3
Connected bodies
A system that consists of more than one particle, is refered to as connected particles. A system
that involve involve particles moving a long a asraight line, is treated as a single particle, where
as a system in which the particles are not moving along a staright line, the motion of each
particle is treated separately.
Examples
1. Two particles connected by a string
Two particles of mass m1 = 4 kg and m2 = 7 kg are connected by a light inextensible string
which is pulled taut. The heavier particle is pulled by a horizontal force P= 50 N along a rough
horizonatl plane. The coeffiecient of friction between each particle and the plane is µ = 0.3.
(a) What is the acceleration of each particle?
(b) What is the tension in the string?
Solution
R2
om
a ms−2
R1
l.c
T
µR1
50 N
gm
ai
µR1
7g
@
4g
rt
m
uj
un
gu
Figure 1:
he
rb
e
(a) Resolving forces on the 4 kg mass perpendicular to the direction of motion.
R1 = 4g N
.
Obtaining the Net force along the direction of motion on the 4 kg-particle and then equating
to Newton’s 2nd law.
Net Force = T − µR1 But, Net force = 4a
⇒ T − 0.3 ∗ 4g = 4a
⇒ T = 1.2g + 4a
(23)
Resolving forces on the 7 kg mass perpendicular to the direction of motion.
R2 = 7g N
.
Obtaining the Net force along the direction of motion on the 7 kg-particle and then applying
Newton’s 2nd law.
Net Force = 50 − T − µR2 But Net force = 7a
⇒ 50 − T − 0.3 ∗ 7g = 7a
⇒ T = 50 − 2.1g − 7a
18
(24)
Equating Equations 23 and 24,
1.2g + 4a = 50 − 2.1g − 7a
4a + 7a = 50 − 2.1g − 1.2g And taking g = 9.8 ms−2
a = 1.6055 ms−2
(25)
∴ the acceleration a of each particle is 1.6055 ms−2 .
(b) Substituting Equation 25 into 23
T = 1.2g + 4a
T = 1.2 × 9.8 + 4 × 1.6055
T = 18.182 N
∴ the tension T in the string is 18.182 N .
2. Two particles hanging on a single fixed pulley
Two particles A and B of respective masses 2 kg and 5 kg are attached to the end of a light
inextensible string which passes over a smooth pulley P. The two particles are held at rest,
at the same level above a horizontal floor with the portions of strings not in contact with the
pulley vertical. The system is then released from rest.
l.c
om
(a) Calculate the acceleration of each particle?
gm
ai
(b) Calculate the tension in the string?
un
gu
@
Solution
rb
e
rt
m
uj
P
he
T
a ms−2
T
a ms−2
B
A
2g
5g
(a) Obtaining the Net force along the direction of motion on the 2 kg-particle and then applying
Newton’s 2nd law.
Net Force = T − 2g, But, Net force = ma
⇒ T − 2g = 2a
⇒ T = 2g + 2a
(26)
Obtaining the Net force on the 5 kg-particle and then applying Newton’s 2nd law.
Net Force = 5g − T, But, Net force = ma
⇒ 5g − T = 5a
⇒ T = 5g − 5a
19
(27)
Equating Equations 26 and 27,
2g + 2a = 5g − 5a
2a + 5a = 5g − 2g And taking g = 9.8 ms−2
a = 4.2 ms−2
(28)
∴ the acceleration a of each particle is 4.2 ms−2 .
(b) Substituting Equation 28 into 26
T = 2g + 2a
T = 2 × 9.8 + 2 × 4.2
T = 28.0 N
∴ the tension T in the string is 28.0 N .
3. One particle hanging freely and another on an inclined plane
A
0.8 m
gm
ai
l.c
Q 2 kg
30o
om
P (3 kg)
un
gu
@
Figure 2: a figure showing One particle hanging freely and another on an inclined plane
he
rb
e
rt
m
uj
The diagram in Figure 2 shows two partyicles P and Q of mass 3kg and 2 kg respectively
conncetd by a light inextensible string. Initially P is held at rest on a fixed smooth plane
inclined at 30o to the horizontal . The string passes over a small smooth light pulley A fixed at
the top of the plane. The part of the string from P to A is parallel to a line of greatest slope
of the plane. The particle Q hangs freely below A . The system is realesd from rest with the
string taut.
(a) Write down an equation of motion for P and an equation of motion for Q.
(b) Hence show that the acceleration of Q is 9.8 ms− 2.
(c) Find the tension in the string.
Solution
R
−2
s
A
am
a ms−2
T
T
P
o
0
3
sin
Q
30o
3g
30o
2g
0.8 m
3g
cos
3g
o
30
(a) Resolving forces on the 3 kg mass perpendicularly to the inclined plane.
R = 3g cos 30o
20
Obtaining the Net force along the direction of motion on the 3 kg-particle and then applying
Newton’s 2nd law.
Net Force = T − 3g sin 30o , But Net force = ma
⇒ T − 3g sin 30o = 3a
⇒ T = 3g sin 30o + 3a
(29)
Obtaining the Net force on the 2 kg-particle and then applying Newton’s 2nd law.
Net Force = 2g − T, But, Net force = ma
⇒ 2g − T = 2a
⇒ T = 2g − 2a
(30)
Equations 29 and 30 are the equations of motion of P and Q, respectively.
(b) Equating Equations 29 and 30,
l.c
gm
ai
∴ the acceleration a of each particle is 0.98 ms−2 .
