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WORK POWER ENERGY
WORK POWER
ENERGY
WORK POWER ENERGY
WORK POWER ENERGY
COLLISIONS
 During collisions, there will be an exchange of momentum between
the bodies.
 The kinetic energy of the bodies may remain constant or change.
 Accordingly, the collisions are classified into two kinds, they are
1) Elastic collisions.
2) Inelastic collisions.
Collision is an isolated event in which a strong force acts between two or
more bodies for a short time as a result of which the energy and
momentum of the interacting particle change.
WORK POWER ENERGY
Types of collision :
(i) On the basis of conservation of kinetic energy.
Definition of elastic collision:
The collision in which both momentum and kinetic energy are
conserved is called elastic collision.
WORK POWER ENERGY
Perfectly elastic collision:
If in a collision, kinetic energy after collision is equal to kinetic energy
before collision, the collision is said to be perfectly elastic.
Examples :
(1) Collision between atomic particles.
(2) Bouncing of ball with same velocity after the collision with earth.
WORK POWER ENERGY
Inelastic collision:
The collision only momentum is conserved but not kinetic energy is
called inelastic collision.
Collision between two automobile on a road. In fact all majority of
collision belongs to this category.
Examples :
WORK POWER ENERGY
Perfectly inelastic collision:
If in a collision two bodies stick together or move with same velocity
after the collision, the collision is said to be perfectly inelastic.
Collision between a bullet and a block of wood into which it is fired
when the bullet remains embedded in the block.
Example :
WORK POWER ENERGY
(ii) On the basis of the direction of colliding bodies
Types of collision :
In a collision if the motion of colliding particles before and after the
collision is along the same line the collision is said to be head on or one
dimensional.
Example :
Collision of two gliders on an air track.
Head on or one dimensional collision:
WORK POWER ENERGY
Oblique collision:
 If two particle collision is ‘glancing’ i.e. such that their directions of
motion after collision are not along the initial line of motion, the
collision is called oblique.
 If in oblique collision the particles before and after collision are in
same plane, the collision is called 2-dimensional otherwise
3-dimensional.
Example :
Collision of billiard balls.
WORK POWER ENERGY
1) The quantities remaining constant in a collision are.......
a) momentum, K.E and temperature
b) momentum, K.E but not temperature.
c) momentum and temperature but not K.E
d) momentum but neither K.E nor
temperature
WORK POWER ENERGY
2) In an inelastic collision, the kinetic energy after collision......
a) is same as before collision
b) is always less than that before collision
c) is always greater than that before
collision
d) may be less or greater than that before
collision
WORK POWER ENERGY
WORK POWER ENERGY
One dimensional elastic collision:
Let two spheres of masses m1 and m2 moving with initial velocities u1
and u2 in the same direction and they collide such that after collision
their final velocities are v1 and v2 respectively.
m1 m2
u1 u2
F1 F2
m1 m2
v1 v2
before collision during
collision
after collision
WORK POWER ENERGY
Momentum of the system before collision = momentum of the system
after the collision.
m1u1+m2u2= m1v1+m2v2
m1u1-m1v1= m2v2-m2u2
m1(u1-v1) = m2(v2-u2) (1)
From the law of conservation of linear momentum.
WORK POWER ENERGY
K.E of the system before collision = kinetic energy of the system after
collision
m1u1
2
1
2
+ m2u2
2
1
2
m1v1
2
1
2
+ m2v2
2
1
2
=
m1u1
2
1
2
- m1v1
2
1
2
m2v2
2
1
2
- m2u2
2
1
2
=
According to law of conservation of kinetic energy
m1(u1
2- v1
2) = m2(v2
2- u2
2) (2)
WORK POWER ENERGY
Dividing Eqn (2) by Eqn (1)
m1(u1
2-v1
2)
m1(u1-v1)
=
m2(v2
2-u2
2)
m2(v2-u2)
(u1+v1) (u1-v1)
(u1-v1)
=
(v2+u2)(v2-u2)
v2-u2
u1+v1 = v2+u2
u1-u2 = v2-v1 (3)
WORK POWER ENERGY
The relative velocity of approach before collision is equal to relative
velocity of separation after collision.
