CONSERVATION OF MOMENTUM

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CONSERVATION OF MOMENTUM
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• Two trolleys A and B are moving along a smooth
  horizontal runway.
• A is moving faster than B. It eventually collides
  with B.
• If the masses of the trolleys and their
  velocities before collision are known, can we
  predict the velocities vA and vB after collision
  ?

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• Two objects are moving along a straight line.
• The mass of A is mA while that of B is mB.
• Before collision, the velocity of A is uA while
  that of B is uB.
• During collision, the force exerted on A by B is
  FA while that exerted by B on A is FB. These are
  the only forces responsible for changing the
  velocity of the objects.
• After collision, the velocity of A is vA while
  that of B is vB.

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By Newtons 2nd Law,
where Dt is the collision time during which FA
and FB exist.
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By Newtons 3rd Law, the interaction forces
between A and B during collision form an
action-reaction pair. They are opposite in
directions, but equal in magnitude. FA FB
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FA FB
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FA FB
The object with smaller mass has an acceleration
of larger magnitude during collision.
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• mAvA mAuA is the change of momentum of A due
  to the collision.
• mBvB mBuB is the change of momentum of B due
  to the collision.
• The change of momentum of A and that of B has
  equal magnitude.
• The change of momentum of A and the change of
  momentum of B are in opposite direction.

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• mAuA mBuB is the total momentum of A and B
  before collision.
• mAvA mBvB is the total momentum of A and B
  after collision.
• The total momentum of A and B is not changed by
  the collision.
• The total momentum of A and B is conserved.

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• The Law of Conservation of Momentum is true if
  the only forces appearing during collision are
  the action and reaction appearing between A and
  B.
• The total momentum of A and B will not be
  conserved if there is a third object C exerting
  forces on A and B during collision. C may be a
  rough runway exerting friction on A and B. C can
  also be a wall against B obstructing its motion
  during collision.

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• Two trolleys A and B are moving along a smooth
  horizontal runway.
• Can we predict the velocities vA and vB after
  collision ?

By applying the Law of Conservation of
Momentum, 2 x 10 6 x 5 2 x vA 6 x vB There
is one equation with two unknowns. So it is
impossible to predict the velocities after
collision.
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• Two trolleys A and B are moving along a smooth
  horizontal runway.
• After collision, they stick together and move
  with a common velocity v. Can we predict v ?

By applying the Law of Conservation of
Momentum, 2 x 10 6 x 5 2 x v 6 x v v 6.25
m s-1
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Total K.E. before collision 0.5 x 2 x 102 0.5
x 6 x 52 100 J 75 J
Total K.E. after collision 0.5 x 2 x 6.252
0.5 x 6 x 6.252 39.0625 J 117.1875 J
There is a loss in K.E. of A but a gain in K.E.
of B.
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Total K.E. before collision 0.5 x 2 x 102 0.5
x 6 x 52 175 J
Total K.E. after collision 0.5 x 2 x 6.252
0.5 x 6 x 6.252 156.25 J
There is a loss in the total K.E. of A and B
175 156.25 18.75 J
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There is a loss in the total K.E. of A and B
175 156.25 18.75 J
The energy is lost as heat, sound, light, and
eventually becomes the internal energy of the
objects and the surroundings. The objects may
deform permanently and their temperatures and
that of the surroundings will increase.
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If the collision time Dt is 0.08 s, find the
average collision force on A during collision.
Find also the average collision force on B.
Average collision force on A 2 (6.25 10) /
0.08 -93.75 N
Average collision force on B 93.75 N
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Perfectly Inelastic Collision The two objects
stick and move together after collision.
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Perfectly Inelastic Collision The two objects
stick and move together after collision.
Total K.E. lost
gt 0
There must be a lost in total K.E. in perfectly
inelastic collision.
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Perfectly Inelastic Collision
By applying the Law of Conservation of
Momentum, 2 x 10 6 x (-2) (2 6) x v v 1
m s-1
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Perfectly Inelastic Collision
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Perfectly Inelastic Collision
By applying the Law of Conservation of
Momentum, 2 x 10 6 x (-5) (2 6) x v v
-1.25 m s-1
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Perfectly Inelastic Collision
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Perfectly Inelastic Collision
By applying the Law of Conservation of
Momentum, 2 x 15 6 x (-5) (2 6) x v v 0 m
s-1
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Perfectly Inelastic Collision
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Perfectly Inelastic Collision Practical setup in
the laboratory
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Elastic Collision

• Two trolleys A and B are moving along a smooth
  horizontal runway.
• Suppose the total K.E. of A and B does not change
  due to the collision
• Can we predict the velocities vA and vB after
  collision ?

