Teach A Level Maths
Connected Particles
and Newton’s 3rd Law
Volume 4: Mechanics 1
Connected Particles and
Newton’s 3rd Law
e.g.1. The diagram shows 2 particles, A and B,
each of mass 3 kg, connected by a string which runs
over a fixed pulley.
The particles are held at rest.
What will happen if the
particles are released.
The particles remain at rest.
A
3g
B
3g
( The extra weight due to the longer string on the
side of B is very small and balanced by the
friction at the pulley. )
e.g.1. The diagram shows 2 particles, A and B,
each of mass 3 kg, connected by a string which runs
over a fixed pulley.
We make the following modelling
assumptions when we solve
problems:
• The string is light (no weight),
and inextensible (it doesn’t
stretch).
T
A
3g
• The pulley is smooth so the
tension in the string is the same throughout.
T
B
3g
Since the forces on A and B are the same, it is not
possible for B to accelerate downwards and A upwards.
Suppose now that A has mass 4 kg
whilst B still has mass 3 kg.
A has the greater mass so it will
accelerate downwards.
The particles are joined
together so the magnitude of
the acceleration is the same
for B.
T
A
a
4g
3g
T a
B
3g
To find the values of a and T, we use Newton’s 2nd
Law on each particle separately.
N2L:
Resultant force = mass  acc.
A:
4g - T = 4a - - - - (1)
- 3g2resolve
=
3a to
-in- -so
- (2)
B: There
We
Tip:
It’s Talways
usually
easier
are
unknowns
theneed
direction
the
substitute
for
g atof
the
end of
we
another
equation
Adding
the
equations
will
acceleration.
the
question.
before
wewe
cancan
solve.
eliminate T and
then
solve for a.
T
A
a
4g
T a
B
3g
N2L:
Resultant force = mass  acc.
A:
4g - T = 4a - - - - (1)
T - 3g = 3a - - - - (2)
B:
(1) + (2):
T
A
a
4g
T a
B
3g
N2L:
Resultant force = mass  acc.
A:
4g - T = 4a - - - - (1)
T - 3g = 3a - - - - (2)
B:
g
(1) + (2):
T
A
a
4g
T a
B
3g
N2L:
Resultant force = mass  acc.
A:
4g - T = 4a - - - - (1)
T - 3g = 3a
B:
g = 7a
(1) + (2):
g
 a = 7
g = 9·8  a = 1·4
Substitute in (2): T
 T
 T
T
A
- - - - (2)
a
4g
m s -2
= 3a + 3g
= 3  1·4 + 3  9·8
= 33·6 newtons
T a
B
3g
T
The force on the pulley due to
the string is 2T downwards.
 T = 33·6 newtons
T
A
a
4g
T
T a
B
3g
 Force on pulley = 2T = 67.2 newtons
Connected particles hanging over a smooth peg
act in the same way as over a pulley.
e.g.2. A particle, A, of mass 0·6 kg, is held at rest
on a smooth table. A is connected by a light,
inextensible string, which passes over a smooth fixed
pulley at the edge of the table, to another particle,
B, of mass 0·4 kg hanging freely.
The string is horizontal and
at right angles to the edge
of the table.
A
A is released. Find
(a) the magnitude of the
reaction of the table on A,
B (b) the acceleration of the
system
(c) the tension in the string, and
(d) the force on the pulley due to the particles.
A: mass 0·6 kg
B: mass 0·4 kg
R
A
0·6g
a
T
T
B
0·4g
a
Solution:
N2L: Resultant force = mass  acceleration
(a) Find R.
R - 0·6g = 0  R = 0·6g - - - - (1)
A:
R = 5·88 newtons
( 3 s.f.
horizontally,
A accelerates
so there
is )
(b) Find a.
no vertical component0·6a
of acceleration.
- - - - (2)
A:
T =
0·4g - T = 0·4a - - - - (3)
B:
A: mass 0·6 kg
B: mass 0·4 kg
R
A
0·6g
a
T
T
a
R = 5·88
B
N
0·4g
(b) Find a.
T = 0·6a - - - - (2)
0·4g - T = 0·4a - - - - (3)
(2) + (3):
0·4g = a  a = 0·4  9·8
 a = 3·92 m s -2
(c) Find T.
Substitute in (2):
T = 0·6  3·92
 T = 2·35 newtons ( 3 s.f. )
A: mass 0·6 kg
B: mass 0·4 kg
a = 3·92 m s -2
T = 2·35 N
R
A
0·6g
T
a
T
T
T
B
0·4g
a
(d) Find the force on the pulley due to the particles.
Since the 2 forces are equal, the T
45
resultant lies between them, at
F
T
an equal angle to each.
F = Tcos45 + Tcos45
= 2Tcos45  F = 3·33 newtons ( 3 s.f. )
e.g.3. The diagram shows 2 equal scale pans, A and B,
each of mass 200 g, connected by a light inelastic
string hanging over a smooth pulley.
C
A
A pebble, C, of mass
300 g is placed in B and
the system starts to
move.
B
Find
(a) the acceleration of the system
(b) the reaction on B due to C.
A and B: mass 200 g
C: mass 300 g
B and C act as one body
T
T
C
so we can show the weight
a
a
of the two together.
A
B
(a) Find a.
0·2g
0·5g
Solution:
N2L: Resultant force = mass  acceleration
A:
B + C:
Adding:
T - 0·2g = 0·2a
0·5g - T = 0·5a
0·3g = 0·7a
 a = 4·2
The system accelerates at 4·2 m s -2
A and B: mass 200 g
C: mass 300 g
a = 4·2
(b) Find the reaction on
B due to C.
T
a
T
C
A
B
0·2g
0·5g
a
Newton’s 3rd Law says that when 2 bodies
interact the force on each due to the other has
the same magnitude but in the opposite direction.
So, the force on B due to C has the same magnitude
We will draw the forces acting on the pebble, C.
as the force on C due to B.
Solution:
a = 4·2
Forces on C
C
T
a
T
C a
A
B
0·2g
0·5g
a
R
0·3g
To find R: N2L: Resultant force = mass  acc.
C:
0·3g - R = 0·3  4·2
 0·3  9·8 - 0·3  4·2 = R

