Physics - XI - Work Power Energy - Collisions.ppt

Useful for iit preparation

1. WORK POWER ENERGY
WORK POWER
ENERGY

2. 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.
Collisionisanisolatedeventinwhichastrongforceactsbetweentwoor
morebodiesforashorttimeasaresultofwhichtheenergyand
momentumoftheinteractingparticlechange.

4. WORK POWER ENERGY
Types of collision :
(i) On the basis of conservation of kinetic energy.
Definitionofelasticcollision:
The collision in which both momentum and kinetic energy are
conserved is called elastic collision.

5. WORK POWER ENERGY
Perfectly elastic collision:
Ifinacollision,kineticenergyaftercollisionisequaltokineticenergy
beforecollision,thecollisionissaidtobeperfectlyelastic.
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.
Collisionbetweenabulletandablockofwoodintowhichitisfired
whenthebulletremainsembeddedintheblock.
Example :

8. WORK POWER ENERGY
(ii) On the basis of the direction of colliding bodies
Types of collision :
Inacollisionifthemotionofcollidingparticlesbeforeandafterthe
collisionisalongthesamelinethecollisionissaidtobeheadonorone
dimensional.
Example :
Collision of two gliders on an air track.
Head on or one dimensional collision:

9. WORK POWER ENERGY
Oblique collision:
Iftwoparticlecollisionis‘glancing’i.e.suchthattheirdirectionsof
motionaftercollisionarenotalongtheinitiallineofmotion,the
collisioniscalledoblique.
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)Thequantitiesremainingconstantinacollisionare.......
a) momentum, K.Eand temperature
b) momentum, K.Ebut not temperature.
c) momentum and temperature but not K.E
d) momentum but neither K.Enor
temperature

11. WORK POWER ENERGY
2)Inaninelasticcollision,thekineticenergyaftercollision......
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

12. WORK POWER ENERGY

13. WORK POWER ENERGY
One dimensional elastic collision:
Let two spheres of masses m
1and m
2moving with initial velocities u
1
and u
2in the same direction and they collide such that after collision
their final velocities are v
1and v
2respectively.
m
1
m
2
u
1 u
2
F
1F
2
m
1 m
2
v
1 v
2
before collisionduring
collision
after collision

14. WORK POWER ENERGY
Momentum of the system before collision = momentum of the system
after the collision.
m
1u
1+m
2u
2= m
1v
1+m
2v
2
m
1u
1-m
1v
1= m
2v
2-m
2u
2
m
1(u
1-v
1) = m
2(v
2-u
2) (1)
From the law of conservation of linear momentum.

15. WORK POWER ENERGY
K.Eof the system before collision = kinetic energy of the system after
collision
m
1u
1
2
1
2
+ m
2u
2
2
1
2
m
1v
1
2
1
2
+ m
2v
2
2
1
2
=
m
1u
1
2
1
2
- m
1v
1
2
1
2
m
2v
2
2
1
2
-m
2u
2
2
1
2
=
According to law of conservation of kinetic energy
m
1(u
1
2
-v
1
2
) = m
2(v
2
2
-u
2
2
) (2)

16. WORK POWER ENERGY
Dividing Eqn(2) by Eqn(1)
m
1(u
1
2
-v
1
2
)
m
1(u
1-v
1)
=
m
2(v
2
2
-u
2
2
)
m
2(v
2-u
2)
(u
1+v
1) (u
1-v
1)
(u
1-v
1)
=
(v
2+u
2)(v
2-u
2)
v
2-u
2
u
1+v
1 = v
2+u
2
u
1-u
2 = v
2-v
1 (3)

17. WORK POWER ENERGY
The relative velocity of approach before collision is equal to relative
velocity of separation after collision.
From eqn(3)
v
2 = u
1 -u
2+ v
1
(4)
Substitute eqn(4) in eqn(1)

18. WORK POWER ENERGY
m
1 (u
1-v
1) = m
2(u
1-u
2+ v
1-u
2)
m
1u
1 -m
1v
1= m
2u
1 -m
2u
2 + m
2v
1 -m
2u
2
m
1u
1-m
2u
1+ 2m
2u
2 = m
1v
1+ m
2v
1
u
1(m
1-m
2) + 2m
2u
2= v
1( m
1+ m
2)
v
1=
m
1-m
2
m
1+ m
2
u
1+
2m
2
m
1+ m
2
u
2

