Transcript System of particles

Physics
Session
System of Particle - 4
Session Objectives
Session Objectives
1. Collision-elastic and inelastic
2. Elastic collisions in one and two dimension
3. Coefficient of restitution
4. Explosions
5. Explosion in a projectile
Collision between two particles
 Since Fext=0 so Pi=Pf
 KEi  KEf
 if KEi=KEf : Perfectly Elastic
 if KEi > KEf : Inelastic
m m m mm m M
M
M
M
All elastic collisions are perfectly elastic
M
M
Elastic collisons
 Since Pi=Pf
mu1  Mu2  mv1  Mv 2
 Since KEi=KEf
1
1
1
2 1
2
mu12  Mu2

mv

Mv
2
1
2
2
2
2
2
NOTE: Solving these two equations you can
get the values of final velocities
Inelastic collisons
 Since Pi=Pf
mu1  Mu2  mv1  Mv 2
 Since KEi >KEf
1
1
1
1
2
2
mu12  Mu2
 mv12  Mv2
2
2
2
2
NOTE: You need one more equality
to solve for final velocities.
Coefficient of restitution
u1
m
u2
M m M
v1
m
v2
M
Velocity of separation =e (velocity of approach)
(v2  v1)  e(u1  u2 )
When e=1: elastic collision
e=0: completely inelastic collision
0  e  1:Inelastic collision
General speed relations:
(Remember)
v1 
(m1  em2 )
(1  e)m2
u1 
u2
(m1  m2 )
(m1  m2 )
(1  e)m1
m2  em1
v2 
u1 
u2
(m1  m2 )
(m1  m2 )
Final speeds (special cases
Elastics collision)
(a) Heavy body and light body:
m1  m2
v1  u1
v2  2u1  u2
(b) If m2>>m1(a light body hits a
heavy body from behind)
m1  m2
v1  u1  u2
v2  u1
(c) Bodies of equal mass (m1=m2)
m1  m2
v1  u2 ,
v2  u1
Questions
Illustrative problem
If a neutron with velocity v collides
elastically with an  particle, what is
its resultant velocity?
1
v
5
3
(c) v
5
(a)
2
v
5
4
(d) v
5
(b)
Solution
v1 
(m1  m2 )u1
2m2

u2
(m1  m2 )
(m1  m2 )
v1 
(m  4m)v
2.4m

.0
(m  4m)
(m  4m)
3
 v
5
3
v  v
5
Special case: perfectly inelastic
collision (e=0)
m1u1  m2u2  (m1  m2 )v
u1
m
u2
M m M
v
m M
v
m M
Two Dimensional Collisions
Elastic collision in 2D
Conservation of momentum:
m1u1x  m2u2x  m1v1x  m2v2x
(x axis)    (1)
m1u1y  m2u2y  m1v1y  m2v2y
(y axis)    (2)
Conservation of energy:
2
2
2
1
1
1
1
2
m1u1  m2u2  m1v1  m2v2    (3)
2
2
2
2
NOTE: There are 3 equations but 4 variables. So one
physical condition is required to solve the problem.
Illustrative Problem
A ball of mass 1kg moving with
velocity V=5i collides with
another ball of mass 2 kg
initially at rest and at origin.
The first ball comes to rest
after collision and second breaks into two equal pieces.
one piece starts moving with a velocity V=3j m/s. Then
the velocity of other piece is:
4m/s
3m/s

v
Solution
In x-direction
1x4+2x0 = 1x v cos
In y-direction
0 = 1x3 –1xvsin
Using given equation:
tan=3/4 and v = 5m/s
Class Test
Class Exercise - 1
A light spring of spring constant K =
1,000 N/m is kept compressed between
two blocks of masses m = 10 kg and M
= 15 kg on a smooth horizontal surface.
When released the blocks acquire
velocities in opposite directions. The
spring loses contact with the blocks
when it acquires natural length. If the
spring was initially compressed through
a distance x = 15 m, find the final
speeds of the two blocks.
K
m
M
Solution
When spring acquired its natural
length, let the velocities of two blocks
be v and u respectively. As no external
force acts on the system,
m
mv  Mu  0  u  v ..........(i)
M
Also initial energy 
1 2
kx
2
1
1
1
Final energy  mv2  Mu2  kx2
2
2
2
2
m
 mv 2  M
v 2  kx 2
M2
Solution
m M  m  2