From eqn (3)
v2 = u1 - u2+ v1
(4)
Substitute eqn (4) in eqn (1)
WORK POWER ENERGY
m1 (u1-v1) = m2 (u1- u2 + v1- u2)
m1u1 - m1v1 = m2u1 - m2u2 + m2v1 - m2u2
m1u1- m2u1 + 2m2u2 = m1v1 + m2v1
u1(m1-m2) + 2m2u2= v1 ( m1+ m2)
v1 =
m1- m2
m1+ m2
u1 +
2m2
m1+ m2
u2
WORK POWER ENERGY
From eqn. (3)
v1 = v2 - u1+ u2
Substituting the value of v1 in eqn. (1)
m1(u1-[v2-u1+u2]) = m2 (v2-u2)
m1u1- m1v2 + m1u1-m1u2 = m2v2 - m2u2
WORK POWER ENERGY
2m1u1-m1v2-m1u2=m2v2-m2u2
2m1u1+m2u2-m1u2=m2v2+m1v2
2m1u1 + u2 (m2-m1 ) = v2 (m1+m2 )
v2
=
2m1
m1+m2
u1 +
m2-m1
m1+m2
u2
WORK POWER ENERGY
Case - I
If two bodies have equal mass m1 = m2 = m
v1 =
m1- m2
m1+m2
u1
2m2
m1+m2
u2
+
Special cases:
v1 = 0 + u2
v1 = u2
WORK POWER ENERGY
v1 =
2m1
m1+m2
u1 +
m2-m1
m1+m2
u2
v2 = u1+0
v2 = u1
Hence, in case of head-on elastic collisions of two bodies of equal
masses, the bodies simply exchange their velocities after collisions.
WORK POWER ENERGY
Case - II
Before collision first body is moving with u1 and second body is at
rest (u2=0).
m1 m2
u1
m1 m2
v2
u2=0
before collision after collision
v1=0 v2= u1
v1 =
m1-m2
m1+m2
2m2
m1+m2
u1 u2
+
WORK POWER ENERGY
v1 =
m1-m2
m1+m2
u1 + 0
v1 =
m1-m2
m1+m2
u1
v2 =
2m1
m1+m2
u1 +
m2-m1
m1+m2
u2
WORK POWER ENERGY
v2 =
2m1
m1+m2
u1 + 0
Note :
if m1= m2
then v1=0
v2 = u1
v2 =
2m1
m1+m2
u1
WORK POWER ENERGY
Case - III
When a smaller body collides with a heavier body (m2>>m1) at rest u2=0.
m1
m2
u1 m1
m2
u2=0
before collision after collision
v1-u1 v20
WORK POWER ENERGY
v1 =
m1-m2
m1+m2
2m2
m1+m2
u1 u2
+
Here we can neglect mass m1
v1 =
-m2
m2
u1
v1 = -u1
and v2 = 0
WORK POWER ENERGY
Case - IV
When a heavy body collides (m1>>m2) with a much lighter body at
rest u2=0
Here we can neglect mass m2
m2
m1
u1 m2
m1
u2=0
before collision after collision
v22u1
v1  u1
WORK POWER ENERGY
v1 =
m1-m2
m1+m2
2m2
m1+m2
u1 u2
+
v1 =
m1
m1
u1
v1 = u1
v1 =
2m1
m1+m2
u1 +
m2-m1
m1+m2
u2
v2 = 2u1
WORK POWER ENERGY
 A body of mass m1 collides head on with another body of mass m2
at rest .The collision is perfectly elastic .
 Then fraction of kinetic energy transferred by the first body to
second body is given by
4m1m2
(m1+m2 )2
 Friction of kinetic energy retained by first body is
m1-m2
m1+m2
2
WORK POWER ENERGY
1) A particle moving with certain velocity collides
elastically with another particle at rest. If it were head
on collision, the K.E. transferred by the colliding
particle is 100% when its mass is equal to.....
a) The mass of the stationary particle
b) Twice the mass of the stationary particle
c) Half the mass of the stationary particle
d) ¼ times the mass of the stationary particle
WORK POWER ENERGY
2) A heavier sphere moving eastward with a certain velocity ‘v’
collides with a lighter sphere at rest. If it is perfectly elastic
head on collision, then after collision...........
a) Heavier sphere moves west ward with same speed
b) Heavier sphere comes to rest
c) Lighter sphere moves east ward with
velocity 2v approximately
d) Lighter sphere moves eastward with
velocity approximately
WORK POWER ENERGY
3) A mass m1 moves with a greater velocity. It strikes another
mass m2 at rest in a head on collision. It comes back along its
path with low speed after collision. Then........
a) m1 > m2
b) m1 < m2
c) m1 = m2
d) data insufficient
WORK POWER ENERGY
WORK POWER ENERGY
Perfectly inelastic collision
 Consider two bodies of masses m1 and m2 moving with velocities u1
and u2 (u1>u2) along a straight line under go perfect inelastic
collision.