(31)
om
3g sin 30o + 3a = 2g − 2a
3a + 2a = 2g − 3g sin 30o And taking g = 9.8 ms−2
a = 0.98 ms−2
@
(c) Substituting Equation 31 into 29
rb
e
rt
m
uj
un
gu
T = 3g sin 30o + 3a
T = 3 × 9.8 ∗ 0.5 + 3 × 0.98
T = 17.64 N
he
∴ the tension T in the string is 17.64 N .
More Examples
4. A block C, of mass 4 kg , is placed on a rough horizontal table, where the coefficient of
Figure 3: Three body system with two particles hanging and one on a horizontal plane.
friction between the table and C is 0.65. C is connected by two light inextensible strings to two
more blocks, A and B , of respective masses 3 kg and 7 kg . Each of the strings passes over
two smooth pulleys, each of the pulleys located at the edge of the table, with A and B hanging
freely at each of the two ends of the table, as shown in the Figure 3. The system is released
from rest with the strings taut. By modelling the three blocks as particles, determine in any
order the acceleration of the system and the tension in each of the two strings. Solution
21
R
C
T2
T1
µR
−2
a ms
a ms−2
T2
0.6
5
T1
4g
µ
=
A
B
3g
7g
Obtaining the Net force along the direction of motion on the 3 kg-particle and then
applying Newton’s 2nd law.
Net Force = T1 − 3g, But Net force = ma
⇒ T1 − 3g = 3a
⇒ T1 = 3g + 3a
(32)
Resolving forces perpendicularly to the plane;
R = 4g
Obtaining the Net force on the 4 kg-particle and then applying Newton’s 2nd law.
l.c
om
T2 − T1 − µR, But, Net force = ma
T2 − T1 − µR = 4a
T2 = T1 + 0.65 ∗ 4g + 4a
T2 = T1 + 2.6g + 4a
gm
ai
Net Force =
⇒
⇒
⇒
(33)
gu
@
Obtaining the Net force on the 7 kg-particle and then applying Newton’s 2nd law.
rb
e
rt
m
uj
un
Net Force = 7g − T2 , But Net force = ma
⇒ 7g − T2 = 7a
⇒ T2 = 7g − 7a
(34)
he
Equating Equations 33 and 34, we obtain;
T1 + 2.6g + 4a = 7g − 7a
T1 = 4.4g − 11a
(35)
Equating Equations 32 and 35, we obtain;
3g + 3a = 4.4g − 11a
14a = 1.4g ⇒ a = 0.98 ms−2
Substituting 36 in Equation 32
T1 = 3g + 3a
T1 = 3 × 9.8 + 3 × 0.98
T1 = 32.34 N
∴ the tension T1 in the string is 32.34 N .
Substituting 36 in Equation 34
T2 = 7g − 7a
T2 = 7 × 9.8 − 7 × 0.98
T2 = 61.74 N
∴ the tension T1 in the string is 32.34 N .
22
(36)
5. The diagram in Figure 3 shows a fixed pulley carrying a string
which has a mass of 4 kg attached at one end and a light pulley A
attached at the other. Another string passes over pulley A and carries
a mass of 3 kg at one end and a mass 1 kg at the other end. Find:
(a) the acceleration of pulley A
(b) the acceleration of the 1 kg, 3 kg and 4 kg masses.
(c) the tensions in the strings.
Solution
Let a1 and a2 , and T1 and T2 be the accelerations and tensions of
the strings aver the pulleys as shown in Figure 3.
Obtaining the Net force on the 4 kgparticle and then applying Newton’s 2nd
law.
Net Force = 4g − T1 , But Net force = ma
⇒ 4g − T1 = 4a1
⇒ T1 = 4g − 4a1
(37)
Obtaining the Net force on pulley A and
then applying Newton’s 2nd law.
om
T1
@
T2
1 kg
a2 ms
a1 ms−2
A
T2
4 kg
−2
g
rt
m
uj
un
gu
Net Force = T1 − 2T2 , But Net force = ma
⇒ T1 − 2T2 = 0 × a1
1
(38)
⇒ T2 = T1
2
gm
ai
l.c
a1 ms−2
T1
3 kg
a2 ms−2
4g
3g
he
rb
e
Obtaining the Net force acting on the 1 kgparticle and then applying Newton’s 2nd
law.
Net Force = T2 − g, But Net force = ma
⇒ T2 − g = 1 ∗ (a2 + a1 )
⇒ T2 = g + a2 + a1
(39)
Obtaining the Net force on the 3 kg-particle and then applying Newton’s 2nd law.
Net Force = 3g − T2 , But, Net force = ma
⇒ 3g − T2 = 3(a2 − a1 )
⇒ T2 = 3g − 3a2 + 3a1
(40)
At this step, we are required to sovle simultaneously the four Equations 37, 38,39 and 40 to
obtain a1 , a2 , T1 , & T2 by applying any method.