19. WORK POWER ENERGY
From eqn. (3)
v
1 = v
2 -u
1+ u
2
Substituting the value of v
1 in eqn. (1)
m
1(u
1-[v
2-u
1+u
2]) = m
2 (v
2-u
2)
m
1u
1-m
1v
2 + m
1u
1-m
1u
2= m
2v
2 -m
2u
2

20. WORK POWER ENERGY
2m
1u
1-m
1v
2-m
1u
2=m
2v
2-m
2u
2
2m
1u
1+m
2u
2-m
1u
2=m
2v
2+m
1v
2
2m
1u
1 + u
2 (m
2-m
1) = v
2 (m
1+m
2)
v
2
=
2m
1
m
1+m
2
u
1+
m
2-m
1
m
1+m
2
u
2

21. WORK POWER ENERGY
Case -I
If two bodies have equal massm
1= m
2= m
v
1=
m
1-m
2
m
1+m
2
u
1
2m
2
m
1+m
2
u
2+
Special cases:
v
1 = 0 + u
2
v
1 = u
2

22. WORK POWER ENERGY
v
1=
2m
1
m
1+m
2
u
1+
m
2-m
1
m
1+m
2
u
2
v
2 = u
1+0
v
2 = u
1
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 u
1and second body is at
rest (u
2=0).
m
1
m
2
u
1
m
1 m
2
v
2
u
2=0
before collisionafter collision
v
1=0 v
2= u
1
v
1=
m
1-m
2
m
1+m
2
2m
2
m
1+m
2
u
1 u
2+

24. WORK POWER ENERGY
v
1=
m
1-m
2
m
1+m
2
u
1+0
v
1=
m
1-m
2
m
1+m
2
u
1
v
2=
2m
1
m
1+m
2
u
1+
m
2-m
1
m
1+m
2
u
2

25. WORK POWER ENERGY
v
2=
2m
1
m
1+m
2
u
1+0
Note :
ifm
1= m
2
thenv
1=0
v
2 = u
1
v
2=
2m
1
m
1+m
2
u
1

26. WORK POWER ENERGY
Case -III
When a smaller body collides with a heavier body (m
2>>m
1) at rest u
2=0.
m
1
m
2
u
1
m
1
m
2
u
2=0
before collision
after collision
v
1-u
1v
20

27. WORK POWER ENERGY
v
1=
m
1-m
2
m
1+m
2
2m
2
m
1+m
2
u
1 u
2+
Here we can neglect mass m
1
v
1=
-m
2
m
2
u
1
v
1= -u
1
and v
2= 0

28. WORK POWER ENERGY
Case -IV
When a heavy body collides (m
1>>m
2) with a much lighter body at
rest u
2=0
Here we can neglect mass m
2
m
2
m
1
u
1 m
2
m
1
u
2=0
before collision after collision
v
22u
1v
1u
1

29. WORK POWER ENERGY
v
1=
m
1-m
2
m
1+m
2
2m
2
m
1+m
2
u
1 u
2+
v
1=
m
1
m
1
u
1
v
1= u
1
v
1=
2m
1
m
1+m
2
u
1+
m
2-m
1
m
1+m
2
u
2
v
2= 2u
1

30. WORK POWER ENERGY
A body of mass m
1collides head on with another body of mass m
2
at rest .The collision is perfectly elastic .
Then fraction of kinetic energy transferred by the first body to
second body is given by
4m
1m
2
(m
1+m
2)
2
Friction of kinetic energy retained by first body is
m
1-m
2
m
1+m
2
2

31. WORK POWER ENERGY
1)Aparticlemovingwithcertainvelocitycollides
elasticallywithanotherparticleatrest.Ifitwerehead
oncollision,theK.E.transferredbythecolliding
particleis100%whenitsmassisequalto.....
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)Aheavierspheremovingeastwardwithacertainvelocity‘v’
collideswithalightersphereatrest.Ifitisperfectlyelastic
headoncollision,thenaftercollision...........
a) Heavier sphere moves west ward with same speed
b) Heavier sphere comes to rest
c) Lighter sphere moves east ward with
velocity 2vapproximately
d) Lighter sphere moves eastward with
velocity approximately

33. WORK POWER ENERGY
3)Amassm
1moveswithagreatervelocity.Itstrikesanother
massm
2atrestinaheadoncollision.Itcomesbackalongits
pathwithlowspeedaftercollision.Then........
a) m
1> m
2
b) m
1< m
2
c) m
1= m
2
d) data insufficient