 Let they move with a common velocity ‘v’ after the collision as
shown in figure.
m1
u1
m2
u2 v
Before collision After collision
WORK POWER ENERGY
According to the law of conservation of linear momentum
m1u1+m2u2=m1v1+m2v2
After colllision
m1u1+m2u2 = (m1+m2 )v
v1 = v2 = v
Common velocity (v) =
m1u1+m2u2
m1+m2
initial total K.E of the system =
1
2
m1u𝟏
𝟐
+
1
2
m2u𝟐
𝟐
Final total K.E of the system =
1
2
m1+m2 v2
WORK POWER ENERGY
Total loss of K.E in perfectly inelastic collision
KE =
1
2
m1u𝟏
𝟐
+
1
2
m2u𝟐
𝟐
-
1
2
(m1+m2)v2
KE =
1
2
m1m2
m1+m2
(u1-u2 )2
If second body is at rest before collision then
u 2= 0 then K.E2= 0
K=
1
2 m1u1
2
m2
m1+m2
∆𝑲
𝑲𝒊
=
𝒎𝟐
𝒎𝟏 + 𝒎𝟐
𝑲𝒇
𝑲𝒊
=
𝒎𝟏
𝒎𝟏 + 𝒎𝟐
WORK POWER ENERGY
collisions in two dimensions:
v1
x-axis
y-axis
v2
1
2
m
m
u
Consider two
bodies each of
mass m.
Before collision, Let u
be the initial velocity of
the first body and the
second body is at rest.
Let v1 and v2 be the
final velocities of the
bodies after collision.
Before collision After collision
WORK POWER ENERGY
In elastic collision momentum is conserved.
So conservation of momentum along x-axis gives
mu = mv1 cos1 + mv2 cos2
u = v1 cos 1 + v2 cos 2 (1)
And along y-axis gives
0 = v1 sin1 - v2 sin  2 (2)
Squaring and adding eqn. (1) and (2) we get
u2=v1
2+v2
2+2v1v2cos (1+2) (3)
As the collision is elastic, apply law of
conservation of K.E
WORK POWER ENERGY
1
2
1
2
mu2
= mv1
2 + 1
2
mv2
2
u2
= v1
2 + v2
2 (4)
Substituting eqn. (4) in eqn. (3)
v1
2 + v2
2 = v1
2 + v2
2 + 2 v1 v2 cos (1+2)
2v1v2cos (1+2)=0
cos (1+2)=0
1+2=900
Hence, two equal masses undergo oblique elastic collision will move at
right angles after collisions, if the second body is initially at rest.
WORK POWER ENERGY
1) The collision in which the relative velocity is zero
after collision is.........
a) perfectly elastic
b) perfectly inelastic
c) partially elastic
d) sometimes elastic and sometimes
inelastic
WORK POWER ENERGY
2) A body of mass m moving with a constant velocity V hits
another body of the same mass moving with the same velocity
V but in opposite direction and sticks to it. The velocity of the
compound body after the collision is............
a) 2V
b) V
c) V/2
d) zero
WORK POWER ENERGY
3) Which of the following is not a perfectly inelastic collision?
a) striking of two glass balls
b) bullet striking a bag of sand
c) an electron captured by a proton
d) a man jumping onto a moving cart
WORK POWER ENERGY
WORK POWER ENERGY
Head on inelastic collision:
Velocity after collision:
Let two bodies A and B collide inelastically and coefficient of
restitution is e.
e =
v2−v1
u1−u2
=
Relative velocity of seperation
Relative velocity of approach
 v2-v1 = e(u1-u2)
 v2 = v1+ e(u1-u2)(1)
WORK POWER ENERGY
From the law of conservation of linear momentum
m1u1+m2u2 = m1v1+m2v2(2)
By solving (i) and (ii) we get
v1 =
m1−em2
m1+m2
u1+
(1+e)m2
m1+m2
u2
similarly v2 =
(1+e)m1
m1+m2
u1+
m2−em1
m1+m2
u2
By substituting e=1, we get value of v1 and v2 for perfectly
elastic head on collision
WORK POWER ENERGY
Ratio of velocities after inelastic collision:
A sphere of mass moving with velocity u hits inelastically with another
stationary sphere of same mass.
e =
v2−v1
u1−u2
 v2-v1 = eu (1)
=
v2−v1
u−0
Momentum before collision = Momentum after collision
By conservation of momentum,
WORK POWER ENERGY
mu = mv1+mv2
 v1+v2 = u(ii)
 Solving equation (i) and (ii) we get, v1=
u
2
(1-e) and v2=
u
2
(1+e)
v1
v2
=
1−e
1+e
Loss is kinetic energy:
Loss(K) = Total initial kinetic energy – Total final kinetic energy
=
1
2
m1u𝟏
𝟐
+
1
2
m2u𝟐
𝟐
-
1
2
m1v𝟏
𝟐
+
1
2
m2v𝟐
𝟐
WORK POWER ENERGY
Substituting the value of v1 and v2 from the above expression
Loss (K)=
1
2
m1m2
m1+m2
(1-e2)(u1-u2)2
By substituting e=1 we get K=0 i.e. for perfectly elastic collision loss
of kinetic energy will be zero or kinetic energy remains constant
before and after collision.