Let us begin by eliminating a2 , by multiply Eqn 39 by 3 and then adding it to Eqn 40;
3× (T2 = g + a2 + a1 )
+ T2 = 3g − 3a2 + 3a1
4T2 = 4g + 6a1
3
3
T2 = g + a1
2
2
23
(41)
Equating equations 41 and 38
3
3
= g + a1
2
2
= 3g + 3a1
1
T
2 1
T2 =
T1
(42)
Finally to obtain a1 we equate Equations 37 and 42,
T1 = 3g + 3a1 = 4g − 4a1
7a1
=g
g
a1
= ms−2
7
Therefore pulley A is accelerating with an acceleration of
b). Substituting a1 =
g
7
ms−2 upwards.
ms−2 in equation 42’
T1 = 3g + 3a1
T1 = 3g + 3 ×
24g
N
7
l.c
N into Eqn 38 to obtain T2 ,
=
1
T1
2
1
1 24g
= T1 = ∗
2
2
7
12g
N
7
rt
m
uj
un
T2
gu
@
T2
gm
ai
12g
7
g
7
om
T1 =
Substituting T1 =
g
7
12g
7
N and a1 =
he
Substituting T1 =
rb
e
T2
g
7
ms−2 into Eqn 39 to obtain a2 ,
T2 = g + a2 + a1
12g
= g + a2 + g7
7
a2 = 12g
− g − g7
7
4g
7
a2 =
ms−2
The 4 kg mass is accelerating at a1 = g7 ms−2 downwards , The 1 kg mass is accelerating
at a1 + a2 = 5g
ms−2 upwards and the 3 kg mass is accelerating at a2 − a1 = 3g
ms−2
7
7
downwards
c). The tension in the string over passing over pulley A is
Pulley A is 24g
N.
7
24
12g
7
N and in the one carrying
PhysicsAndMathsTutor.com
M1 Dynamics - Connected particles
gm
ai
l.c
om
1.
(a)
he
rb
e
rt
m
uj
un
gu
@
Two particles A and B have mass 0.4 kg and 0.3 kg respectively. The particles are attached to
the ends of a light inextensible string. The string passes over a small smooth pulley which is
fixed above a horizontal floor. Both particles are held, with the string taut, at a height of 1m
above the floor, as shown in the diagram above. The particles are released from rest and in the
subsequent motion B does not reach the pulley.
Find the tension in the string immediately after the particles are released.
(6)
(b)
Find the acceleration of A immediately after the particles are released.
(2)
When the particles have been moving for 0.5 s, the string breaks.
(c)
Find the further time that elapses until B hits the floor.
(9)
(Total 17 marks)
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M1 Dynamics - Connected particles
2.
gu
Show that the tension in the string as A descends is
15
mg .
4
(3)
(b)
Find the value of k.
he
rb
e
rt
m
uj
un
(a)
@
gm
ai
l.c
om
Two particles A and B have masses 5m and km respectively, where k < 5. The particles are
connected by a light inextensible string which passes over a smooth light fixed pulley. The
system is held at rest with the string taut, the hanging parts of the string vertical and with A and
B at the same height above a horizontal plane, as shown in Figure 4. The system is released
1
from rest. After release, A descends with acceleration g .
4
(3)
(c)
State how you have used the information that the pulley is smooth.
(1)
After descending for 1.2 s, the particle A reaches the plane. It is immediately brought to rest by
the impact with the plane. The initial distance between B and the pulley is such that, in the
subsequent motion, B does not reach the pulley.
(d)
Find the greatest height reached by B above the plane.
(7)
(Total 14 marks)
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M1 Dynamics - Connected particles
3.
A car of mass 800 kg pulls a trailer of mass 200 kg along a straight horizontal road using a light
towbar which is parallel to the road. The horizontal resistances to motion of the car and the
trailer have magnitudes 400 N and 200 N respectively. The engine of the car produces a
constant horizontal driving force on the car of magnitude 1200 N. Find
(a)
the acceleration of the car and trailer,
(3)
(b)
the magnitude of the tension in the towbar.
(3)
The car is moving along the road when the driver sees a hazard ahead. He reduces the force
produced by the engine to zero and applies the brakes. The brakes produce a force on the car of
magnitude F newtons and the car and trailer decelerate. Given that the resistances to motion are
unchanged and the magnitude of the thrust in the towbar is 100 N,
om
find the value of F.
(7)
(Total 13 marks)
he
rb
e
rt
m
uj
un
gu
@
gm
ai
l.c
(c)
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M1 Dynamics - Connected particles
4.
(ii)
the tension in the string,
gu
the acceleration of the scale pan,
rt
m
uj
un
(i)
(8)
he
rb
e
(a)
@
gm
ai
l.c
om
One end of a light inextensible string is attached to a block P of mass 5 kg. The block P is held
at rest on a smooth fixed plane which is inclined to the horizontal at an angle α , where
3
sin α = . The string lies along a line of greatest slope of the plane and passes over a smooth
5
light pulley which is fixed at the top of the plane. The other end of the string is attached to a
light scale pan which carries two blocks Q and R, with block Q on top of block R, as shown in
Figure 3. The mass of block Q is 5 kg and the mass of block R is 10 kg. The scale pan hangs at
rest and the system is released from rest. By modelling the blocks as particles, ignoring air
resistance and assuming the motion is uninterrupted, find
(b)
the magnitude of the force exerted on block Q by block R,
(3)
(c)
the magnitude of the force exerted on the pulley by the string.
(5)
(Total 16 marks)
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M1 Dynamics - Connected particles
5.
A (m)
P
B (2m)
Two particles A and B, of mass m and 2m respectively, are attached to the ends of a light
inextensible string. The particle A lies on a rough horizontal table. The string passes over a
small smooth pulley P fixed on the edge of the table. The particle B hangs freely below the
pulley, as shown in the diagram above. The coefficient of friction between A and the table is µ.
The particles are released from rest with the string taut. Immediately after release, the
4
magnitude of the acceleration of A and B is g . By writing down separate equations of motion
9
for A and B,
find the tension in the string immediately after the particles begin to move,
(3)
gu
@
2
.