v  kx2 ...............(ii)
M
v
15  1000
 15
10  25
 15  60 m
= 1.16 m/s
and u 
m k
x 
M M  m 
10  1000
 15  0.24 m/s
15  25
Class Exercise - 2
Two equal spheres A and B lie on a
smooth horizontal circular groove at
opposite ends of a diameter. A is
projected along the groove and at time t
it impinges on B. If e is coefficient of
restitution, the second impact will occur
after a time
2t
e
 t
(c)
e
(a)
t
e
2 t
(d)
e
(b)
Solution
Let r be the radius of the groove, then
arc AB = r
r
speed of A : u 
t
r
Velocity of A before collision  u 
t
Velocity of B before collision = 0
Velocity of A after collision = v
Velocity of B after collision = v'
A
•
r
B
Solution
Then v – v' = e(–u)
 v  v  e u
Let t' be the time after next impact
takes place, then
t 
Relative dis tance
Relative velocity
2r
2t


eu
e
Class Exercise - 3
A neutron travelling with a velocity v and
kinetic energy E collides elastically head-on
with the nucleus of an atom of mass
number A at rest. What fraction of total
energy is retained by the neutron?
Solution
1 V = 1V1 + AVA
1
1
1
 1V 2   1V 12 AVA 2
2
2
2
 V2 – V12 = AVA2
Using (i),
V1  V
A –1
A 1
1 2
2
1V1
A
–
1


 2

1 2  A  1 
1V
2
Class Exercise - 4
A moving body with a mass m1 strikes
a stationary body of mass m2. The
masses m1 and m2 should be in the
ratio m1/m2 so as to decrease the
velocity of the first body 0.5 times
assuming a perfectly elastic impact.
What is the ratio of m1/m2 ?
Solution
Before Collision
Vi = u
Vi = O
m1
m2
After Collision
Vf =
V
2
m1
Vf = V
m2
Applying momentum conservation,
u
m1u m2(o)  m1    m2v
2
 m1  u
m1u

 m2v  
 v

2
 m2  2
Solution
Also,
u
v   e(u)  u
2
3u
 v
2
 m1  u
3u
 


2
 m2  2

m1
3
m2

e  1
Class Exercise - 5
A ball of mass m moving with a
speed u undergoes a head on
elastic collision with a ball of mass
nm initially at rest. What fraction of
the incident energy is transferred to
the heavier ball?
Solution
Before Collision
m
After Collision
nm
u
m
Vi =O
nm
V1
V2
Applying momentum conservation,
mu  mv1  nmv 2  u  v1  nv 2
Also, v2 – v1 = u
2u
 (n  1)v 2  2u  v 2 
(n  1)

nv 22
2
u

n
2
u

4u2
(n  1)
2

4n
(n  1)2
1
 nm  v 22
 Fraction  2
1
mu2
2
Class Exercise - 6
A body of mass m moving with a
velocity v1 in the x-direction collides
with another body of mass M moving in
y-direction with a velocity v2. They
coalesce into one body during collision.
What is the direction and magnitude of
the momentum of the final body?
Solution
From conservation of momentum in
x- and y-directions, we get
mv1  m  M v cos .............(1)
and Mv 2  m  M v sin  ..............(ii)
mv1
1  Mv 2 

