation in t, u, g,  .

Eg

• Using vertical distance =0 to find expression for time.

Answer	Marks
• Find expression for time to max height and × 2	M1

Correct unsimplified equation in t, u, g,  .

1 usint− gt2 =0

Answer	Marks
2	A1
Solve to obtain the distance in terms of u, g and 	DM1

2 u s i n u 2 s i n 2  

X Y u c o s  =  = *

Answer	Marks
g g	A1*

(6)

5(a)

Answer	Marks
M1	Use horizontal motion to find an expression for the horizontal

distance.

Answer	Marks
A1	Correct unsimplified equation.
M1	Method using vertical motion with relevant suvat to find an

equation in t, u, g,  (condone use of 9.8).

2 u s i n  usin

M0 if t = or t = is quoted (ie if it appears without

g g

any evidence of method)

Answer	Marks
A1	Correct equation, (condone use of 9.8)

Eg

1 • usint− gt2 =0

2 • u s i n u s i n g t   − = −

• u s i n g t 0  − = double the time.

Answer	Marks
DM1	Dependent on the preceding M marks. Solve to obtain the distance

in terms of u, g and .

Answer	Marks
A1*	Obtain given answer from correct working. Must recover g if 9.8 is

used. Must have XY = at this stage.

Answer	Marks
5(b)	2 0 2  s i n 1 2 0 

C B = − 1 2

Answer	Marks
g	M1
C B = 2 3 ( m ) o r C B = 2 3 . 3 ( m )	A1

(2)

5(b)

Answer	Marks
M1	Complete method using result in (a), or using horizontal motion

from first principles, to find the horizontal distance CB.

Answer	Marks
A1	Correct answer, 2sf or 3sf

Answer	Marks	Guidance
5(c)	Horizontal motion	M1
1 2 = 2 0 c o s 6 0  t	A1

Vertical motion

1 y = 2 0 s i n 6 0   1 .2 −  g  1 .2 2 ( − 5 )

Answer	Marks
2	M1

( ) ( )

Answer	Marks
Height above D = 8 .7 m o r 8 .7 3 m	A1

(4)

(12)

Notes

Over-accuracy or under-accuracy is penalised max once per

complete question. Penalise final A mark in the appropriate part.

Use of g = 9.81 or 10 is penalised max once per complete question.

Penalise final A mark in the appropriate part.

If both errors occur they could lose the final two A marks.

5(c)

Answer	Marks
M1	Complete method for horizontal motion to obtain an equation in t. If

t =1.2 o.e. is seen in earlier working, it must be used in (c) to earn

the marks.

Answer	Marks
A1	Correct unsimplified equation (t = 1.2)
M1	Method using suvat with vertical motion to obtain a relevant vertical

distance. Height of pole not required.

Answer	Marks	Guidance
A1	Correct answer, 2 sf or 3 sf
Question	Scheme	Marks

Question 5:
--- 5(a) ---
5(a) | Horizontal motion to find expression for the distance | M1
( X Y ) u c o s t  = | A1
Method using vertical motion to find relevant equation in t, u, g,  .
Eg
• Using vertical distance =0 to find expression for time.
• Find expression for time to max height and × 2 | M1
Correct unsimplified equation in t, u, g,  .
1
usint− gt2 =0
2 | A1
Solve to obtain the distance in terms of u, g and  | DM1
2 u s i n u 2 s i n 2  
X Y u c o s  =  = *
g g | A1*
(6)
5(a)
M1 | Use horizontal motion to find an expression for the horizontal
distance.
A1 | Correct unsimplified equation.
M1 | Method using vertical motion with relevant suvat to find an
equation in t, u, g,  (condone use of 9.8).
2 u s i n  usin
M0 if t = or t = is quoted (ie if it appears without
g g
any evidence of method)
A1 | Correct equation, (condone use of 9.8)
Eg
1
• usint− gt2 =0
2
• u s i n u s i n g t   − = −
• u s i n g t 0  − = double the time.
DM1 | Dependent on the preceding M marks. Solve to obtain the distance
in terms of u, g and .
A1* | Obtain given answer from correct working. Must recover g if 9.8 is
used. Must have XY = at this stage.
--- 5(b) ---
5(b) | 2 0 2  s i n 1 2 0 
C B = − 1 2
g | M1
C B = 2 3 ( m ) o r C B = 2 3 . 3 ( m ) | A1
(2)
5(b)
M1 | Complete method using result in (a), or using horizontal motion
from first principles, to find the horizontal distance CB.
A1 | Correct answer, 2sf or 3sf
--- 5(c) ---
5(c) | Horizontal motion | M1
1 2 = 2 0 c o s 6 0  t | A1
Vertical motion
1
y = 2 0 s i n 6 0   1 .2 −  g  1 .2 2 ( − 5 )
2 | M1
( ) ( )
Height above D = 8 .7 m o r 8 .7 3 m | A1
(4)
(12)
Notes
Over-accuracy or under-accuracy is penalised max once per
complete question. Penalise final A mark in the appropriate part.
Use of g = 9.81 or 10 is penalised max once per complete question.
Penalise final A mark in the appropriate part.
If both errors occur they could lose the final two A marks.
5(c)
M1 | Complete method for horizontal motion to obtain an equation in t. If
t =1.2 o.e. is seen in earlier working, it must be used in (c) to earn
the marks.
A1 | Correct unsimplified equation (t = 1.2)
M1 | Method using suvat with vertical motion to obtain a relevant vertical
distance. Height of pole not required.
A1 | Correct answer, 2 sf or 3 sf
Question | Scheme | Marks

Show LaTeX source

\begin{enumerate}
  \item The fixed points $X$ and $Y$ lie on horizontal ground.
\end{enumerate}

At time $t = 0$, a particle $P$ is projected from $X$ with speed $u \mathrm {~ms} ^ { - 1 }$ at angle $\theta$ to the ground.

Particle $P$ moves freely under gravity and first hits the ground at $Y$.\\
(a) Show that $X Y = \frac { u ^ { 2 } \sin 2 \theta } { g }$

The points $A$ and $B$ lie on horizontal ground. A vertical pole $C D$ has length 5 m .\\
The end $C$ is fixed to the ground between $A$ and $B$, where $A C = 12 \mathrm {~m}$.\\
At time $t = 0$, a particle $Q$ is projected from $A$ with speed $20 \mathrm {~ms} ^ { - 1 }$ at $60 ^ { \circ }$ to the ground.\\
Particle $Q$ moves freely under gravity, passes over the pole and first hits the ground at $B$, as shown in Figure 4.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{3e78f951-041d-4227-aa4b-e67a6ab5b4cd-14_335_1179_1032_443}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}

(b) Find the distance $C B$.\\
(c) Find the height of $Q$ above $D$ at the instant when $Q$ passes over the pole.

\hfill \mbox{\textit{Edexcel M2 2024 Q5}}

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