WORK POWER ENERGY
1) In an inelastic collision, the kinetic energy after
collision.....
a) is same as before collision
b) is always less than before collision
c) is always greater than that before collision
d) may be less or greater than that before collision
WORK POWER ENERGY
2) One sphere collides with another sphere of same mass at rest
inelastically. If the value of coefficient of restitution is ½, the
ratio of their speeds after collision shall be......
a) 1:2
b) 1:1
c) 1:3
d) 3:1
v1
v2
=
1−e
1+e
WORK POWER ENERGY
WORK POWER ENERGY
Coefficient of restitution:
The coefficient of restitution between two bodies in a collision is defined
as the ratio of the relative velocity of separation after collision to the
relative velocity of approach before collision.
e =
relative velocity of separation
relative velocity of approach
e =
v2−v1
u1−u2
Definition:
WORK POWER ENERGY
 The value of coefficient of restitution is independent of masses and
the velocities of the colliding bodies. It depends on the nature of
colliding bodies.
 For perfectly elastic collision, e = 1
 For perfectly inelastic collision, e = 0
 For inelastic collisions, the value of e lies between 0 to 1.
WORK POWER ENERGY
Determination of coefficient of restitution
 To determine the coefficient of restitution
between two materials, One of them is taken in
the form of a very heavy plate and the other in
the form of a small sphere.
 The small sphere is dropped on to the plate from a height h. It hits
the plate with velocity u1.
 Let the sphere rebound to a height h1 after collision with the plate.
h
h1
u1
u2 = v2 = 0
Assuming the velocity
of the plate before and
after collision to be
zero (u2=v2=0)
v1
WORK POWER ENERGY
e =
u1-u2
v2-v1
e =
−v1
u1
When the ball rebounds with a velocity v1 it reaches a height h2,
so that
v1 = − 2gh1
But u1 = 2gh
-ve sign indicates that v1 is in opposite direction to u1
WORK POWER ENERGY
e =
u1
-v1
=
- (- 2gh1 )
2gh
2gh1
2gh
=
e
e =
h1
h
 h1 = e2 h = e21 h
Height raised after nth bounce (hn = e2n h)
WORK POWER ENERGY
Total distance travelled by the sphere before it stops bouncing:
If the ball is dropped from height h and the ball keeps on bouncing
rising to height h1, h2, h3….. eventually coming to rest.
The distance travelled by the ball before coming to rest is,
d = h+2h1+2h2+2h3 +……..
∵ hn = (e2)nh)
d = h+2 e2 h + 2 e 4 h+2 e6 h+…..
= h+2e2h [1+e2+e4+…. ]
WORK POWER ENERGY
=h+2e2h
1-e2
1
= h
1-e2
2e2
1+
=h
1-e2
1-e2+2e2
 d = h
1+e2
1−e2
WORK POWER ENERGY
Total time taken by the ball to stop bouncing:
T = t + 2t 1+ 2t2+ 2t3 + …….
T= +
+ +
𝟐𝐡
𝐠
𝟐
𝟐𝐡𝟏
𝐠
𝟐
𝟐𝐡𝟐
𝐠
𝟐
𝟐𝐡𝟑
𝐠
+…..
T= +
+ +
𝟐𝐡
𝐠
𝟐
𝟐𝐞𝟐𝐡
𝐠
𝟐
𝟐𝐞𝟒𝐡
𝐠
𝟐
𝟐𝐞𝟔𝐡
𝐠
+….
T=
𝟐𝐡
𝐠
[1+2 𝒆𝟐 +2 𝒆𝟒 + 2 𝒆𝟔 +………]
WORK POWER ENERGY
T=
𝟐𝐡
𝐠
1+2e [1+e1 +e2+e3 +……]
T=
𝟐𝐡
𝐠
1+2e
1-e
1
T =
𝟐𝐡
𝐠 1-e
2e
1+ = 𝟐𝐡
𝐠
1-e+2e
1-e
T =
2h
g
1+e
1−e
WORK POWER ENERGY
Average speed of the ball during its entire journey:
Average speed =
Total time taken
Total distance travelled
=
h
1-e2
1+e2
1+e
1-e
𝟐𝐡
𝐠
(1+e2)
(1+e)2
=
𝐠𝐡
𝟐
WORK POWER ENERGY
Average velocity of the ball during its entire journey:
Average velocity =
Total time taken
Net displacement
=
h
1-e
1+e
𝟐𝐡
𝐠
(1-e)
(1+e)
=
𝐠𝐡
𝟐
WORK POWER ENERGY
1) The coefficient of restitution (e) for a perfectly
elastic collision is....
a) -1
b) 0
c) 
d) 1
WORK POWER ENERGY
2) Coefficient of restitution depends upon, .........
a) The relative velocities of approach and separation
b) The masses of the colliding bodies
c) The materials of the colliding bodies
d) all the above
WORK POWER ENERGY
3) The coefficient of restitution is......
a) a number which varies from -1 to 1
b) a number which varies from 0 to 1
c) a number which varies from 0 to -1
d) a positive number
WORK POWER ENERGY
Thank you…

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Physics - XI - Work Power Energy - Collisions.ppt

  • 1. WORK POWER ENERGY WORK POWER ENERGY
  • 3. WORK POWER ENERGY COLLISIONS  During collisions, there will be an exchange of momentum between the bodies.  The kinetic energy of the bodies may remain constant or change.  Accordingly, the collisions are classified into two kinds, they are 1) Elastic collisions. 2) Inelastic collisions. Collision is an isolated event in which a strong force acts between two or more bodies for a short time as a result of which the energy and momentum of the interacting particle change.
