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Elastic Collisions
Momentum and Kinetic Energy

An object in motion has a momentum based on its
mass and velocity.
• p = mv

The object also has kinetic energy.
• K = ½ mv2 = p2 / 2m
Kinetic Energy at Collision
m1

Energy is conserved only for
conservative forces.
v1i
v2i
m2
Before: K i  12 m1v12i  12 m2 v22i


Internal forces may be
nonconservative.
The force at the collision is
not always conservative.
energy lost
to heat
v1f
After:
v2f
K f  12 m1v12f  12 m2v22 f
Elastic Collision

For conservative forces the
energy is conserved.
Elastic

After the collision of contact
the potential energy is zero.

The total kinetic energy is
conserved – equal before
and after the collision.

This an elastic collision.
Pi  Pf
Ki  K f
Double Conservation


Elastic collisions conserve
both momentum and kinetic
energy.
Two equations govern all
elastic collisions.
m1
v1i
m1
m2
v2i
before




m1v1i  m2 v2i  m1v1 f  m2 v2 f
1
2
m1v12i  12 m2 v22i  12 m1v12f  12 m2 v22 f
m2
v1f
v2f
after
Head-on Collision

An elastic head-on collision
takes place in one
dimension.
v1i
v2i

If the collision is not headon, the force pair is in a
different direction.
v1i
v2i
m1
m2
force and velocity in a line
m1
m2
force and velocity on
different lines
Related Velocities
momentum in a line
solve for velocities
m1v1i  m1v1 f  m2v2 f  m2v2i
v1i  v1 f  v2 f  v2i
m1 (v1i  v1 f )  m2 (v2 f  v2i )
kinetic energy conservation
1
2
m1v12i  12 m1v12f  12 m1v22 f  12 m2 v22i
m1 (v12i  v12f )  m2 (v22 f  v22i )
m1 (v1i  v1 f )(v1i  v1 f )  m2 (v2 f  v2i )(v2 f  v2i )
v1i  v2i  v2 f  v1 f
v1i
v2i
m1
m2
Equal Masses

A 150 g ball moves at 1.4 m/s.
• The momentum is 0.21 kg m/s

v1i
m1
It strikes an equal mass ball at
rest.
•
•
•
•
v1i = 1.4 m/s
v2i = 0
Therefore, v1f = 0
and v2f = v1i
m2
m1
m2
v2f
momentum:
v1i  v2i  v2 f  v1 f
kinetic energy:
v1i  v2i  v2 f  v1 f
Striking a Heavy Mass


Let m1 << m2, when a golf
ball bounces off the floor.
The floor is at rest.
• v2i = 0

v1i
m1v1i  m2 v2 f  m1v1 f
kinetic energy:
v1i  v2 f  v1 f
The final velocity is equal
and opposite the initial
velocity
m1
momentum:
v1f
combined:
v1 f 
m1  m2
v1i  v1i
m1  m2
v2 f  0
Striking a Light Mass


Let m1 >> m2, when a car
strikes a ball.
The ball is at rest.
v1i  v2 f  v1 f
For a very heavy m1 , the
final velocity of m2 is twice
the initial velocity of m1 .
v2f
m2
v1i
m1v1i  m2 v2 f  m1v1 f
kinetic energy:
• v2i = 0

momentum:
combined:
m1  m2
v1 f 
v1i  v1i
m1  m2
v2 f  2v1i