Chapter 6: Momentum

and Collisions

Objectives

• Understand the concept of momentum.• Use the impulse-momentum theorem to solve

problems.• Understand how time and force are related in

collisions.

Momentum

momentum: inertia in motion;the product of mass and velocity

p = m · vHow much momentum does a 2750 kg Hummer H2 moving at 31 m/s possess?

Note: momentum is a vector; units are kg·m/s

Impulse Changes Momentum

Newton actually wrote his second law in this form:

F · t = m · v

The quantity F·t is called impulse.

The quantity m·v represents a change in momentum.

Thus, an impulse causes a change in momentum

“impulse-momentum theorem”

F·t = m·v = pF

Highway Safety and ImpulseWater-filled highwaybarricades increasethe time it takes to stopa car. Why is this safer?

Seatbelts andairbags alsoincrease the stopping timeand reduce theforce of impact.

They reduce the forceduring impact!

F = (m · v) / t

Impulse ProblemA car traveling at 21 m/s hits a concrete wall. If the 72 kg passenger is not wearing a seatbelt, he hits the dashboard and stops in 0.13 s.• What is the p?• How much impulse is applied to the passenger?• How much force does the dashboard apply to the passenger?

What is the force applied to the passenger if he is wearing a seatbelt takes 0.62 s to stop?

Impulse Problem

The face of a golf club applies an average force of5300 N to a 49 gram golf ball. The ball leaves the clubface with a speed of 44 m/s. How much time is the ball in contact with the clubface?

F·t = m·v

F

Bouncing

Which collision involves more force: a ball bouncingoff a wall or a ball sticking to a wall? Why?The ball bouncing because there is a greater v.

F · t = m · v F ~ v so

Pelton wheel

Objectives

• Understand the concept of conservation of momentum.

• Understand why momentum is conserved in an interaction.

• Be able to solve problems involving collisions.

Conservation of Momentum

conservation of momentum: in any interaction (such as a collision) the total combined momentum of the objects remains unchanged (as long as no external forces are present).

system: all of the objects involved in an interaction

mamb

+F-F tp = -F · t p = +F · t

pTOTAL = ( -F·t ) + ( +F·t ) = 0

vai vbi

mamb

vaf vbf

pi = pf

ma·vai + mb·vbi = pi

ma·vaf + mb·vbf = pf

ma·vai + mb·vbi = ma·vaf + mb·vbf

system

Law of Conservation of Momentum:

Conservationof

Momentum

Slingshot Manuever

Jupiter

The spacecraft is pulledtoward Jupiter by gravity, but as Jupiter moves alongits orbit, the spacecraft justmisses colliding with the planet and speeds up.

The spacecraft substantiallyincreased its momentum(as speed) and Jupiterlost the same amountof momentum, but becauseJupiter is so massive,its overall speed remainedvirtually unchanged.

pi = pf

Conservation of Momentum Problem

A 0.85 kg bocce ball rolling at 3.4 m/s hits a stationary 0.17 kg target ball. The bocce ball slows to 2.6 m/s. How fast does the target ball (“pallino”) move? Assume all motion is in one dimension.

ma·vai + mb·vbi = ma·vaf + mb·vbf

Objectives

• Understand the difference between elastic and inelastic collisions.

• Solve problems involving conservation of momentum during an inelastic collision.

Collisions• elastic: objects collide and rebound,

maintaining shape• both KE and p are conserved (DEMO—

Newton spheres)• perfectly inelastic: objects collide, deform,

and combine into one mass• KE is not conserved (becomes sound, heat,

etc.)• real collisions are usually somewhere in

between

Types of Collisions

elastic

perfectly inelastic

ma·vai + mb·vbi = ma·vaf + mb·vbf

ma·vai + mb·vbi = (ma+ mb) ·vf

Conservation of Momentum ProblemVictor, who has a mass of 85 kg, is trying to make a “get-away” in his 23-kg canoe. As he is leaving thedock at 1.3 m/s, Dakota jumps into the canoe and sits down. If Dakota has a mass of 64 kg and she jumpsat a speed of 2.7 m/s, what is the final speed of the the canoe and its passengers?

Conservation of Momentum in Two-Dimensions

Collisions in 2-D involve vectors.

ma mbpi

ma

mb

paf

pbf

initial

final

Equal Mass Collision

A cue ball (m = 0.16 kg) rolling at 4.0 m/s hits a stationary eight ball of the same mass. If the cue ball travels 25o above its original path and the eight ball travels 65o below the original path, what is the speed of each ball after the collision?

Unequal Mass CollisionA 0.85 kg bocce ball rolling at 3.4 m/s hits a stationary 0.17 kg target ball. The bocce ball slows to 2.8 m/s and travels at a 15o angle above its original path. What is the speed of the target ball it travels at a 75o below the original path?