  • 4. WORK POWER ENERGY Types of collision : (i) On the basis of conservation of kinetic energy. Definition of elastic collision: The collision in which both momentum and kinetic energy are conserved is called elastic collision.
  • 5. WORK POWER ENERGY Perfectly elastic collision: If in a collision, kinetic energy after collision is equal to kinetic energy before collision, the collision is said to be perfectly elastic. Examples : (1) Collision between atomic particles. (2) Bouncing of ball with same velocity after the collision with earth.
  • 6. WORK POWER ENERGY Inelastic collision: The collision only momentum is conserved but not kinetic energy is called inelastic collision. Collision between two automobile on a road. In fact all majority of collision belongs to this category. Examples :
  • 7. WORK POWER ENERGY Perfectly inelastic collision: If in a collision two bodies stick together or move with same velocity after the collision, the collision is said to be perfectly inelastic. Collision between a bullet and a block of wood into which it is fired when the bullet remains embedded in the block. Example :
  • 8. WORK POWER ENERGY (ii) On the basis of the direction of colliding bodies Types of collision : In a collision if the motion of colliding particles before and after the collision is along the same line the collision is said to be head on or one dimensional. Example : Collision of two gliders on an air track. Head on or one dimensional collision:
  • 9. WORK POWER ENERGY Oblique collision:  If two particle collision is ‘glancing’ i.e. such that their directions of motion after collision are not along the initial line of motion, the collision is called oblique.  If in oblique collision the particles before and after collision are in same plane, the collision is called 2-dimensional otherwise 3-dimensional. Example : Collision of billiard balls.
  • 10. WORK POWER ENERGY 1) The quantities remaining constant in a collision are....... a) momentum, K.E and temperature b) momentum, K.E but not temperature. c) momentum and temperature but not K.E d) momentum but neither K.E nor temperature
  • 11. WORK POWER ENERGY 2) In an inelastic collision, the kinetic energy after collision...... a) is same as before collision b) is always less than that before collision c) is always greater than that before collision d) may be less or greater than that before collision
  • 13. WORK POWER ENERGY One dimensional elastic collision: Let two spheres of masses m1 and m2 moving with initial velocities u1 and u2 in the same direction and they collide such that after collision their final velocities are v1 and v2 respectively. m1 m2 u1 u2 F1 F2 m1 m2 v1 v2 before collision during collision after collision
  • 14. WORK POWER ENERGY Momentum of the system before collision = momentum of the system after the collision. m1u1+m2u2= m1v1+m2v2 m1u1-m1v1= m2v2-m2u2 m1(u1-v1) = m2(v2-u2) (1) From the law of conservation of linear momentum.
  • 15. WORK POWER ENERGY K.E of the system before collision = kinetic energy of the system after collision m1u1 2 1 2 + m2u2 2 1 2 m1v1 2 1 2 + m2v2 2 1 2 = m1u1 2 1 2 - m1v1 2 1 2 m2v2 2 1 2 - m2u2 2 1 2 = According to law of conservation of kinetic energy m1(u1 2- v1 2) = m2(v2 2- u2 2) (2)
  • 16. WORK POWER ENERGY Dividing Eqn (2) by Eqn (1) m1(u1 2-v1 2) m1(u1-v1) = m2(v2 2-u2 2) m2(v2-u2) (u1+v1) (u1-v1) (u1-v1) = (v2+u2)(v2-u2) v2-u2 u1+v1 = v2+u2 u1-u2 = v2-v1 (3)
  • 17. WORK POWER ENERGY The relative velocity of approach before collision is equal to relative velocity of separation after collision. From eqn (3) v2 = u1 - u2+ v1 (4) Substitute eqn (4) in eqn (1)
  • 18. WORK POWER ENERGY m1 (u1-v1) = m2 (u1- u2 + v1- u2) m1u1 - m1v1 = m2u1 - m2u2 + m2v1 - m2u2 m1u1- m2u1 + 2m2u2 = m1v1 + m2v1 u1(m1-m2) + 2m2u2= v1 ( m1+ m2) v1 = m1- m2 m1+ m2 u1 + 2m2 m1+ m2 u2
  • 19. WORK POWER ENERGY From eqn. (3) v1 = v2 - u1+ u2 Substituting the value of v1 in eqn. (1) m1(u1-[v2-u1+u2]) = m2 (v2-u2) m1u1- m1v2 + m1u1-m1u2 = m2v2 - m2u2
  • 20. WORK POWER ENERGY 2m1u1-m1v2-m1u2=m2v2-m2u2 2m1u1+m2u2-m1u2=m2v2+m1v2 2m1u1 + u2 (m2-m1 ) = v2 (m1+m2 ) v2 = 2m1 m1+m2 u1 + m2-m1 m1+m2 u2
  • 21. WORK POWER ENERGY Case - I If two bodies have equal mass m1 = m2 = m v1 = m1- m2 m1+m2 u1 2m2 m1+m2 u2 + Special cases: v1 = 0 + u2 v1 = u2
  • 22. WORK POWER ENERGY v1 = 2m1 m1+m2 u1 + m2-m1 m1+m2 u2 v2 = u1+0 v2 = u1 Hence, in case of head-on elastic collisions of two bodies of equal masses, the bodies simply exchange their velocities after collisions.