3
un
show that µ =
(5)
rb
e
rt
m
uj
(b)
gm
ai
l.c
om
(a)
he
When B has fallen a distance h, it hits the ground and does not rebound. Particle A is then a
1
distance h from P.
3
(c)
Find the speed of A as it reaches P.
(6)
(d)
State how you have used the information that the string is light.
(1)
(Total 15 marks)
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M1 Dynamics - Connected particles
6.
P(0.5 kg)
3.15 m
Q(m kg)
gm
ai
Show that the acceleration of P as it descends is 2.8 m s–2.
(3)
un
gu
@
(a)
l.c
om
Two particles P and Q have mass 0.5 kg and m kg respectively, where m < 0.5. The particles are
connected by a light inextensible string which passes over a smooth, fixed pulley. Initially P is
3.15 m above horizontal ground. The particles are released from rest with the string taut and the
hanging parts of the string vertical, as shown in the diagram above. After P has been descending
for 1.5 s, it strikes the ground. Particle P reaches the ground before Q has reached the pulley.
rt
m
uj
Find the tension in the string as P descends.
(3)
he
rb
e
(b)
(c)
Show that m =
5
.
18
(4)
(d)
State how you have used the information that the string is inextensible.
(1)
When P strikes the ground, P does not rebound and the string becomes slack. Particle Q then
moves freely under gravity, without reaching the pulley, until the string becomes taut again.
(e)
Find the time between the instant when P strikes the ground and the instant when the
string becomes taut again.
(6)
(Total 17 marks)
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M1 Dynamics - Connected particles
7.
A
Q(2 kg)
P(3 kg)
0.8 m
30 °
gm
ai
Write down an equation of motion for P and an equation of motion for Q.
(4)
rt
m
Hence show that the acceleration of Q is 0.98 m s–2.
(2)
he
rb
e
(b)
uj
un
gu
@
(a)
l.c
om
The diagram above shows two particles P and Q, of mass 3 kg and 2 kg respectively, connected
by a light inextensible string. Initially P is held at rest on a fixed smooth plane inclined at 30° to
the horizontal. The string passes over a small smooth light pulley A fixed at the top of the plane.
The part of the string from P to A is parallel to a line of greatest slope of the plane. The particle
Q hangs freely below A. The system is released from rest with the string taut.
(c)
Find the tension in the string.
(2)
(d)
State where in your calculations you have used the information that the string is
inextensible.
(1)
On release, Q is at a height of 0.8 m above the ground. When Q reaches the ground, it is brought
to rest immediately by the impact with the ground and does not rebound. The initial distance of
P from A is such that in the subsequent motion P does not reach A. Find
(e)
the speed of Q as it reaches the ground,
(2)
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M1 Dynamics - Connected particles
(f)
the time between the instant when Q reaches the ground and the instant when the string
becomes taut again.
(5)
(Total 16 marks)
8.
A car is towing a trailer along a straight horizontal road by means of a horizontal tow-rope. The
mass of the car is 1400 kg. The mass of the trailer is 700 kg. The car and the trailer are modelled
as particles and the tow-rope as a light inextensible string. The resistances to motion of the car
and the trailer are assumed to be constant and of magnitude 630 N and 280 N respectively. The
driving force on the car, due to its engine, is 2380 N. Find
(a)
the acceleration of the car,
(3)
(b)
the tension in the tow-rope.
gm
ai
l.c
om
(3)
find the distance moved by the car in the first 4 s after the tow-rope breaks.
uj
(c)
un
gu
@
When the car and trailer are moving at 12 m s–1, the tow-rope breaks. Assuming that the driving
force on the car and the resistances to motion are unchanged,
State how you have used the modelling assumption that the tow-rope is inextensible.
(1)
(Total 13 marks)
he
(d)
rb
e
rt
m
(6)
9.
A (3m)
(smooth)
30°
B (m)
(rough)
30°
A fixed wedge has two plane faces, each inclined at 30° to the horizontal. Two particles A and
B, of mass 3m and m respectively, are attached to the ends of a light inextensible string. Each
particle moves on one of the plane faces of the wedge. The string passes over a small smooth
light pulley fixed at the top of the wedge. The face on which A moves is smooth. The face on
which B moves is rough. The coefficient of friction between B and this face is µ. Particle A is
held at rest with the string taut. The string lies in the same vertical plane as lines of greatest
slope on each plane face of the wedge, as shown in the figure above.
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M1 Dynamics - Connected particles
The particles are released from rest and start to move. Particle A moves downwards and B
moves upwards. The accelerations of A and B each have magnitude 101 g .
(a)
By considering the motion of A, find, in terms of m and g, the tension in the string.
(3)
(b)
By considering the motion of B, find the value of µ.
(8)
(c)
Find the resultant force exerted by the string on the pulley, giving its magnitude and
direction.
l.c
om
(3)
(Total 14 marks)
uj
un
gu
@
gm
ai
10.
rb
e
rt
m
15°
he
This figure shows a lorry of mass 1600 kg towing a car of mass 900 kg along a straight
horizontal road. The two vehicles are joined by a light towbar which is at an angle of 15° to the
road. The lorry and the car experience constant resistances to motion of magnitude 600 N and
300 N respectively. The lorry’s engine produces a constant horizontal force on the lorry of
magnitude 1500 N. Find
(a)
the acceleration of the lorry and the car,
(3)
(b)
the tension in the towbar.
(4)
When the speed of the vehicles is 6 m s–1, the towbar breaks. Assuming that the resistance to the
motion of the car remains of constant magnitude 300 N,
(c)
find the distance moved by the car from the moment the towbar breaks to the moment
when the car comes to rest.
