Transcript Momentum and Impulse

Principles of Physics
- property of an object related to its mass and
velocity.
- “mass in motion” or “inertia in motion”
p = momentum (vector)
p = mv
m = mass (kg)
v = velocity (m/s) (vector)
units for momentum are kg m/s
**If an object is not moving, it has no momentum
Momentum of an individual object
p = mv
Example: Calculate the momentum of a 250
kg cart with a velocity of 25 m/s.
p = mv
p = 250 kg (25 m/s)
p = 6250 kg m/s
Momentum of a system
p = m 1 v 1 + m 2 v2 + m 3 v3 + …
Example: A 300 kg car is travelling to east at
45 m/s and a 500 kg truck is travelling
west at 30 m/s. Calculate the total
momentum of the system.
p = m 1 v 1 + m 2 v2
p = 300 kg(45 m/s) + 500 kg(-30 m/s)
p = -1500 kg m/s

Conservation – to keep constant even though
changes occur
◦ whatever you had before an event you will still have
after the event
The total momentum of a system is the same
before and after a collision
ptotal before = ptotal after
ptotal before = ptotal after
ptotal before= m1v1i + m2v2i + m3v3i + …
**remember v has direction
ptotal after = m1v1f + m2v2f + m3v3f + …
Newton’s Cradle Demo:
Elastic - bounce
◦ objects hit and bounce off from each other
Inelastic – stick
◦ multiple objects hit and stick together
or
◦ one objects separates into 2 or more (explosion)
Example 1
A cue ball with mass 0.050 kg moves at a velocity of
0.20 m/s. It collides with the 8 ball (0.050 kg)
moving in the same direction with velocity 0.10 m/s.
After the collision, the cue ball is moving with a
velocity of 0.08 m/s in the same direction. What is
the 8 ball’s final velocity?
Write down given information
Givens:
m1 = 0.050 kg
m2 = 0.050 kg
v1i= 0.20 m/s
v2i= 0.10 m/s
v1f = 0.08 m/s
v2f = ?
Example 1
A cue ball with mass 0.050 kg moves at a velocity of
0.20 m/s. It collides with the 8 ball (0.050 kg)
moving in the same direction with velocity 0.10 m/s.
After the collision, the cue ball is moving with a
velocity of 0.08 m/s in the same direction. What is
the 8 ball’s final velocity?
Givens:
m1 = 0.050 kg
m2 = 0.050 kg
v1i= 0.20 m/s
v2i= 0.10 m/s
v1f = 0.08 m/s
v2f = ?
Determine the momentum of the cue ball
before and after the collision
p1i = m1v1i
= 0.050 kg(0.20 m/s)
= 0.01 kg m/s
p1f = m1v1f
= 0.050 kg(0.08 m/s)
= 0.004 kg m/s
Example 1
A cue ball with mass 0.050 kg moves at a velocity of
0.20 m/s. It collides with the 8 ball (0.050 kg)
moving in the same direction with velocity 0.10 m/s.
After the collision, the cue ball is moving with a
velocity of 0.08 m/s in the same direction. What is
the 8 ball’s final velocity?
Givens:
m1 = 0.050 kg
m2 = 0.050 kg
v1i= 0.20 m/s
v2i= 0.10 m/s
v1f = 0.08 m/s
v2f = ?
Determine the momentum of the 8 ball before
the collision
p2i = m2v2i
= 0.050 kg(0.10 m/s)
= 0.005 kg m/s
Example 1
A cue ball with mass 0.050 kg moves at a velocity of
0.20 m/s. It collides with the 8 ball (0.050 kg)
moving in the same direction with velocity 0.10 m/s.
After the collision, the cue ball is moving with a
velocity of 0.08 m/s in the same direction. What is
the 8 ball’s final velocity?
Set up a conservation equation
pbefore = pafter
p1i + p2i = p1f + p2f
Example 1
A cue ball with mass 0.050 kg moves at a velocity of
