Chapter 3 - BITS Pilani

Chapter 3
Momentum
1. Dynamics of a System of Particles &
Conservation of Momentum
2. Concept of Centre of Mass
3. Motion of Systems with Variable Mass
Dynamics of a System of Particles &
Conservation of Momentum
The Two Particles System
Two masses connected by a mass less
spring and falling in the gravitational field

Fint
k x
k x
m1g
m 2g
Earth

Fext
Forces on First Mass

dp1  ext  int
 F1  F1
dt

dp 2  ext  int
 F2  F2
dt
Adding the two

 
 int  int
 ext  ext

d 
(p1  p 2 )  F1  F2  F1  F2
dt
We have

 int  int 
F1  F2  0

dp tot  ext
 Ftot
dt

Dynamics of a System of Indefinite
Number of Particles

Fj 

dp j
dt

 int  ext
Fj  Fj  Fj
 int
 Fj 
j
 int 
 Fj  0
j
 ext
 Fj 
j

j

dp j
dt
 ext 
 Fj  Fext
j


dp j
j
dt

 Fext

dP 
Or,
 Fext
dt
Where,

P

 pj
j
is the total momentum of
the system
Conservation of Momentum

dP 
Since
 Fext
dt



Fext  0  P  Const.
In the absence of any net external force,
the total momentum of a system of
particles is conserved, although
individual momenta may change with
time
What if the internal forces do not
add up to zero?
Electrostatics
 q2
f 21

f12
q1


f12  - f 21
Electrodynamics
q1

f12
 q2
f 21


f12   f 21
In electrodynamics, momentum may
not be conserved!
Centre of Mass

r2



 m1 r1 + m 2 r2 ...... + m N rN
R=
m1 + m 2 ....... + m N
m1x1 + m 2 x 2 ....... + m N x N
X=
m1 + m 2 ...... + m N
m1 y1 + m 2 y 2 ....... + m N y N
Y=
m1 + m 2 ...... + m N
m1z1 + m 2 z 2 ....... + m N z N
Z=
m1 + m 2 ...... + m N

r1
m1
m2

R CM

r3 m 3 
r4
m4
O
Motion of CM

R

 mi ri

MR 
i
Μ

 mi ri 
i


 pi  P
i


2
d R dP 
M 2 
 Fext Or
dt
dt


M a CM  Fext
The centre of mass of a system of
particles moves as though it is particle,
carrying the total mass of the system,
and is acted upon by the net external
force on the system.

Fext


 dR
0 V
 Const.
dt



 R  R0  V t

R 0 is the position of CM at t = 0
CM
Spring-mass system tossed into the air
Prob. 3.3
Suppose that a system consists of
several bodies, and the position of the
CM of each body is known. Prove that the
CM of the combined system can be found
by treating each body as a particle
concentrated at its own CM.
CM of sub-system 1
CM of sub-system 2
CM of entire system
M2
M1

R

R1

R2
O


 M1 R 1 + M 2 R 2
To prove : R =
M1 + M 2
M1 : Mass of Sub-system 1
M2 : Mass of Sub-system 2

 mi R i

R1 

i 1
M1
;

R2 


M1R 1  M 2 R 2

M1  M 2

 mi R i
M N

i 1
M tot

 mi R i
M N
M

R
i  M 1
M2

 mi R i 

 m1R i
M
M N
i 1
i  M 1
M1  M 2
Prob. 3.7
Two masses, m1 & m2 are connected by
a spring of spring constant k and
unstretched length L. The entire system
is pushed against a wall so that the
spring is compressed to length L/2.
Mass m2 is then released at t = 0.
Find the motion of the CM as function of
time.
m1
m2
O
/2

The mass m1 will not leave the wall until m 2
reaches position 
Motion of m 2 from x   2 to x   is
SHM of amplitude  2 and frequency   k m 2
1. For time

0 t 
2

x1 ( t )  0 ; x 2 ( t )    cos t
2
m x  m2 x 2
 X( t )  1 1
m1  m 2
2. For time

1

 1  cos t 
2
2

(Assuming
equal masses)

t
2
CM moves uniformly forward

 X(t )   2  V( 2) t   2 
t
4
CM Coordinates and Reduced Mass
 
m r F
1 1
1
 
m 2 r2  F2
  
Define : r  r2 - r1


 F2
F1

r
m2
m1


F2
 1 / m1  1 / m2  F2 

m1

r

r1

R

r2
O
m2
 
or,  r  F2
μ
Where,
m1m 2

m1  m 2

r
m1
is the reduced mass of the system
 
r1  R 
m2 
r
m1  m 2


; r2  R 
m1 
r
m1  m 2
Prob. 2.4
Two masses m & M respectively,
undergo uniform circular motion about
each other at a separation of R, under
the influence of their gravitational
attraction. Find the time period of
rotation.
M
m
CM
Motion observed by an Outside Inertial Observer
μ
R
Motion of ‘m’
seen from ‘M’
M
GMm
R 
R2
2


GMm
R 3

G ( M  m)
R3
R3 2
 T  2
G ( M  m)
m
1
For the earth-moon system,

M 80
Thus, the correction due to the motion of the
earth is a reduction in time period of the
moon by  0.62 %
Impulse and Momentum
Impulse of a force over the time interval
t1 to t 2
 t2 
I   F dt
t1
 dp
Putting F 
dt