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Elastic Collision
By applying the Law of Conservation of
Momentum, 2 x 10 6 x 5 2 x vA 6 x vB
Total K.E. is conserved, 0.5 x 2 x 102 0.5 x 6
x 52 0.5 x 2 x vA2 0.5 x 6 x vB2
Solution vA 2.5 m s-1 vB 7.5 m s-1
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Elastic Collision
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Elastic Collision
By applying the Law of Conservation of
Momentum, 2 x 10 4 x 0 2 x vA 4 x vB
Total K.E. is conserved, 0.5 x 2 x 102 0.5 x 4
x 02 0.5 x 2 x vA2 0.5 x 4 x vB2
Solution vA -3.33 m s-1 vB 6.67 m
s-1
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Elastic Collision
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Elastic Collision
By applying the Law of Conservation of
Momentum, 2 x 10 1 x 0 2 x vA 1 x vB
Total K.E. is conserved, 0.5 x 2 x 102 0.5 x 1
x 02 0.5 x 2 x vA2 0.5 x 1 x vB2
Solution vA 3.33 m s-1 vB 13.33 m
s-1
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Elastic Collision
By applying the Law of Conservation of
Momentum, 2 x 10 2 x 0 2 x vA 2 x vB
Total K.E. is conserved, 0.5 x 2 x 102 0.5 x 2
x 02 0.5 x 2 x vA2 0.5 x 2 x vB2
Solution vA 0 m s-1 vB 10 m s-1
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Elastic Collision
By applying the Law of Conservation of
Momentum, 2 x 10 2 x (-10) 2 x vA 2 x vB
Total K.E. is conserved, 0.5 x 2 x 102 0.5 x 2
x 102 0.5 x 2 x vA2 0.5 x 2 x vB2
Solution vA -10 m s-1 vB 10 m s-1
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Elastic Collision
By applying the Law of Conservation of
Momentum, 2 x 10 2 x (-4) 2 x vA 2 x vB
Total K.E. is conserved, 0.5 x 2 x 102 0.5 x 2
x 42 0.5 x 2 x vA2 0.5 x 2 x vB2
Solution vA -4 m s-1 vB 10 m s-1
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Elastic Collision
By applying the Law of Conservation of
Momentum, 2 x 10 2 x 4 2 x vA 2 x vB
Total K.E. is conserved, 0.5 x 2 x 102 0.5 x 2
x 42 0.5 x 2 x vA2 0.5 x 2 x vB2
Solution vA 4 m s-1 vB 10 m s-1
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Elastic Collision The total K.E. of A and B is
not changed by the collision.
By applying the Law of Conservation of Momentum,
mA uA mB uB mA vA mB vB
Total K.E. is conserved, 0.5mA uA2 0.5mB uB2
0.5mA vA2 0.5mB vB2
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Elastic Collision The total K.E. of A and B is
not changed by the collision.
Prove that it is impossible to have vA vB . It
is impossible for A and B moving together after
an Elastic collision.
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Elastic Collision Practical setup in the
laboratory
During collision, the spring is first being
compressed. Kinetic energy of the trolleys is
converted to elastic potential energy stored in
the compressed spring. At the last stage of the
collision, the spring extends again and the
elastic potential energy is converted back to the
kinetic energy of the trolleys.
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Elastic Collision Practical setup in the
laboratory
During collision, the interaction forces between
the trolleys are the repulsive magnetic forces
between the magnets. At the first stage of the
collision, the kinetic energy of the trolleys is
converted to magnetic field energy stored in the
magnetic field between the magnets. At the last
stage of the collision, the magnetic field energy
is converted back to the kinetic energy of the
trolleys. During collision, the trolleys do not
touch each other, resulting in no loss of energy
to sound, heat etc.
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• Inelastic Collision
• The objects separate after collision.
• The total K.E. of A and B decreases after
  collision. There is a loss of K.E. to heat, sound
  energy etc.

By applying the Law of Conservation of
Momentum, 2 x 10 6 x 5 2 x vA 6 x 7 vA 4
m s-1
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• Inelastic Collision
• The objects separate after collision.
• The total K.E. of A and B decreases after
  collision. There is a loss of K.E. to heat, sound
  energy etc.

Total K.E. lost due to collision 0.5(2)(102)
0.5(6)(52) - 0.5(2)(42) - 0.5(6)(72) 175 J
-163 J 12 J
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Possible Collisions Which of the following
collisions are possible ? Explain.
possible
possible
possible
impossible
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Possible Collisions Which of the following
collisions are possible ? Explain.
possible
impossible
impossible
impossible
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• Possible Collisions
• The total momentum of A and B is conserved. It is
  the same before and after collision.
• AND
• The total kinetic energy of A and B after
  collision is less than or equal to the total
  kinetic energy before collision.

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• Two objects stick together and are moving along a
  straight line.
• The mass of A is mA while that of B is mB.
• Before explsoion, the velocity of A and B is u.
• During explosion, the force exerted on A by B is
  FA while that exerted by B on A is FB. These are
  the only forces responsible for changing the
  velocity of the objects.
• After explosion, the velocity of A is vA while
  that of B is vB.

