Conservation of Momentum

• For a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision

• The momentum changes are equal in magnitude but opposite in direction

(m1·v1) + (m2·v2 ) = -(m1·v1 ) - (m2·v2)

Conservation of MomentumConservation of Momentum

(m1·v1) + (m2·v2 ) = -(m1·v1 ) - (m2·v2)

Conservation of MomentumConservation of Momentum

The left side of the equation

represents the INITIAL /BEFORE

information

The right side of the equation

represents the FINAL/AFTER

information

Example: Conservation of Momentum

A 76 kg boater steps off a canoe (which is at rest) to the right onto a dock. If the canoe has a mass of 45 kg and the boater steps out of the canoe with a velocity of 2.5 m/s and then stops, what is the final velocity of the boat?

(m1·v1) + (m2·v2 ) = -(m1·v1 ) - (m2·v2)

(76·2.5) + (45·0 ) = -(76·0 ) - (45·v2)

190 = -45v2

-4.2 m/s = v2

190 = -45v2

-45 -45

Similar conceptual picture…think about how the idea of Conservation of Momentum relates to Newton’s 3rd Law

Inelastic CollisionsInelastic Collisions• Collisions in which two objects stick together

after the collision so that they share the final velocity

• Formula: m1v1 + m2v2 = (m1 + m2)vf

Example: Inelastic Collision

A 0.500 kg cart moving east with a speed of .928 m/s collides with a 1.50 kg cart moving west with a speed of .216 m/s. If the two carts stick together and move as a single object after the collision then determine the post-collision speed of the two carts.

Answer

m1v1 + m2v2 = (m1 + m2)vf

(.500 ·.928) + (1.50 · -.216) = (.500 + 1.50)vf

(.464) + (-.324) = (2.00)vf

.14 = (2.00)vf

2.00 2.00

.07 m/s (to the right) = vf

Elastic CollisionsElastic Collisions

Collision in which the two objects bounce off each other after the collision so that they continue to move separately

Formula: (m1·v1) + (m2·v2 ) = (m1·v1 ) + (m2·v2)

Example: Elastic Collision

A .015 kg marble moving to the right at .225 m/s collides head-on with a .030 kg marble moving to the left at .180 m/s. After the collision, the smaller marble moves to the left at .315 m/s. What is the velocity of the larger marble after the collision?

Answer(m1·v1) + (m2·v2 ) = (m1·v1 ) + (m2·v2)

(.015·.225) + (.030·-.180) = (.015·-.315) + (.030·v2)

(.0034) + (-.0054) = (-.0047) + (.030)v2

-.002 = -.0047 + .030v2

.0027 = .030v2

.09 m/s (to the right) = v2