 cot     tan 

Mv 2
mv

1
y
(m
and mv1   Mv2   m  M v 2
2
2
2
+
V1

m
 v
mv1 2  Mv 2 2
 m  M
V2
M
M)
V
x
Class Exercise - 7
Two equal spheres A and B of masses 2
g and 30 g respectively lie at rest on a
smooth floor, so that the line joining
their centres is perpendicular to a fixed
vertical wall, sphere A being nearer to
the wall. A is projected towards B. Show
that A is brought to rest after its second
collision with B, if the coefficient of
restitution between the two spheres
3
and that between A and the wall is
.
5
Solution
B
A
Let A be projected with a velocity u.
Then 2u  2v A  30vB  u  v A  15vB .....(i)
Also, vB  v A 
3
u ........(ii)
5
Using (i) and (ii), we get
vB 
u
u
and v A  –
10
2
Now, ball A moves towards the wall and
returns back, say with a velocity VA’, then
VA 
3
3u
VA 
5
10
Solution
u
1
0
3
u
1
0
Now vB 
u
3u
, v A 
10
10
 2
3u 3u

 v A  15vB
10
2
3u 30u

 2v A  30vB
10 10
3u  15u  10v A  150vB
 10v A  150vB  18u


 3u u  3 2u 3u
Also, vB  v A  e 
  

10
10
5
10
25


Solving (iii) and (iv), we get
3u
vB 
and v A  0
25
Class Exercise - 8
A ball moving translation ally collides
elastically with another, stationary ball
of the same mass. At the moment of
impact the angle between the straight
line passing through the centers of
the balls and the direction of the
initial motion of the striking ball is
equal to a = 45°. Assuming the balls
to be smooth, find the fraction
 of the kinetic energy of the
striking ball that turned into potential
energy at the moment of the
maximum deformation.
Solution
Along X-axis,
mu cos = mv1x + mv2x
When maximum deformation takes place,
v1x = v2x
y
u cos = v1x + v2x = 2v1x
ucos 
 v1x  v 2x 
2
initial KE 
1
mu2
2
u

•
m
x
m
Solution
2
1
1 2
 u cos  
2
2
Final KE  mv1x
 mv2x
 mv1x
 m

2
2
 2 
mu2 cos2 

2
mu2 cos2 
Final KE
cos2 
4
Now  


1
Initial KE
2
mu2
2
cos2 45 1

  0.25
2
4
Class Exercise - 9
A ball moving with a speed of 9 m/s
strikes an identical stationary ball
such that after the collision, the
direction of each ball makes an
angle of 30° with the original line of
motion. Find the speeds of the two
balls after the collision if the kinetic
energy is conserved in the collision
process.
9 m/s
v3
30o
Stationary
ball
mp
30o
v2
Solution
Applying momentum of conservation
in x- and y-directions, we get
9  m  m  0  mv1 cos30  mv 2 cos30
 3
 9   v1  v2  
  v1  v2  6 3 ........(i)
2


Also, in y-direction,
0  mv1 sin30  mv 2 sin30
 v1  v 2 ......................(ii)
Using equations (i) and (ii), we get
Solution
2v1  2v 2  6 3 m/s
 v1  v 2  3 3 m/s
KEfinal 
1
1
mv12  mv 22  mv12   m  27  J
2
2
KEinitial 
1
81m
m  (9)2 
J
2
2
 KEinitial  KEfinal
Kinetic energy is not conserved.
Class Exercise - 10
Two blocks of masses m1 = 2 kg and m2
= 5 kg are moving in the same direction
along a frictionless surface with speeds
10 m/s and 3 m/s respectively, m2 being
ahead of m1. An ideal spring with K =
1,120 N/m is attached to the back side
of m2. Find the maximum compression
of the spring when the blocks collide.
10 m/s
m1=2 kg
3 m/s
m2=5 kg
Solution
10 m/s
3 m/s
m1=2 kg
m2=5 kg
Let V be the velocity of the blocks of the time of
maximum compression. Note that both the blocks
will move with equal speeds at the time of
maximum compression. Hence,
2  10  3  5  (2  5) v  v 
From energy conservation,
35
 5 m/s
7
Solution
1
1
1
1
 2  102   5  32   7  52   1120  x 2
2
2
2
2
 100 
45 175

 560 x 2
2
2
 100 – 65 = 560x2
1
1
x x 
 x   0.25 m
16
4
2
2
Thank you