34. WORK POWER ENERGY

35. WORK POWER ENERGY
Perfectly inelastic collision
Considertwobodiesofmassesm
1andm
2movingwithvelocitiesu
1
andu
2(u
1>u
2)alongastraightlineundergoperfectinelastic
collision.
Lettheymovewithacommonvelocity‘v’afterthecollisionas
showninfigure.
m
1
u
1
m
2
u
2 v
Before collision
After collision

36. WORK POWER ENERGY
According to the law of conservation of linear momentum
m
1u
1+m
2u
2=m
1v
1+m
2v
2
After colllision
m
1u
1+m
2u
2= (m
1+m
2)v
v
1= v
2= v
Common velocity (v) =
m
1u
1+m
2u
2
m
1+m
2
initial total K.Eof the system =
1
2
m
1u
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+
1
2
m
2u
&#3627409360;
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Final total K.E of the system =
1
2
m
1+m
2v
2

37. WORK POWER ENERGY
Total loss of K.E in perfectly inelastic collision
KE =
1
2
m
1u
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+
1
2
m
2u
&#3627409360;
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-
1
2
(m
1+m
2)v
2
KE =
1
2
m
1m
2
m
1+m
2
(u
1-u
2)
2
If second body is at rest before collision then
u
2= 0 then K.E
2= 0
K=
1
2
m
1u
1
2
m
2
m
1+m
2
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??????
=
??????
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??????
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=
??????
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??????
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38. WORK POWER ENERGY
collisions in two dimensions:
v
1
x-axis
y-axis
v
2

1

2
mm
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 v
1and v
2be 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 = mv
1cos
1+mv
2cos
2
u = v
1cos
1 +v
2cos
2 (1)
And along y-axis gives
0 = v
1sin
1 -v
2sin 
2 (2)
Squaring and adding eqn. (1) and (2) we get
u
2
=v
1
2
+v
2
2
+2v
1v
2cos (
1+
2) (3)
As the collision is elastic, apply law of
conservation of K.E

40. WORK POWER ENERGY
1
2
1
2
mu
2
=mv
1
2
+
1
2
mv
2
2
u
2
=v
1
2
+v
2
2
(4)
Substituting eqn. (4) in eqn. (3)
v
1
2
+ v
2
2
= v
1
2
+ v
2
2
+ 2 v
1 v
2 cos(
1+
2)
2v
1v
2cos (
1+
2)=0
cos(
1+
2)=0

1+
2=90
0
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)Thecollisioninwhichtherelativevelocityiszero
aftercollisionis.........
a) perfectly elastic
b) perfectly inelastic
c) partially elastic
d) sometimes elastic and sometimes
inelastic

42. WORK POWER ENERGY
2)AbodyofmassmmovingwithaconstantvelocityVhits
anotherbodyofthesamemassmovingwiththesamevelocity
Vbutinoppositedirectionandstickstoit.Thevelocityofthe
compoundbodyafterthecollisionis............
a) 2V
b) V
c) V/2
d) zero

43. WORK POWER ENERGY
3)Whichofthefollowingisnotaperfectlyinelasticcollision?
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

44. WORK POWER ENERGY

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 =
v
2−v
1
u
1−u
2
=
Relativevelocityofseperation
Relativevelocityofapproach
v
2-v
1= e(u
1-u
2)
v
2= v
1+ e(u
1-u
2)(1)

46. WORK POWER ENERGY
From the law of conservation of linear momentum
m
1u
1+m
2u
2= m
1v
1+m
2v
2(2)
By solving (i) and (ii) we get
v
1=
m
1−em
2
m
1+m
2
u
1+
(1+e)m
2
m
1+m
2
u
2
similarly v
2=
(1+e)m
1
m
1+m
2
u
1+
m
2−em
1
m
1+m
2
u
2
By substituting e=1, we get value of v
1and v
2for 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 =
v
2−v
1
u
1−u
2
v
2-v
1= eu(1)
=
v
2−v
1
u−0
Momentum before collision = Momentum after collision
By conservation of momentum,

48. WORK POWER ENERGY
mu = mv
1+mv
2
v
1+v
2= u(ii)
Solving equation (i) and (ii) we get, v
1=
u
2
(1-e) and v
2=
u
2
(1+e)
v
1
v
2
=
1−e
1+e
Loss is kinetic energy:
Loss(K) = Total initial kinetic energy –Total final kinetic energy
=
1
2
m
1u
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+
1
2
m
2u
&#3627409360;
&#3627409360;
-
1
2
m
1v
&#3627409359;
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+
1
2
m
2v
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49. WORK POWER ENERGY
Substituting the value of v
1and v
2from the above expression
Loss (K)=
1
2
m
1m
2
m
1+m
2
(1-e
2
)(u
1-u
2)
2
Bysubstitutinge=1wegetK=0i.e.forperfectlyelasticcollisionloss
ofkineticenergywillbezeroorkineticenergyremainsconstant
beforeandaftercollision.

