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Physics
Session
Particle dynamics - 4
Session Objectives
Session Objective
1. Motion of varying mass
system (rockets)
2. Problems
Session Opener
Rocket moves when fuel is ejected as
gases (carrying some momentum)
 Rocket gets a momentum in
opposite direction.
 Mass of rocket reduces.
M  M
V
M
M
 Ve (w.r.t. rocket)
M  M V  M  V  V   M  V  Ve 
 V  V
Motion of Varying System
(Rockets)
V e M  V  V   M  V  Ve 
MMV MMV
t  o : MdV  VedM
M
V
M
dMi
 fVf  Vi  Ve f n
  dV  Ve  M
f
Vi
Mi M
dV
dM
Thrust  M
 Ve
dt
dt
Illustrative Problem
A rocket moving in free space has a
speed of 3.0  103 m/s relative to
earth. Its engines are turned on and
fuel is ejected in a direction opposite
the rocket’s motion at a speed of 5.0
 103 m/s relative to the rocket.
(a) What is the speed of the rocket relative to
Earth once its mass is reduced to one half its
mass before ignition ?
(b) What is the thrust on the rocket if it burns
fuel at the rate of 50 kg/s ?
Solution
(a)
 Mi 
vf  vi  ve ln 

M
 f
 Mi 
 3.0  10  5.0  10 ln 

0.5M
i

3
3
 6.5  103 m / s
(b)
Thrust  ve
dM 
m   kg 
  5.0  103   50
dt
s 
s 

 2.5  105 N
Class Exercise
Class Exercise - 1
A block of mass 2.5 kg rests on a
smooth, horizontal surface. It is
struck by a horizontal jet of water
at the flow rate of 1kg/s and a
speed of 10m/s. What will be the
starting acceleration of the block?
Jet
A
a
Solution
Water jet supplies the force as
mass changes over time and
velocity is constant.
F
dm
dv dm
vm

v ( v const.)
dt
dt
dt
= 1kg/s × 10m/s = 10N
This provides the acceleration to the block.
a
F
10

 4m / s2
m
2.5
Class Exercise - 2
Two forces f and F are acting on
objects m and M respectively as
shown. The horizontal surface is
smooth and the spring joining the
masses is light. What is the
extension of the spring if its force
constant is K? (F > f)
f
m
T
T
M
F
Solution
Total force providing the motion: F – f.
Total mass accelerated: (m + M)
a 
F – f 
m  M
For M: F – T = Ma
For m: T – f = ma
Substituting for ‘a’ in any of equations
T
x
mF  Mf
mM
T mF  Mf

k k(m  M)
f
m
T
T
M
F
Class Exercise - 3
A light spring, hanging from a
rigid support, supports a light
pulley. A light and unstretchable
string goes around the pulley.
One end of the string is tied to
the floor and the other end
supports a hanging mass m. If
the mass has dropped by a
height h, express k in terms of m
and h.
k
P
m
Solution
Let the spring stretch by x.
Then the string moves by
2x. So when mass moves
by h, the spring stretches
by h
.
2
k
T
T
h
2T
m
g
T
T
Solution
Tension (force) on the spring = force
on the pulley = 2T.
T = mg as the mass is at rest
 Spring force = 2T = 2mg =
k(extension).
h
2mg = k.
2
k
4mg
h
Class Exercise - 4
An insect inside a hemispherical
bowl is trying to climb out. The
radius of the bowl is r and the
coefficient of friction between the
bowl and the insect is m . How high
can the insect climb?
r
• Insect
Solution
FBD of insect gives [at the point
where it can reach. Beyond that
point mgsinq > f and insect will
slip down.]
N = mg cosq, f = mgsinq
f = mN = mmg cosq
O
r
N
 tanq = m
q
j
y
h
mg
Tangent
Solution
r 2 – y2
IJ
tan q 

OJ
y
m
r 2 – y2
y
y
r
f
q
1  m2

 h  r – y  r 1 –


N
mg cos q


1  m2 
1
q
mg sin q
Class Exercise - 5
A mass m is sliding down an
equally inclined right angled
trough, inclined to the horizontal
by an angle q. m is the coefficient
of kinetic friction between the
trough and m. What is the
acceleration of the mass?
90o
M
q
Solution
The block has two surfaces of
contact.
So mgsinq – mN1 – mN2 = ma ... (i)
|N1| = |N2| (surfaces identical)
y
N = 2N1 cos45 =
N
2 N1
q
x
mg
M
Solution
 N1  N2 
N
2
 N 
mN1  mN2  m  2
  m 2N
 2
N2
But N = mgcosq