R = 1·68
Using Newton’s 3rd Law we can say that the force on
B due to C is 1·68 newtons downwards.
EXERCISE
1. Two particles, A and B, of mass 0·5 kg and 0·9 kg
respectively, are connected by a light
inextensible string passing over a fixed, smooth
pulley.
The particles are released from rest. Find
(a) the acceleration of the particles,
(b) the magnitude of the tension in the string, and
(c) the magnitude of the force on the pulley due to
the particles.
(a) Find the acceleration
A: mass 0·5 kg
B: mass 0·9 kg
Solution:
T
A
a
0·5g
T a
B
0·9g
N2L: Resultant force = mass  acceleration
A:
T - 0·5g = 0·5a - - - - - (1)
B:
0·9g - T = 0·9a - - - - - (2)
0·4g = 1·4a  a = 2·8 m s -2
(1) + (2): 
(b) Find the magnitude of
the tension in the
string
A: mass 0·5 kg
B: mass 0·9 kg
a = 2·8 m s -2
Solution:
(b)
T
T
A
a
0·5g
T
T a
B
0·9g
T - 0·5g = 0·5a - - - - - (1)
Substitute for a:
T = 0·5  2·8 + 0·5  9·8
 T = 6·3 newtons
(c) Find the force on the pulley due to the particles.
F = 2T = 12·6 newtons
EXERCISE
2. A particle, A, of mass 2 kg rests on a smooth
plane inclined at 30 to the horizontal.
The particle is connected by a
light inextensible string
passing over a smooth, fixed
pulley at the top of the plane
to a particle B of mass 0·4 kg
hanging freely.
A
B
30
Show that A slides down the plane with an
acceleration of magnitude 2·45 m s -2.
EXERCISE
Show that A slides down the plane with an acceleration
of magnitude 2·45 m s -2.
A: mass 2 kg
B: mass 0·4 kg
a
30
R
A
T
2g
T
B
a
0·4g
Solution:
N2L: Resultant force = mass  acceleration
A:
2g sin 30 - T = 2a - - - - - (1)
B:
T - 0·4g = 0·4a - - - - - (2)
(1) + (2):