(4)
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M1 Dynamics - Connected particles
(d)
State whether, when the towbar breaks, the normal reaction of the road on the car is
increased, decreased or remains constant. Give a reason for your answer.
(2)
(Total 13 marks)
11.
A (0.5 kg)
B (0.8 kg)
uj
un
gu
@
gm
ai
l.c
om
P
he
rb
e
rt
m
A block of wood A of mass 0.5 kg rests on a rough horizontal table and is attached to one end of
a light inextensible string. The string passes over a small smooth pulley P fixed at the edge of
the table. The other end of the string is attached to a ball B of mass 0.8 kg which hangs freely
below the pulley, as shown in the diagram above. The coefficient of friction between A and the
table is µ. The system is released from rest with the string taut. After release, B descends a
distance of 0.4 m in 0.5 s. Modelling A and B as particles, calculate
(a)
the acceleration of B,
(3)
(b)
the tension in the string,
(4)
(c)
the value of µ.
(5)
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M1 Dynamics - Connected particles
(d)
State how in your calculations you have used the information that the string is
inextensible.
(1)
(Total 13 marks)
12.
m kg
gm
ai
l.c
om
3 kg
(b)
the tension in the string immediately after the particles are released,
he
(a)
rb
e
rt
m
uj
un
gu
@
The particles have mass 3 kg and m kg, where m < 3. They are attached to the ends of a light
inextensible string. The string passes over a smooth fixed pulley. The particles are held in
position with the string taut and the hanging parts of the string vertical, as shown in the diagram
above. The particles are then released from rest. The initial acceleration of each particle has
magnitude 73 g. Find
(3)
the value of m.
(4)
(Total 7 marks)
13.
P (4 kg)
Edexcel Internal Review
Q (6 kg)
40 N
11
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M1 Dynamics - Connected particles
Two particles P and Q, of mass 4 kg and 6 kg respectively, are joined by a light inextensible
string. Initially the particles are at rest on a rough horizontal plane with the string taut. The
coefficient of friction between each particle and the plane is 72 . A constant force of magnitude
40 N is then applied to Q in the direction PQ, as shown in the diagram above.
(a)
Show that the acceleration of Q is 1.2 m s–2.
(4)
(b)
Calculate the tension in the string when the system is moving.
(3)
(c)
State how you have used the information that the string is inextensible.
(1)
om
After the particles have been moving for 7 s, the string breaks. The particle Q remains under the
action of the force of magnitude 40 N.
l.c
Show that P continues to move for a further 3 seconds.
gm
ai
(d)
uj
Calculate the speed of Q at the instant when P comes to rest.
(4)
(Total 17 marks)
he
rb
e
rt
m
(e)
un
gu
@
(5)
14.
A (4 kg)
B (3 kg)
30
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M1 Dynamics - Connected particles
A particle A of mass 4 kg moves on the inclined face of a smooth wedge. This face is inclined at
30° to the horizontal. The wedge is fixed on horizontal ground. Particle A is connected to a
particle B, of mass 3 kg, by a light inextensible string. The string passes over a small light
smooth pulley which is fixed at the top of the plane. The section of the string from A to the
pulley lies in a line of greatest slope of the wedge. The particle B hangs freely below the pulley,
as shown in the diagram above. The system is released from rest with the string taut. For the
motion before A reaches the pulley and before B hits the ground, find
(a)
the tension in the string,
(6)
(b)
the magnitude of the resultant force exerted by the string on the pulley.
(3)
l.c
om
The string in this question is described as being ‘light’.
Write down what you understand by this description.
(ii)
State how you have used the fact that the string is light in your answer to part (a).
gm
ai
(i)
uj
un
gu
@
(c)
he
rb
e
rt
m
(2)
(Total 11 marks)
15.
P
1.4 m
B (0.4kg)
A (m kg)
1m
30
Edexcel Internal Review
13
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M1 Dynamics - Connected particles
The diagram above shows two particles A and B, of mass m kg and 0.4 kg respectively,
connected by a light inextensible string. Initially A is held at rest on a fixed smooth plane
inclined at 30° to the horizontal. The string passes over a small light smooth pulley P fixed at
the top of the plane. The section of the string from A to P is parallel to a line of greatest slope of
the plane. The particle B hangs freely below P. The system is released from rest with the string
taut and B descends with acceleration 15 g.
(a)
Write down an equation of motion for B.
(2)
(b)
Find the tension in the string.
(2)
(c)
Prove that m =
16
35
.
State where in the calculations you have used the information that P is a light smooth
pulley.
@
(d)
gm
ai
l.c
om
(4)
rt
m
uj
un
gu
(1)
he
rb
e
On release, B is at a height of one metre above the ground and AP = 1.4 m. The particle B
strikes the ground and does not rebound.
(e)
Calculate the speed of B as it reaches the ground.
(2)
(f)
Show that A comes to rest as it reaches P.
(5)
(Total 16 marks)
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M1 Dynamics - Connected particles
16.
A car which has run out of petrol is being towed by a breakdown truck along a straight
horizontal road. The truck has mass 1200 kg and the car has mass 800 kg. The truck is
connected to the car by a horizontal rope which is modelled as light and inextensible. The
truck’s engine provides a constant driving force of 2400 N. The resistances to motion of the
truck and the car are modelled as constant and of magnitude 600 N and 400 N respectively. Find
(a)
the acceleration of the truck and the car,
(3)
(b)
the tension in the rope.