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By Newtons 2nd Law,
where Dt is the explosion time during which FA
and FB exist.
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By Newtons 3rd Law, the interaction forces
between A and B during explosion form an
action-reaction pair. They are opposite in
directions, but equal in magnitude. FA FB
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FA FB
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FA FB
The object with smaller mass has an acceleration
of larger magnitude during collision.
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• mAvA mAu is the change of momentum of A due to
  the explosion.
• mBvB mBu is the change of momentum of B due to
  the explosion.
• The change of momentum of A and that of B has
  equal magnitude.
• The change of momentum of A and the change of
  momentum of B are in opposite direction.

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• (mA mB )uB is the total momentum of A and B
  before explosion.
• mAvA mBvB is the total momentum of A and B
  after explosion.
• The total momentum of A and B is not changed by
  the explosion.
• The total momentum of A and B is conserved.

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• The Law of Conservation of Momentum is true if
  the only forces appearing during explosion are
  the action and reaction appearing between A and
  B.
• The total momentum of A and B will not be
  conserved if there is a third object C exerting
  forces on A and B during explosion. C may be a
  rough runway exerting friction on A and B. C can
  also be a wall against B obstructing its motion
  during explosion.

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If only mA , mB and u are known, vA and vB
cannot be determined. There is only one equation
but two unknowns.
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There is energy released during explosion which
becomes the kinetic energy of A and B. There is a
gain in the total kinetic of A and B due to the
explosion.
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Let the total kinetic energy gain be E.
gt 0
There must be a gain in total K.E. in explosion.
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By applying the Law of Conservation of
Momentum, (2 6) x 5 2 x vA 6 x 5.5 vA 3.5
m s-1
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Total K.E. before explosion 0.5 x 2 x 52 0.5
x 6 x 52 100 J
Total K.E. after explosion 0.5 x 2 x 3.52 0.5
x 6 x 5.52 103 J
The total K.E. gain due to explosion 3 J
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Average explosion force on A 2 (3.5 5) /
0.08 -37.5 N
Average explosion force on B 37.5 N
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By applying the Law of Conservation of
Momentum, (2 6) x 5 2 x vA 6 x 8 vA -4 m
s-1
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Total K.E. before explosion 0.5 x 2 x 52 0.5
x 6 x 52 100 J
Total K.E. after explosion 0.5 x 2 x 42 0.5 x
6 x 82 208 J
The total K.E. gain due to explosion 108 J
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Explosion Practical setup in the laboratory

• The two trolleys are held at rest with the
  compressed spring in between. Elastic potential
  energy is stored in the compressed spring.

• The two trolleys are released. The spring
  explodes. It stretches out and pushes the
  trolleys apart. The elastic potential energy is
  converted to the kinetic energy of the trolleys.

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Before explosion
After explosion
Conservation of momentum mA(-vA) mBvB 0 mAvA
mBvB
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After explosion
Conservation of momentum mAvA mBvB
Kinetic Energy K.E. gained by A EA
0.5mAvA2 K.E. gained by B EA 0.5mAvB2
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After explosion
The trolley with smaller mass gains more kinetic
energy. In the case of explosion between a gun
and a bullet, the gun recoils backward while the
bullet moves forward. Most of the energy released
becomes the kinetic energy of the bullet because
its mass is much smaller than that of the gun.
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Case 1
Before explosion
After explosion
Find vB and the elastic potential energy
originally stored in the compressed spring.
Ans 1 m s-1 , 12 J
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Case 2
Before explosion
After explosion
The same trolleys are used as in Case 1. The
same elastic spring is used and it is compressed
to the same extent as in Case 1. Find VB.
Ans 2 m s-1
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Case 2
Before explosion
After explosion
Total momentum of A and B before explosion 0
kg m s-1 Total momentum of A and B after
explosion gt 0 kg m s-1 Momentum is not conserved
in this case. Why ?
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Comparison of different types of collisions and
explosion
Total momentum is conserved only if there is no
friction and no obstruction along the line of
motion of the objects.
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Estimation of the speed of a bullet
A lump of plasticine of mass M is suspended at
rest by a light inextensible string. A bullet of
mass m is shot horizontally with speed u
towards the plasticine.
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Estimation of the speed of a bullet
The bullet is embedded into the plasticine and
they moves off horizontally with speed v after
collision. Assume the movement during collision
is negligible and the string remains vertical
after collision.
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Estimation of the speed of a bullet
The plasticine, with the bullet embedded, swings
upward through a vertical height h before
coming to rest instantaneously.
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Estimation of the speed of a bullet
Conservation of energy 0.5(M m) v2 (M m)
gh v2 2gh (2)
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Estimation of the speed of a bullet
The speed of the bullet u can be estimated by
measuring m, M and h.