Chapter 9

Linear momentum: kg∙m/s

Stress vector nature of momentum and importance of coordinate system.Change in momentum: Δp = pf -pi

Beanbag: Δp = pf -pi =0 – (-mv) = mvRubber ball: Δp = pf -pi = +mv – (-mv) = 2 mv

What about a collection of several objects? Since momentum is a vector, the momentum of the SYSTEM is the sum of the individual momenta of each object.

mDucks = 4.00 kg mGoose = 9.00 kg.

So far, we’ve learned Newton’s 2nd Law as: , but this is a generalization restricted to constant mass!

Newton actually wrote: (Well, sort of…)

Yes, there ARE cases when the mass varies, more about those later.

For now, let’s multiply the equation through by Δt:

If we let the force be the average force, then the quantity on the left is called the IMPULSE.

9-18. To make a bounce pass, a player throws a 0.60-kg basketball toward the floor. The ball hits the floor with a speed of 5.4 m/s at an angle of 65° to the vertical. If the ball rebounds with the same speed and angle, what was the impulse delivered to it by the floor?

The impulse is equal to the change in the y-component of the momentum of the ball, because the change in the x-component is zero.

CONSERVATION OF LINEAR MOMENTUM:

If the net force acting on an object is zero, momentum is conserved.

Internal vs. External forces:

Internal forces (action-reaction pairs) sum to zero.

Only a net external force invalidates conservation of momentum.

ALSO APPLIES TO A SYSTEM OF OBJECTS!!

65°

Michael Jordan

A honeybee lands on a popsicle stick. P=0. The honeybee walks to the other end of the stick. The stick moves the other way.

Inelastic collisions: Two objects come together and stick.

1. Energy is NOT conserved2. Momentum IS conserved.

Elastic collisions: Two objects collide and bounce off of each other.

1. Energy IS conserved3. Momentum IS conserved.

Inelastic collisions:

9-30. • Two 72.0-kg hockey players skating at 5.45 m/s collide and stick together. If the angle between their initial directions was 115°, what is their velocity after the collision?

+y

+x#1

#2

115 °

Elastic collisions: Initial momentum = Final momentumKINETIC ENERGY IS CONSERVED!!Objects do NOT stick together!

37. • A 732-kg car stopped at an intersection is rear-ended by a 1720-kg truck moving with a speed of 15.5 m/s. If the car was in neutral and its brakes were off, so that the collision is approximately elastic, find the final speed of both vehicles after the collision.

Use conservation of momentum to find an equation for the final speed of the truck.

There is one equation and two unknowns. Use conservation of energy to find a second equation relating v1f and v2f.

Substitute for v1f and solve for v2f.

Substitute for v2f in the equation for v1f.

Using the given information, m1 = 1720 kg,

m2 = 732 kg, and m/s, the final speeds of the truck and car are: and

In 2 dimensions, remember to conserve momentum in the x and y-directions separately, but the kinetic energy is a scalar.

Center of Mass

In 2 dimensions:

IMPORTANT: You must set up a coordinate system!!Easy tip: Put one mass at the origin if you can!

For uniform distributions of mass, the center of mass is located at the geometric center of the object.

37. • You are holding a shopping basket at the grocery store with two 0.56-kg cartons of cereal at the left end of the basket. The basket is 0.71-m long. Where should you place a half gallon of milk (1.8 kg) so that the center of mass of your groceries is at the center of the basket?

+y

+x

Xcm

Motion of the center-of-mass.

Systems with variable mass.

Recall:

The first term is called “thrust”.

Saturn V engines eject fuel @ 13,800 kg/s and v=2440 m/s (of the ejecting fuel).

Thrust = (13,800 kg/s)(2440 m/s)=33.7 x 106 N (7.6 million pounds). Rocket weighs 6.30 million pounds, so “Houston, we have liftoff”!

Fnet= thrust – mg =33.7 x 106 N – 28.0 x 106 N = 5.7 x 106 N

Note that as the fuel is ejected, W decreases and so a is increasing.

62. •• A 0.540-kg bucket rests on a scale. Into this bucket you pour sand at the constant rate of 56.0 g/s. If the sand lands in the bucket with a speed of 3.20 m/s, (a) what is the reading of the scale when there is 0.750 kg of sand in the bucket? (b) What is the weight of the bucket and 0.750 kg of sand?

(a) The reading of the scale is given by the total weight of the sand and bucket plus the force of impact due to the pouring sand.

(b)