  • 23. WORK POWER ENERGY Case - II Before collision first body is moving with u1 and second body is at rest (u2=0). m1 m2 u1 m1 m2 v2 u2=0 before collision after collision v1=0 v2= u1 v1 = m1-m2 m1+m2 2m2 m1+m2 u1 u2 +
  • 24. WORK POWER ENERGY v1 = m1-m2 m1+m2 u1 + 0 v1 = m1-m2 m1+m2 u1 v2 = 2m1 m1+m2 u1 + m2-m1 m1+m2 u2
  • 25. WORK POWER ENERGY v2 = 2m1 m1+m2 u1 + 0 Note : if m1= m2 then v1=0 v2 = u1 v2 = 2m1 m1+m2 u1
  • 26. WORK POWER ENERGY Case - III When a smaller body collides with a heavier body (m2>>m1) at rest u2=0. m1 m2 u1 m1 m2 u2=0 before collision after collision v1-u1 v20
  • 27. WORK POWER ENERGY v1 = m1-m2 m1+m2 2m2 m1+m2 u1 u2 + Here we can neglect mass m1 v1 = -m2 m2 u1 v1 = -u1 and v2 = 0
  • 28. WORK POWER ENERGY Case - IV When a heavy body collides (m1>>m2) with a much lighter body at rest u2=0 Here we can neglect mass m2 m2 m1 u1 m2 m1 u2=0 before collision after collision v22u1 v1  u1
  • 29. WORK POWER ENERGY v1 = m1-m2 m1+m2 2m2 m1+m2 u1 u2 + v1 = m1 m1 u1 v1 = u1 v1 = 2m1 m1+m2 u1 + m2-m1 m1+m2 u2 v2 = 2u1
  • 30. WORK POWER ENERGY  A body of mass m1 collides head on with another body of mass m2 at rest .The collision is perfectly elastic .  Then fraction of kinetic energy transferred by the first body to second body is given by 4m1m2 (m1+m2 )2  Friction of kinetic energy retained by first body is m1-m2 m1+m2 2
  • 31. WORK POWER ENERGY 1) A particle moving with certain velocity collides elastically with another particle at rest. If it were head on collision, the K.E. transferred by the colliding particle is 100% when its mass is equal to..... a) The mass of the stationary particle b) Twice the mass of the stationary particle c) Half the mass of the stationary particle d) ¼ times the mass of the stationary particle
  • 32. WORK POWER ENERGY 2) A heavier sphere moving eastward with a certain velocity ‘v’ collides with a lighter sphere at rest. If it is perfectly elastic head on collision, then after collision........... a) Heavier sphere moves west ward with same speed b) Heavier sphere comes to rest c) Lighter sphere moves east ward with velocity 2v approximately d) Lighter sphere moves eastward with velocity approximately
  • 33. WORK POWER ENERGY 3) A mass m1 moves with a greater velocity. It strikes another mass m2 at rest in a head on collision. It comes back along its path with low speed after collision. Then........ a) m1 > m2 b) m1 < m2 c) m1 = m2 d) data insufficient
  • 35. WORK POWER ENERGY Perfectly inelastic collision  Consider two bodies of masses m1 and m2 moving with velocities u1 and u2 (u1>u2) along a straight line under go perfect inelastic collision.  Let they move with a common velocity ‘v’ after the collision as shown in figure. m1 u1 m2 u2 v Before collision After collision
  • 36. WORK POWER ENERGY According to the law of conservation of linear momentum m1u1+m2u2=m1v1+m2v2 After colllision m1u1+m2u2 = (m1+m2 )v v1 = v2 = v Common velocity (v) = m1u1+m2u2 m1+m2 initial total K.E of the system = 1 2 m1u𝟏 𝟐 + 1 2 m2u𝟐 𝟐 Final total K.E of the system = 1 2 m1+m2 v2
  • 37. WORK POWER ENERGY Total loss of K.E in perfectly inelastic collision KE = 1 2 m1u𝟏 𝟐 + 1 2 m2u𝟐 𝟐 - 1 2 (m1+m2)v2 KE = 1 2 m1m2 m1+m2 (u1-u2 )2 If second body is at rest before collision then u 2= 0 then K.E2= 0 K= 1 2 m1u1 2 m2 m1+m2 ∆𝑲 𝑲𝒊 = 𝒎𝟐 𝒎𝟏 + 𝒎𝟐 𝑲𝒇 𝑲𝒊 = 𝒎𝟏 𝒎𝟏 + 𝒎𝟐