g - 0·4g = 2·4a
0·6g = a 
2·4
a = 2.45 m s -2
When connected particles hanging over a peg or pulley
start to move they have a constant acceleration
( provided the forces remain constant ).
We can use the equations of motion for constant
acceleration to find, for example, the velocity when
the time or displacement is given.
We can use any of the equations but the most useful
are:
v = u + at and v2 = u2 + 2as
e.g.4. Two small blocks, A of mass 2 kg and B of
mass 3 kg, are connected by a light, inextensible
string which runs over a smooth pulley as shown in
the diagram. The blocks are held at rest.
A
B
1
30
B rests on a smooth
plane at 30 to the
horizontal and A is
1 m above the ground.
(a) Find the acceleration when the blocks are released
from rest.
A: mass 2 kg
B: mass 3 kg
a
a
1
T
A
T
2g
3g
R
B
30
(a) Find the acceleration when the blocks are released
from rest.
We aren’t told which direction
Solution: N2L: Resultant force = mass  acceleration
the blocks move. If I choose
- - - - (1)- the
2g - Tit =won’t
2a -matter
A:
wrongly
- - -negative.
- (2)
3g sin 30 = 3a
T - acceleration
will- be
B:
(1) + (2):


2g - 1·5g = 5a
0·1g = a  a = 0·98 m s -2
a
A: mass 2 kg
B: mass 3 kg
a
a = 0·98 m s -2
1
T
A
T
2g
3g
R
B
30
(b) Find the velocity of A when it reaches the ground.
= u 2 + 2as
Decide with yourv 2partner
which of the equations
for from
constant
acceleration
Blocks start
rest,
so u = 0 you would use.


v 2 = 2  0·98  1
v = 1·4 m s -1
A: mass 2 kg
B: mass 3 kg
a = 0·98 m s -2
T
A
1
A
1·4 R
T
B
1
3g
30
(b) Find the velocity of A when it reaches the ground.
v 2 = u 2 + 2as
Blocks start from rest, so u = 0
v 2 = 2  0·98  1

v = 1·4 m s -1

T
A: mass 2 kg
B: mass 3 kg
a = 0·98 m s -2
v = 1·4 m s -1
R
B
1
T
A
3g
1·4
a1
1
30
(c) Assuming B does not reach the pulley, find the
total distance up the plane that B travels before it
first comes to rest.
Decide
with goes
your slack
partner
whattension
factorsdisappears.
determine
The string
so the
how B moves after A reaches the ground.
B will slow down with a negative acceleration
caused by the component of its weight.
R
A: mass 2 kg
B: mass 3 kg
B
3g
1
u
v = 1·4 m s -1
A
1·4
a1
1
30
(c) Assuming B does not reach the pulley, find the
total distance up the plane that B travels before it
first comes to rest.
N2L: Resultant force = mass  acceleration
- 3g sin 30 = 3a1 - - - - - (4)
B:



a1 = -0·5g
v2 = u2 + 2as
0 = 1·4 2 + 2(-0·5g)s
s = 0·2 m
Be careful here. u is now 1·4 and v is 0.
B travels a total of 1·2 m up the plane.
SUMMARY
 For particles connected by a string passing over
a pulley or peg, we assume that
• the pulley or peg is smooth which makes the
tensions equal on either side,
• the string is light and inextensible giving a
constant acceleration.
 We solve problems by using Newton’s 2nd Law for
each particle separately.
 The equations of motion for constant acceleration
can be used.
 If one particle reaches the ground, or the string
breaks, the tension disappears and the other
particle continues to move but with a different
acceleration ( which may be negative ).
The following page contains the summary in a form
suitable for photocopying.
TEACH A LEVEL MATHS – MECHANICS 1
CONNECTED PARTICLES AND NEWTON’S 3RD LAW
Summary

For particles connected by a string passing over a pulley or peg, we
assume that
•
the pulley or peg is smooth which makes the tensions equal on either
side,
•
the string is light and inextensible giving a constant acceleration.

We solve problems by using Newton’s 2nd Law for each particle
separately.

The equations of motion for constant acceleration can be used.

If one particle reaches the ground, or the string breaks, the tension
disappears and the other particle continues to move but with a different
acceleration ( which may be negative ).