So (i)  a  g sin q – 2m cos q

N
N1
Class Exercise - 6
A block A of mass 2 kg rests on
top of a block B of mass 3 kg
which is on the floor of a lift. The
lift accelerates upward with an
acceleration of 2m/s2. Find the
normal reaction on B by the floor.
(g = 10m/s2)
A
B
a
Solution
Using FBD of A
F
A
 mA g – NA  – mA a
 NA  mA  g  a
Using FBD of B
F
B
NA
 NB – mBg – NA  mBa
A a
 NB – mBg –mA g  a  mBa
NB
 NB  mA  mB   g  a
= 5 × 12 = 60N
NA
mg
mBg
B a
Class Exercise - 7
In the given figure the weight of A,
hanging from a fixed, light pulley is
200 N and that of B, hanging from a
free pulley is 300 N. Find the
tensions T1 and T2 (g = 10m/s2)
N
P1
T1
A
cT
P2
2
200 N
B
300 N
Solution
From FBD of P2, 2T1 = T2
B will rise as upward force on it
(T2) is more.
Noting that when A falls by y, B rises
y
by and in time t
2
y 1
1
2
 aBt and y  aA t2, aA  2aB
2 2
2
T1
T1
P2
T2
Solution
From FBD of A and B,
T2 – mBg = mBaB  2T1 – 300  mBaB
mAg – T1 = mAaA
 200 – T1 
200
.2aB ... (ii)
10
aB
300
(i)  2T1 – 300 
aB
10
100 = (80 + 30)aB aB 
100
m / s2
110
T2
T1
B
A
mBg
mAg
aA
Solution
Putting aB in (ii)
200 10
T1  200 –
.
2
10 11
2

 200 1 – 
 11
= 163.6N
T2 = 2T1 = 327.2N
Class Exercise - 8
Three blocks A, B and C, weighing
3kg, 4kg and 8kg respectively are
arranged one on top of other as
shown. Top block A is attached to a
rigid wall by a rigid, light rod.
Blocks B and C are connected to
each other by a light, unstretchable
string passing around a light pulley.
Find the force F to drag block C with
constant speed. All surfaces have
coefficient of kinetic friction equal to
0.25(g = 10m/s2)
rod
A
B
C
F
Solution
f1  mNA
 m mA g

NA  mAg
NA
R
f1
NB
MAg
NA
f1
T
f2  mNB m mA  mB  g
For B: T – f2 – f1  0
f2
MBg
Solution
f3  mNC  m mA  mB  mC  g
For C: F – T – f3 – f2 = 0
F – f3 – 2f2 – f1  0
NC
f2
NB
T
 F  f3  2f2  f1  mg  4mA 3mB  mC 
F
f3
MCg
= 0.25 × 10 × 32
= 80.0N
Class Exercise - 9
Block A has a mass of 200 gm and
block B has a mass of 700 gm in
the arrangement shown. Coefficient
of friction is 0.2. How large should
be the value of force F so that block
B has an acceleration of 0.5m/s2? (g
= 10m/s2). Friction is present at all
the contact surfaces.
A
B
F
Solution
f2  mmA g
For mA: T – f2 = mAa
T
f2
a
f2
NAMAg
f1  m mA  mB  g
For mB F – T – f1 – f2  mBa
NA
T
f1
F
a
NB
MBg
Solution
F –  f1  2f2   mA  mB  a
F=
mA  mB  a  m mB  3mA  g
= (0.2 + 0.7)0.5 + 0.2 (0.7 + 0.6)10
= 0.45 + 2.60 = 3.05N
Class Exercise - 10
In the arrangement shown, when
m is 3.0 kg its acceleration is
0.6m/s2 and when m is 4.0kg its
acceleration is 1.6 m/s2. Find the
frictional force on M and the mass
of M. Assume the pulleys to be
massless and strings to be light (g
= 10m/s2)
M
m
Solution
mg – 2T1 = ma
... (i)
2T1
m
m
T1 – f = M.2a
2T1 – 2f = 4Ma
... (ii)
T1
Using (i) and (ii)
T2 = 2T1
mg – 2f = (m + 4M)a
Solution
as given for m = 3kg a = 0.6m/s2
3 × 10 – 2f = (3 + 4M)(0.6)
... (iii)
and for m = 4kg a = 1.6 m/s2
Using (iv) and (iii)
40 – 30 = 6.4 + 6.4M – 1.8 – 2.4M
 4M = 5.4
M = 1.35 kg
Substituting M in (iii),
2f = 30 – (3 × 4 × 1.35) (0.6)= 12.48N
Thank you