I

dp
t dt dt 
1
t2
t2
 


 dp  p(t 2 )  p(t1 )  p
t1
Systems With Variable Mass
1. Mass Continually Added to System :
Loading of a moving wagon, Conveyer belts
2. Mass Continually Leaving The System :
Rocket, Leaking wagon

v′

Fext
m
System : M + Δm
M

v
Time : t

Fext
Δm
M
Time : t + Δt


v  v
Change in the momentum over time t :





P  (M  m) (v  v)  (M v  m v)

 

 M v  m (v  v)  m  v
Equating this to the impulse
of the external

force on the system, Fext t , and dividing out
by t



v m  
v

Fext  M

( v  v )  m 
t
t
t
In the limit as t  0 , the last term drops out, and
 
dm 
M a  Fext 
v rel
dt



Where, v rel  v  v , is the relative velocity of
the mass entering/leaving the system w.r.t. the
main mass
Prob. 3.20
A rocket ascends from rest in a uniform
gravitational field by ejecting exhaust.
Speed of exhaust : u
Rate of exhaust : dm/dt = γm (γ < 0)
(m is instantaneous mass of the rocket)
Retarding force : bmv (v is instantaneous speed
of rocket)
Find : velocity of rocket as function of time
Equation of motion
dv
m
  mg  bvm  um
dt
dv
  (bv  g  u )
dt
dv
0 bv  g  u   0 dt
v
t
1  bv  g  u 
ln 
t
b  g  u 


g  u  bt
v( t ) 
e 1
b

u  g
b
1  e 
 bt
Example*
Change of Course by a Spaceship :

v

u

v0
Rate of Exhaust = b
 
uv  0

u  u0

M(t )  M0  bt
Equation of Motion


dv dm 
M

v rel   bu
dt
dt

Taking dot product with v

 dv
 
M v
  bv  u  0
dt


v  Const.  v0

Taking dot product with v 0

 dv
 
 M v0 
  bv 0  u
dt

v

u
 
v0  v  v02 cos 

M v 02


v0
 
v0  u  v0 u 0 sin 
d
(cos )   b v0 u 0 sin 
dt
bu 0
d
1

dt
v 0 M 0  bt

u 0  M0 
( t ) 
n 

v 0  M 0  bt 
Prob. 3.10
An empty freight car of mass M starts from
rest under an applied force F. At the same
time sand begins to run into the car at a
steady rate b from a hopper at rest along the
track.
Find the speed of the car after a mass of
sand m has been transferred.
F
Here,

dm
i)
b
dt
ii ) v rel   v
Equation of Motion
dv
M
 F  bv
dt
dv
0 F  bv 
v

mb

o
dt
M 0  bt
1  F  bv  1  M 0  m 
 ln 
 ln 


b  F  b  M0 
iii ) M  M0  bt
Prob. 3.18
Raindrop of initial mass M0 falls from rest
under the influence of gravity. The drop gains
mass at a rate proportional to the product of
its instantaneous mass and instantaneous
velocity :
dM
 kMv
dt
Show that the speed of the drop eventually
becomes effectively constant and give an
expression for the terminal speed. Neglect air
resistance
Moister
Raindrop
v
Equation of Motion
dv
M
 Mg  kMv2
dt
dv
o g  kv2 
v

t
 dt
0
( v rel   v)
v

0
dv

g  k v
dv
 2 gt
g  k v
v

0
Integrating and solving for v,
v
2

g e
 2
k  e
 v term 
gk t
gk t
g
k
 1

 1
1
v rel

Fext
2
v rel
dm1
dt
dm 2
dt
M

v
Equation of Motion
 
dv
 dm1   1
 dm 2   2
M
 Fext  
 v rel  
 v rel
dt
 dt 
 dt 
Prob. 3.13
A ski tow consists of a long belt of rope
around two pulleys, one at the bottom of a
slope and the other at the top.
A motor drives the pulleys to move the belt
at constant speed of 1.5 m/s.
Length of belt : 100 m
A skier of mass 70 kg takes the tow every 5 s.
Find the average force required to pull the belt
Fpull
v
v : 1.5 m/s
θ
L
L : 100 m
θ : 200
Mass input to the tow at the bottom
 dm 

  14 kg / s
 dt in
vinrel  1.5 m / s
Mass output from the tow at the top
 dm 

  14 kg / s
 dt out
v out
rel  0
100
 13
No. of skiers on the tow at any time =
5  1.5
Equation of motion of the tow
dv
 dm  in
 dm  out
0
M
 Fpull  Mgsin 20  
 v rel  
 v rel
dt
 dt in
 dt out
 Fpull
 dm  in
 Mgsin 20  
 v rel
 dt in
0
M  13  70  910 kg
Example :The Water Jet Boat
v
v out
rel
v inrel
vinrel   v
vout
rel   u 0
Equation of Motion
dv
 dm  in  dm  out
M
  bv  
 v rel  
 v rel
dt
 dt in
 dt out
  b v   (v)  () ( u 0 )   (b  ) v   u 0
Integration gives
u 0
v( t ) 
b
v term
 u0

b
b 

t

M
1  e




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