Question 5 - A-Level Maths

CAIE M1 2019 November — Question 5 7 marks

Exam Board	CAIE
Module	M1 (Mechanics 1)
Year	2019
Session	November
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Projectiles
Type	Two projectiles meeting - vertical motion only
Difficulty	Moderate -0.3 This is a standard two-particle projectile problem requiring setting up SUVAT equations for both particles with a time offset, then solving simultaneously. The collision condition is straightforward (equal heights), and finding speeds involves direct substitution. While it requires careful bookkeeping of the 1-second delay and multiple equations, the techniques are routine for M1 students with no novel problem-solving insight needed.
Spec	3.02d Constant acceleration: SUVAT formulae 3.02h Motion under gravity: vector form

Two particles \(A\) and \(B\) move in the same vertical line. Particle \(A\) is projected vertically upwards from the ground with speed 20 m s\(^{-1}\). One second later particle \(B\) is dropped from rest from a height of 40 m.

Find the height above the ground at which the two particles collide. [4]
Find the difference in the speeds of the two particles at the instant when the collision occurs. [3]

Show mark scheme Show mark scheme source

Question 5:

Answer	Marks	Guidance
5(i)	h =20t− 1 ×10t2 or h =± 1 ×10 ( t−1 )2
A 2 B 2	B1	OE

h =20 ( T +1 ) − 1 ×10 ( T +1 )2 or h =± 1 ×10T2

A 2 B 2

[Meet when 20t− 1 ×10t2 + 1 ×10 ( t−1 )2 =40]

Answer	Marks	Guidance
2 2	*M1	Set up an equation using their h , their h and 40

A B

Answer	Marks	Guidance
10t–35= 0	DM1	Solve for t and attempt to find the height at collision.
t=3.5 so height at collision=8.75m	A1	T =2.5 and height at collision=8.75m

Alternative method for question 5(i)

1 h =20×1− ×10×12 =15,v=20−10×1=10

Answer	Marks	Guidance
A 2	B1	Finding distance travelled by A and its speed after 1 second

H +H = 25

A B

 1  1

10T − ×10×T2  + ×10×T2 =25

Answer	Marks	Guidance
 2  2	*M1	T is the time beyond 1s until the particles reach same level H

A

and H are distances travelled by A and B in T seconds.

B

Answer	Marks	Guidance
[10T=25 → T = 2.5]	DM1	Solve for T and attempt to find the height at collision
t=3.5 so height = 8.75 m	A1

4

Answer	Marks	Guidance
Question	Answer	Mark

Answer	Marks	Guidance
5(ii)	v =20–gt = –15 or v 2=202+2(–g)(8.75)
A A	M1	Use of their t or their h ⩽ 20 from 5(i) in a constant acceleration

formula which would lead to finding v

A

v =–g(t – 1) = –25 or v 2=2(g)(40–8.75)

Answer	Marks	Guidance
B B	M1	Use of their t ± 1 or their 40 – h from 5(i) in a constant

acceleration formula which would lead to finding v

B

Answer	Marks	Guidance
Difference = 10 ms–1	A1	CWO

3

Answer	Marks	Guidance
Question	Answer	Mark

Question 5:
--- 5(i) ---
5(i) | h =20t− 1 ×10t2 or h =± 1 ×10 ( t−1 )2
A 2 B 2 | B1 | OE
h =20 ( T +1 ) − 1 ×10 ( T +1 )2 or h =± 1 ×10T2
A 2 B 2
[Meet when 20t− 1 ×10t2 + 1 ×10 ( t−1 )2 =40]
2 2 | *M1 | Set up an equation using their h , their h and 40
A B
10t–35= 0 | DM1 | Solve for t and attempt to find the height at collision.
t=3.5 so height at collision=8.75m | A1 | T =2.5 and height at collision=8.75m
Alternative method for question 5(i)
1
h =20×1− ×10×12 =15,v=20−10×1=10
A 2 | B1 | Finding distance travelled by A and its speed after 1 second
H +H = 25
A B
 1  1
10T − ×10×T2  + ×10×T2 =25
 2  2 | *M1 | T is the time beyond 1s until the particles reach same level H
A
and H are distances travelled by A and B in T seconds.
B
[10T=25 → T = 2.5] | DM1 | Solve for T and attempt to find the height at collision
t=3.5 so height = 8.75 m | A1
4
Question | Answer | Mark | Guidance
--- 5(ii) ---
5(ii) | v =20–gt = –15 or v 2=202+2(–g)(8.75)
A A | M1 | Use of their t or their h ⩽ 20 from 5(i) in a constant acceleration
formula which would lead to finding v
A
v =–g(t – 1) = –25 or v 2=2(g)(40–8.75)
B B | M1 | Use of their t ± 1 or their 40 – h from 5(i) in a constant
acceleration formula which would lead to finding v
B
Difference = 10 ms–1 | A1 | CWO
3
Question | Answer | Mark | Guidance

Show LaTeX source

Two particles $A$ and $B$ move in the same vertical line. Particle $A$ is projected vertically upwards from the ground with speed 20 m s$^{-1}$. One second later particle $B$ is dropped from rest from a height of 40 m.

\begin{enumerate}[label=(\roman*)]
\item Find the height above the ground at which the two particles collide. [4]

\item Find the difference in the speeds of the two particles at the instant when the collision occurs. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2019 Q5 [7]}}

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