Download Conservation of Momentum

Momentum and Impulse
Momentum = mass x velocity
p = mv
units: kg·m/s
***momentum is a vector quantity
Conceptually, momentum is a characteristic of motion that
reflects how difficult it would be to stop the moving object
Impulse(change in momentum) = Force x time:
Δp = FΔt
units: N·s
A net force applied for a given time interval will result in a
change in momentum…this is really Newton’s Second Law:
F = ma  F = mΔv/ Δt  F Δt = Δ mv
Sample Problem
• A 1400 kg car moving westward with a
velocity of 15 m/s collides with a utility pole
and is brought to rest in 0.30 s. Find the force
exerted on the car during the collision.
Conservation of Momentum
• Keep in mind…
– The individual objects in a collision can, and do,
change momentum
– Two objects colliding experience equal force for
equal time, thus the magnitude of impulse, or
change in momentum, must be the same for both
objects
– When there are no external influences,
momentum gained by one object must equal the
momentum lost by another and the total
momentum of the system is constant
Conservation of Momentum
• Whenever objects interact in the absence of
external forces, the net momentum of the
objects before the interaction equals the net
momentum of the objects after the interaction.
Σp before = Σp after
***This general relationship will take on
different appearances when applied to
different situations
Types of collisions
• Perfectly Inelastic Collision (Sticky): A collision in
which two objects stick together and move with a
common velocity after colliding
• For 2 objects colliding inelastically…
Σp before = Σp after
m1v1+m2v2=(m1+m2)vf
In these collisions, some kinetic energy is lost in the
form of heat and sound as the objects interact
during the collision
Types of collisions
• Perfectly Elastic Collision (bouncy) : A collision in
which the total momentum and the total kinetic
energy remain constant
• After the collision, the two objects move
independently
• For 2 objects colliding elastically, both of these
must be true…
Σp before = Σp after
m1v1i + m2v2i = m1v1f + m2v2f
and
ΣKEbefore = ΣKEafter
½ m1v1i2 + ½ m2v2i2 = ½ m1v1f2 + ½ m2v2f2
In Realality…
• Only particle collisions are truly perfectly
elastic, however, many collisions are close
enough to be approximated as elastic
(ie…billiard balls)
• Macroscopic collisions are somewhere along a
continuum of between being elastic and
perfectly inelastic
Summary of Conservation of Momentum
Problem Solving Situations…
• Separate objects before  separate objects after
m1v1i + m2v2i = m1v1f + m2v2f
• Separate objects before  objects combined after
m1v1i + m2v2i = (m1 + m2)vf
• Objects combined before  separate objects after
(m1 + m2)vi = m1v1f + m2v2f
Practice #1
• A 0.015 kg marble moving to the right at 0.225
m/s makes an elastic head-on collision with a
0.030 kg marble moving to the left at 0.180
m/s. After the collision, the smaller marble
moves to the left at 0.315 m/s. What is the
velocity of the 0.030 kg marble after the
collision?
Practice #2
• A 1850 kg luxury sedan stopped at a traffic
light is struck from the rear by a compact car
with a mass of 975 kg. The two cars become
entangled as a result of the collision. If the
compact car was moving at a velocity of 22.0
m/s to the north before the collision, what is
the velocity of the entangled mass after the
collision?
Practice #3
• A 76 kg boater, initially at rest in a stationary
45 kg boat, steps out of the boat and onto the
dock. If the boater moves out of the boat
with a velocity of 2.5 m/s to the right, what is
the final velocity of the boat?