  • 38. WORK POWER ENERGY collisions in two dimensions: v1 x-axis y-axis v2 1 2 m m u Consider two bodies each of mass m. Before collision, Let u be the initial velocity of the first body and the second body is at rest. Let v1 and v2 be the final velocities of the bodies after collision. Before collision After collision
  • 39. WORK POWER ENERGY In elastic collision momentum is conserved. So conservation of momentum along x-axis gives mu = mv1 cos1 + mv2 cos2 u = v1 cos 1 + v2 cos 2 (1) And along y-axis gives 0 = v1 sin1 - v2 sin  2 (2) Squaring and adding eqn. (1) and (2) we get u2=v1 2+v2 2+2v1v2cos (1+2) (3) As the collision is elastic, apply law of conservation of K.E
  • 40. WORK POWER ENERGY 1 2 1 2 mu2 = mv1 2 + 1 2 mv2 2 u2 = v1 2 + v2 2 (4) Substituting eqn. (4) in eqn. (3) v1 2 + v2 2 = v1 2 + v2 2 + 2 v1 v2 cos (1+2) 2v1v2cos (1+2)=0 cos (1+2)=0 1+2=900 Hence, two equal masses undergo oblique elastic collision will move at right angles after collisions, if the second body is initially at rest.
  • 41. WORK POWER ENERGY 1) The collision in which the relative velocity is zero after collision is......... a) perfectly elastic b) perfectly inelastic c) partially elastic d) sometimes elastic and sometimes inelastic
  • 42. WORK POWER ENERGY 2) A body of mass m moving with a constant velocity V hits another body of the same mass moving with the same velocity V but in opposite direction and sticks to it. The velocity of the compound body after the collision is............ a) 2V b) V c) V/2 d) zero
  • 43. WORK POWER ENERGY 3) Which of the following is not a perfectly inelastic collision? a) striking of two glass balls b) bullet striking a bag of sand c) an electron captured by a proton d) a man jumping onto a moving cart
  • 45. WORK POWER ENERGY Head on inelastic collision: Velocity after collision: Let two bodies A and B collide inelastically and coefficient of restitution is e. e = v2−v1 u1−u2 = Relative velocity of seperation Relative velocity of approach  v2-v1 = e(u1-u2)  v2 = v1+ e(u1-u2)(1)
  • 46. WORK POWER ENERGY From the law of conservation of linear momentum m1u1+m2u2 = m1v1+m2v2(2) By solving (i) and (ii) we get v1 = m1−em2 m1+m2 u1+ (1+e)m2 m1+m2 u2 similarly v2 = (1+e)m1 m1+m2 u1+ m2−em1 m1+m2 u2 By substituting e=1, we get value of v1 and v2 for perfectly elastic head on collision
  • 47. WORK POWER ENERGY Ratio of velocities after inelastic collision: A sphere of mass moving with velocity u hits inelastically with another stationary sphere of same mass. e = v2−v1 u1−u2  v2-v1 = eu (1) = v2−v1 u−0 Momentum before collision = Momentum after collision By conservation of momentum,
  • 48. WORK POWER ENERGY mu = mv1+mv2  v1+v2 = u(ii)  Solving equation (i) and (ii) we get, v1= u 2 (1-e) and v2= u 2 (1+e) v1 v2 = 1−e 1+e Loss is kinetic energy: Loss(K) = Total initial kinetic energy – Total final kinetic energy = 1 2 m1u𝟏 𝟐 + 1 2 m2u𝟐 𝟐 - 1 2 m1v𝟏 𝟐 + 1 2 m2v𝟐 𝟐
  • 49. WORK POWER ENERGY Substituting the value of v1 and v2 from the above expression Loss (K)= 1 2 m1m2 m1+m2 (1-e2)(u1-u2)2 By substituting e=1 we get K=0 i.e. for perfectly elastic collision loss of kinetic energy will be zero or kinetic energy remains constant before and after collision.