(3)
When the truck and car are moving at 20 m s−1, the rope breaks. The engine of the truck
provides the same driving force as before. The magnitude of the resistance to the motion of the
truck remains 600 N.
Show that the truck reaches a speed of 28 m s−1 approximately 6 s earlier than it would
have done if the rope had not broken.
(7)
(Total 13 marks)
he
rb
e
rt
m
uj
un
gu
@
gm
ai
l.c
om
(c)
Edexcel Internal Review
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M1 Dynamics - Connected particles
17.
P
A
B
0.6 m
rb
e
the tension in the string before B reaches the ground,
(5)
he
(a)
rt
m
uj
un
gu
@
gm
ai
l.c
om
A particle A of mass 0.8 kg rests on a horizontal table and is attached to one end of a light
inextensible string. The string passes over a small smooth pulley P fixed at the edge of the table.
The other end of the string is attached to a particle B of mass 1.2 kg which hangs freely below
the pulley, as shown in the diagram above. The system is released from rest with the string taut
and with B at a height of 0.6 m above the ground. In the subsequent motion A does not reach P
before B reaches the ground. In an initial model of the situation, the table is assumed to be
smooth. Using this model, find
(b)
the time taken by B to reach the ground.
(3)
In a refinement of the model, it is assumed that the table is rough and that the coefficient of
friction between A and the table is 15 . Using this refined model,
(c)
find the time taken by B to reach the ground.
(8)
(Total 16 marks)
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M1 Dynamics - Connected particles
1.
( ↓ )0.4 g − T =
0.4a
( ↑ )T − 0.3 g =
0.3a
(a)
M1 A1
M1 A1
DM1
A1
solving for T
T = 3.36 or 3.4 or 12g/35 (N)
0.4 g − 0.3 g =
0.7 a
a = 1.4 m s -2 , g/7
(b)
(c)
6
DM1
A1
( ↑ )v =u + at
v = 0.5 x 1.4
= 0.7
2
M1
A1 ft on a
( ↑ )s = ut + 12 at 2
s = 0.5 x 1.4 x 0.52
= 0.175
M1
A1 ft on a
( ↓ )s = ut + 12 at 2
1.175 = − 0.7t + 4.9t 2
4.9t 2 − 0.7t − 1.175 = 0
om
DM1 A1 ft
l.c
0.7 ± 0.7 2 + 19.6 x 1.175
9.8
= 0.5663..or − ...
gm
ai
t=
gu
@
DM1 A1 cao
un
Ans 0.57 or 0.566 s
9
uj
A1 cao
he
rb
e
rt
m
[17]
2.
(a)
N2L
A:
5mg – T = 5m ×
T=
(b)
N2L
B:
1
g
4
15
mg *
4
T – kmg = km ×
cso
1
g
4
k=3
(c)
The tensions in the two parts of the string are the same
Edexcel Internal Review
M1 A1
A1
3
M1 A1
A1
3
B1
1
17
PhysicsAndMathsTutor.com
M1 Dynamics - Connected particles
(d)
1 1
× g × 1.2 2 = 0.18 g (≈ 1.764 )
2 4
Distance of A above ground s1 =
Speed on reaching ground v =
For B under gravity
(0.3g )2
1
g × 1.2 = 0.3 g (≈ 2.94 )
4
= 2 gs 2 ⇒ s 2 =
(0.3)2
g = (≈ 0.441)
2
S = 2s1 + s2=3.969 ≈ 4.0 (m)
M1 A1
M1 A1
M1 A1
A1
7
[14]
3.
(a)
For whole system: 1200 – 400 – 200 = 1000a
M1 A1
a = 0.6 m s–2
For trailer: T – 200 = 200 × 0.6
om
T = 320 N
For car: 1200 – 400 – T = 800 × 0.6
gm
ai
OR:
3
M1 A1 ft
l.c
(b)
A1
OR:
M1 A1 ft
A1
3
For trailer: 200 + 100 = 200f or –200f
M1 A1
rt
m
(c)
uj
un
gu
@
T = 320 N
A1
f = 1.5 m s–2 (–1.5)
rb
e
A1
he
For car: 400 + F – 100 = 800f or –800f
M1 A2
F = 900
A1
7
(N.B. For both: 400 + 200 + F = 1000f)
[13]
4.
(a)
T – 5gsinα = 5a
M1 A1
15g – T = 15a
(b)
Edexcel Internal Review
M1 A1
solving for a
M1
a = 0.6g
A1
solving for T
M1
T = 6g
A1
8
M1 A1
A1 f.t.
3
For Q:
5g –
N = 2g
N = 5a
18
PhysicsAndMathsTutor.com
M1 Dynamics - Connected particles
(c)
90º – α
)
2
M1 A2
= 12gcos 26.56..°
= 105 N
A1 f.t.
A1
F = 2T cos(
5
[16]
3
M1B1A1
DM1A1
5
om
M1A1
A1
A: T – µmg = m × 4g/9
Sub for T and solve: µ = 2/3*
When B hits: v2 = 2 × 4g/9 × h
Deceleration of A after B hits: ma = µmg ⇒ a = 2g/3
Speed of A at P: V2 = 8gh/9 – 2 × 2g/3 × h/3
2
⇒ V = √(gh)
3
un
M1A1
M1A1ft
DM1
he
rb
e
rt
m
uj
(c)
gu
@
(b)
B: 2mg – T = 2m × 4g/9
⇒ T = 10mg/9
l.c
(a)
gm
ai
5.