0.20 m/s. It collides with the 8 ball (0.050 kg)
moving in the same direction with velocity 0.10 m/s.
After the collision, the cue ball is moving with a
velocity of 0.08 m/s in the same direction. What is
the 8 ball’s final velocity?
Substitute momentum values into the conservation equation and
solve for the momentum of the 8 ball after the collision
pbefore = pafter
p1i + p2i = p1f + p2f
0.01 kg m/s + 0.005 kg m/s = 0.004 kg m/s + p2f
p2f = 0.011 kg m/s
Example 1
A cue ball with mass 0.050 kg moves at a velocity of
0.20 m/s. It collides with the 8 ball (0.050 kg)
moving in the same direction with velocity 0.10 m/s.
After the collision, the cue ball is moving with a
velocity of 0.08 m/s in the same direction. What is
the 8 ball’s final velocity?
Determine the final velocity of the 8 ball
p2f = m2v2f
0.011 kg m/s = 0.050 kg (v2f)
v2f = 0.22 m/s
Example 2
A 0.015 kg bullet is shot into a 5.085 kg
wood block at rest on a frictionless surface.
The block and the bullet acquire a speed of 1
m/s as a result of the collision. Calculate the
initial velocity of the bullet.
Write down given information
Givens:
mb = 0.015 kg
mbb = 5.085 kg
vbi= ?
vbbi= 0 m/s
vbf = 1 m/s
vbbf = 1 m/s
Example 2
A 0.015 kg bullet is shot into a 5.085 kg
wood block at rest on a frictionless surface.
The block and the bullet acquire a speed of 1
m/s as a result of the collision. Calculate the
initial velocity of the bullet.
Determine the momentum of the block before
and after the collision
Givens:
m1 = 0.015 kg
m2 = 5.085 kg
v1i= ?
v2i= 0 m/s
v1f = 1 m/s
v2f = 1 m/s
p2i = m2v2i
= 5.085 kg(0 m/s)
= 0 kg m/s
p2f = m2 v2f
= (5.085 kg) (1 m/s)
= 5.085 kg m/s
Example 2
A 0.015 kg bullet is shot into a 5.085 kg
wood block at rest on a frictionless surface.
The block and the bullet acquire a speed of 1
m/s as a result of the collision. Calculate the
initial velocity of the bullet.
Determine the momentum of the bullet after
the collision
Givens:
m1 = 0.015 kg
m2 = 5.085 kg
v1i= ?
v2i= 0 m/s
v1f = 1 m/s
v2f = 1 m/s
p1f = (m1) v1f
= (0.015 kg) (1 m/s)
= 0.015 kg m/s
Example 2
A 0.015 kg bullet is shot into a 5.085 kg
wood block at rest on a frictionless surface.
The block and the bullet acquire a speed of 1
m/s as a result of the collision. Calculate the
initial velocity of the bullet.
Set up a conservation equation
pbefore = pafter
p1i + p2i = p1f + p2f
Example 2
A 0.015 kg bullet is shot into a 5.085 kg
wood block at rest on a frictionless surface.
The block and the bullet acquire a speed of 1
m/s as a result of the collision. Calculate the
initial velocity of the bullet.
Substitute the momentum values and determine the
momentum of the bullet before the collision
pbefore = pafter
p1i + p2i = p1f + p2f
p1i + 0 = 0.015 kg m/s + 5.085 kg m/s
p1i = 5.1 kg m/s
Example 2
A 0.015 kg bullet is shot into a 5.085 kg
wood block at rest on a frictionless surface.
The block and the bullet acquire a speed of 1
m/s as a result of the collision. Calculate the
initial velocity of the bullet.
Determine the initial velocity of the bullet
p1i = m1v1i
5.1 kg m/s = 0.015 kg(v1i)
v1i = 340 m/s

http://www.youtube.com/watch?NR=1&v=wC
XqcBpURZE&feature=endscreen