50. WORK POWER ENERGY
1)Inaninelasticcollision,thekineticenergyafter
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)Onespherecollideswithanothersphereofsamemassatrest
inelastically.Ifthevalueofcoefficientofrestitutionis½,the
ratiooftheirspeedsaftercollisionshallbe......
a) 1:2
b) 1:1
c) 1:3
d) 3:1
v
1
v
2
=
1−e
1+e

52. WORK POWER ENERGY

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 =
v
2−v
1
u
1−u
2
Definition:

54. WORK POWER ENERGY
Thevalueofcoefficientofrestitutionisindependentofmassesand
thevelocitiesofthecollidingbodies.Itdependsonthenatureof
collidingbodies.
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
Todeterminethecoefficientofrestitution
betweentwomaterials,Oneofthemistakenin
theformofaveryheavyplateandtheotherin
theformofasmallsphere.
The small sphere is dropped on to the plate from a height h. It hits
the plate with velocity u
1.
Let the sphere rebound to a height h
1after collision with the plate.
h
h
1
u
1
u
2= v
2= 0
Assuming the velocity
of the plate before and
after collision to be
zero (u
2=v
2=0)
v
1

56. WORK POWER ENERGY
e=
u
1-u
2
v
2-v
1
e =
−v
1
u
1
When the ball rebounds with a velocity v
1it reaches a height h
2,
so that
v
1= −2gh
1
But u
1= 2gh
-vesign indicates that v
1is in opposite direction to u
1

57. WORK POWER ENERGY
e=
u
1
-v
1
=
-(-2gh
1)
2gh
2gh
1
2gh
=e
e =
h
1
h
h
1= e
2
h = e
21
h
Height raised after n
th
bounce (h
n= e
2n
h)

58. WORK POWER ENERGY
Totaldistancetravelledbythespherebeforeitstopsbouncing:
Iftheballisdroppedfromheighthandtheballkeepsonbouncing
risingtoheighth
1,h
2,h
3…..eventuallycomingtorest.
The distance travelled by the ball before coming to rest is,
d = h+2h
1+2h
2+2h
3+……..
∵hn
=(e
2
)
n
h)
d = h+2 e
2
h + 2 e
4
h+2 e
6
h+…..
= h+2e
2
h [1+e
2
+e
4
+….]

59. WORK POWER ENERGY
=h+2e
2
h
1-e
2
1
= h
1-e
2
2e
2
1+
=h
1-e
2
1-e
2
+2e
2
d = h
1+e
2
1−e
2

60. WORK POWER ENERGY
Total time taken by the ball to stop bouncing:
T = t + 2t
1+ 2t
2+ 2t
3+ …….
T= ++ +
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+…..
T= ++ +
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??????
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+….
T=
&#3627409360;&#3627408417;
&#3627408416;
[1+2&#3627408518;
&#3627409360;
+2&#3627408518;
&#3627409362;
+ 2&#3627408518;
??????
+………]

61. WORK POWER ENERGY
T=
&#3627409360;&#3627408417;
&#3627408416;
1+2e [1+e
1
+e
2
+e
3
+……]
T=
&#3627409360;&#3627408417;
&#3627408416;
1+2e
1-e
1
T =
&#3627409360;&#3627408417;
&#3627408416;
1-e
2e
1+ =
&#3627409360;&#3627408417;
&#3627408416;
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-e
2
1+e
2
1+e
1-e
&#3627409360;&#3627408417;
&#3627408416;
(1+e
2
)
(1+e)
2
=
&#3627408416;&#3627408417;
&#3627409360;

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&#3627409360;&#3627408417;
&#3627408416;
(1-e)
(1+e)
=
&#3627408416;&#3627408417;
&#3627409360;

64. WORK POWER ENERGY
1)Thecoefficientofrestitution(e)foraperfectly
elasticcollisionis....
a) -1
b) 0
c) 
d) 1

65. WORK POWER ENERGY
2)Coefficientofrestitutiondependsupon,.........
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)Thecoefficientofrestitutionis......
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

67. WORK POWER ENERGY
Thank you…