(d)
Same tension on A and B
A1
6
B1
1
[15]
6.
(a)
(b)
(c)
1 2
1
9
at ⇒ 3.15 = a ×
2
2
4
–2
a = 2.8 (m s ) *
s = ut +
M1A1
A1
3
N2L for P: 0.5g – T = 0.5 × 2.8
T = 3.5 (N)
M1A1
A1
3
N2L for Q: T – mg = 2.8m
3.5
5
= *
m=
12.6 18
M1A1
Edexcel Internal Review
cso
cso
DM1A1
4
19
PhysicsAndMathsTutor.com
M1 Dynamics - Connected particles
(d)
The acceleration of P is equal to the acceleration of Q.
(e)
v = u + at ⇒ v = 2.8 × 1.5
(or v2 = u2 + 2as ⇒ v2 = 2 × 2.8 × 3.15)
(v2 = 17.64, v = 4.2)
B1
1
M1A1
v = u + at ⇒ 4.2 = –4.2 + 9.8t
6
t = , 0.86, 0.857 (s)
7
DM1A1
DM1A1
6
[17]
7.
(a)
2g – T = 2a
T – 3g sin 30° = 3a
M1 A1
M1 A1
4
cso
M1
A1
2
or equivalent
accept 17.6
M1
A1
2
B1
1
M1
A1
2
2g – 3g sin 30° = 5a
a = 0.98 (m s–2) *
gm
ai
T = 2(g – a)
≈18 (N)
un
gu
@
(c)
l.c
om
(b)
N2L Q
N2L P
The (magnitudes of the) accelerations of P and Q are equal
(e)
v2 = u2 + 2as ⇒ v2 =2 × 0.98 × 0.8 (=1.568)
v ≈ 1.3 (m s–1)
(f)
he
rb
e
rt
m
uj
(d)
N2L for P
accept 1.25
–3g sin 30° = 3a
1
a = (−) g
2
1 2
1
at ⇒ 0 = √1.568t – 4.9t 2
2
2
t = 0.51 (s)
s = ut +
M1 A1
or equivalent
M1 A1
accept 0.511
A1
5
[16]
A maximum of one mark can be lost for giving too great accuracy.
8.
(a)
Car + trailer:
2100a = 2380 − 280 − 630
= 1470 ⇒ a = 0.7 m s−2
M1 A1
A1
3
M1 for a complete (potential) valid method to get a
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20
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M1 Dynamics - Connected particles
(b)
700 × 0.7 = T − 280
e.g. trailer
M1 A1ft
⇒ T = 770 N
A1
If consider car: then get 1400a = 2380 − 630 − T.
Allow M1 A1 for equn of motion for car or trailer wherever
seen (e.g. in (a)).
So if consider two separately in (a), can get M1 A1 from (b) for
one equation; then M1 A1 from (a) for second equation, and
then A1 [(a)] for a and A1 [(b)] for T.
In equations of motion, M1 requires no missing or extra terms
and dimensionally correct (e.g. extra force, or missing mass, is
M0).
If unclear which body is being considered, assume that the body
is determined by the mass used. Hence if ‘1400a’ used, assume
it is the car and mark forces etc accordingly.
But allow e.g. 630/280 confused as an A error.
(c)
Car: 1400a’ = 2380 − 630
3
M1 A1
om
⇒ a′ = 1.25 ms−2
gm
ai
l.c
distance = 12 × 4 + ½ × 1.25 × 42
= 58 m
A1
M1 A1ft
A1
6
Same acceleration for car and trailer
B1
Allow o.e. but you must be convinced they are saying that it is
same acceleration for both bodies.
E.g. ‘acceleration constant’ on its own is B0
Ignore extras, but ‘acceleration and tension same at A and B’ is
B0
he
(d)
rb
e
rt
m
uj
un
gu
@
Must be finding a new acceleration here. (If they get 1.25
erroneously in (a), and then simply assume it is the same acceln
here, it is M0).
1
[13]
9.
A:
T
(a)
3mg – sin 30 – T = 3m.
1
10
g
M1 A1
3mg
⇒
T=
Edexcel Internal Review
6
5
mg
A1
3
21
PhysicsAndMathsTutor.com
M1 Dynamics - Connected particles
T
(b)
R
mg
F: R(perp): R = mg cos 30
T – mg sin 30 – F = m. 101 g
R(//):
M1 A2, 1, 0
Using F = µR
M1
6 mg − 1 mg − µmg 3 = 1 mg
5
2
2 10
M1
→
(c)
M1 A1
µ = 0.693 or 0.69 or
T
2 3
5
A1
8
T
Magn of force on pulley = 2Tcos60 =
6
5
mg
M1 A1ft
B1 (cso)
3
[14]
T
T
600
1500
rt
m
300
un
(a)
uj
10.
gu
@
gm
ai
l.c
om
Direction is vertically downwards
he
rb
e
Lorry + Car: 2500a = 1500 – 300 – 600
a = 0.24 m s–2
(b)
Car: T cos 15 – 300 = 900a
OR Lorry: 1500 – T cos 15 – 600 = 1600a
Sub and solve: T ≈ 534 N
(c)
300
Deceleration of car = 300/900 = 1/3 m s–1
Hence 62 = 2 × 1/3 × s ⇒ s = 54 m
(d)
Vertical component of T now removed
Hence normal reaction is increased
M1 A1
A1
3
M1 A1
↓↓
M1 A1
4
M1 A1
M1 A1
4
M1
A1cso
2
[13]
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22
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M1 Dynamics - Connected particles
11.