  • 50. WORK POWER ENERGY 1) In an inelastic collision, the kinetic energy after collision..... a) is same as before collision b) is always less than before collision c) is always greater than that before collision d) may be less or greater than that before collision
  • 51. WORK POWER ENERGY 2) One sphere collides with another sphere of same mass at rest inelastically. If the value of coefficient of restitution is ½, the ratio of their speeds after collision shall be...... a) 1:2 b) 1:1 c) 1:3 d) 3:1 v1 v2 = 1−e 1+e
  • 53. WORK POWER ENERGY Coefficient of restitution: The coefficient of restitution between two bodies in a collision is defined as the ratio of the relative velocity of separation after collision to the relative velocity of approach before collision. e = relative velocity of separation relative velocity of approach e = v2−v1 u1−u2 Definition:
  • 54. WORK POWER ENERGY  The value of coefficient of restitution is independent of masses and the velocities of the colliding bodies. It depends on the nature of colliding bodies.  For perfectly elastic collision, e = 1  For perfectly inelastic collision, e = 0  For inelastic collisions, the value of e lies between 0 to 1.
  • 55. WORK POWER ENERGY Determination of coefficient of restitution  To determine the coefficient of restitution between two materials, One of them is taken in the form of a very heavy plate and the other in the form of a small sphere.  The small sphere is dropped on to the plate from a height h. It hits the plate with velocity u1.  Let the sphere rebound to a height h1 after collision with the plate. h h1 u1 u2 = v2 = 0 Assuming the velocity of the plate before and after collision to be zero (u2=v2=0) v1
  • 56. WORK POWER ENERGY e = u1-u2 v2-v1 e = −v1 u1 When the ball rebounds with a velocity v1 it reaches a height h2, so that v1 = − 2gh1 But u1 = 2gh -ve sign indicates that v1 is in opposite direction to u1
  • 57. WORK POWER ENERGY e = u1 -v1 = - (- 2gh1 ) 2gh 2gh1 2gh = e e = h1 h  h1 = e2 h = e21 h Height raised after nth bounce (hn = e2n h)
  • 58. WORK POWER ENERGY Total distance travelled by the sphere before it stops bouncing: If the ball is dropped from height h and the ball keeps on bouncing rising to height h1, h2, h3….. eventually coming to rest. The distance travelled by the ball before coming to rest is, d = h+2h1+2h2+2h3 +…….. ∵ hn = (e2)nh) d = h+2 e2 h + 2 e 4 h+2 e6 h+….. = h+2e2h [1+e2+e4+…. ]
  • 59. WORK POWER ENERGY =h+2e2h 1-e2 1 = h 1-e2 2e2 1+ =h 1-e2 1-e2+2e2  d = h 1+e2 1−e2
  • 60. WORK POWER ENERGY Total time taken by the ball to stop bouncing: T = t + 2t 1+ 2t2+ 2t3 + ……. T= + + + 𝟐𝐡 𝐠 𝟐 𝟐𝐡𝟏 𝐠 𝟐 𝟐𝐡𝟐 𝐠 𝟐 𝟐𝐡𝟑 𝐠 +….. T= + + + 𝟐𝐡 𝐠 𝟐 𝟐𝐞𝟐𝐡 𝐠 𝟐 𝟐𝐞𝟒𝐡 𝐠 𝟐 𝟐𝐞𝟔𝐡 𝐠 +…. T= 𝟐𝐡 𝐠 [1+2 𝒆𝟐 +2 𝒆𝟒 + 2 𝒆𝟔 +………]
  • 61. WORK POWER ENERGY T= 𝟐𝐡 𝐠 1+2e [1+e1 +e2+e3 +……] T= 𝟐𝐡 𝐠 1+2e 1-e 1 T = 𝟐𝐡 𝐠 1-e 2e 1+ = 𝟐𝐡 𝐠 1-e+2e 1-e T = 2h g 1+e 1−e
  • 62. WORK POWER ENERGY Average speed of the ball during its entire journey: Average speed = Total time taken Total distance travelled = h 1-e2 1+e2 1+e 1-e 𝟐𝐡 𝐠 (1+e2) (1+e)2 = 𝐠𝐡 𝟐
  • 63. WORK POWER ENERGY Average velocity of the ball during its entire journey: Average velocity = Total time taken Net displacement = h 1-e 1+e 𝟐𝐡 𝐠 (1-e) (1+e) = 𝐠𝐡 𝟐
  • 64. WORK POWER ENERGY 1) The coefficient of restitution (e) for a perfectly elastic collision is.... a) -1 b) 0 c)  d) 1
  • 65. WORK POWER ENERGY 2) Coefficient of restitution depends upon, ......... a) The relative velocities of approach and separation b) The masses of the colliding bodies c) The materials of the colliding bodies d) all the above
  • 66. WORK POWER ENERGY 3) The coefficient of restitution is...... a) a number which varies from -1 to 1 b) a number which varies from 0 to 1 c) a number which varies from 0 to -1 d) a positive number