(a)
‘s = ut + ½at2’ for B: 0.4 = ½ a(0.5)2
M1 A1
a = 3.2 m s–2
(b)
A1
N2L for B:
0.8g – T = 0.8 × 3.2
M1 A1ft
↓
M1 A1
T = 5.28 or 5.3 N
(c)
(d)
A:
F = μ × 0.5g
N2L for A:
T – F = 0.5a
Sub and solve
μ = 0.75 or 0.751
3
4
B1
Same acceleration for A and B.
M1 A1
↓
M1 A1
5
B1
1
[13]
om
M1 A1
12g
or 16.8 N or 17 N
7
A1
3
uj
3g
7
M1 A1
rt
m
m kg: T – mg = m.
↓
rb
e
(b)
un
gu
⇒T=
3g
7
l.c
3 kg: 3g – T = 3 ×
gm
ai
(a)
@
12.
he
12g
3mg
= mg +
(Sub for T and solve)
7
7
⇒ m = 1.2
M1
A1
4
[7]
R1
13.
R2
F2
40
4g
(a)
F1 =
6g
2
7
× 4g (= 11.2) or F2 =
System: 40 –
2
7
× 4g –
⇒ a = 1.2 m s (*)
–2
Edexcel Internal Review
2
7
2
7
× 6g (= 16.8)
× 6g = 10a (equn in a and not T)
B1
M1 A1
A1
4
23
PhysicsAndMathsTutor.com
M1 Dynamics - Connected particles
(b)
P: T –
g = 4 × 1.2 or Q: 40 – T –
8
7
12
7
g = 6 × 1.2
M1 A1
⇒ T = 16 N
A1
3
(c)
Accelerations of P and Q are same
B1
1
(d)
v = 1.2 × 7 = 8.4
8
2
P: (–) g = 4a ⇒ a = (–) g = 2.8
7
7
B1
M1 A1
↓
M1 A1
0 = 8.4 – 2.8t ⇒ t = 3 s (*)
(e)
Q: 40 –
12
7
g = 6a (⇒ a ≈ 3.867)
v = 8.4 + 3.867 × 3 = 20 m s
5
M1 A1
↓
M1 A1
–1
4
a
gm
ai
a
(b)
he
rb
e
rt
m
uj
3mg
4mg
A: T – 4g sin 30 = 4a
B: 3g – T = 3a
18g
= 25.2 N
⇒T=
7
@
T
R
gu
(a)
T
un
14.
l.c
om
[17]
M1 A1
6
M1 A1
A1
3
T
T
R
R = 2T cos 30
≈ 44 or 43.6 N
(c)
M1 A1
M1 A1
(i)
String has no weight/mass
B1
(ii)
Tension in string constant, i.e. same at A and B
B1
2
[11]
Edexcel Internal Review
24
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M1 Dynamics - Connected particles
15.
T
(a)
B
0.2g 0.4g
0.4g – T = 0.4 ×
(b)
8
g
25
T=
(c)
1
g
5
or
3.14
or 3.1 N
M1 A1
2
M1 A1
2
T
A
mg
om
gm
ai
16
(*)
35
M1 A1
M1 A1
4
B1
1
v=
(f)
1
g×1
5
uj
v2 = 2 ×
rt
m
(e)
rb
e
Same T for A & B
he
(d)
un
gu
@
→m=
1
g
5
l.c
T – mg sin 30° = m ×
2g
≅ 1.98 or 2 ms–1
5
1
1
mg = ma ⇒ a = – g
2
2
2g
1
– 2 × g × 0.4
v2 =
5
2
⇒v=0
A: –
M1
A1
2
M1 A1
M1 A1
A1
5
[16]
16.
(a)
Car + truck: 2000a = 2400 – 600 – 400
a = 0.7 m s−2
Edexcel Internal Review
M1 A1
A1
3
25
PhysicsAndMathsTutor.com
M1 Dynamics - Connected particles
(b)
(c)
Car only: T – 400 = 800 × 0.7
[or truck only: 2400 – T – 600 = 1200 × 0.7]
T = 960 N
M1 A1 ft
A1
New acceleration of truck a′ given by
1200 a′ = 2400 – 600
a′ = 2400 – 600 = 1.5 m s−1
28 − 20
Time to reach 28 m s−1 =
= 5.33 s
1.5
3
M1
A1
M1 A1
Time to reach 28 m s−1 if rope had not broken =
Difference = 6.1 s ≈ 6 s (*)
28 − 20
= 11.43 s M1 A1
0.7
A1
7
[13]
R
A
om
T
l.c
(a)
B
0.8g
gm
ai
17.
a
T
R
F
T
un
uj
rt
m
B1
M1 A1
M1 A1
5
M1
M1
A1
3
rb
e
a = 0.6g = 5.88
Hence 0.6 = ½ × 0.6g × t2
t = 0.45 or 0.452 s
he
(b)
gu
A: T = 0.8a
B: 1.2g – T = 1.2a
Solve: T = 0.48g = 4.7 N
@
1.2g
,
T,
0.8g
1.2g
1
× 0.8g
5
A: T ′ – F = 0.8a′
B: 1.2g – T ′ = 1.2a′
M1 A1
B1
Solve:
a′ = 0.52g
0.6 = ½ × 0.52g × t2
t = 0.49 or 0.485 s
M1 A1
M1
A1
F = µR =
B1
8
[16]
Edexcel Internal Review
26
un
uj
rt
m
rb
e
he
l.c
gm
ai
@
gu
om
END
51