Today: Momentum – chapter 911/03 Finish momentum & review for exam11/8 Exam 2 (5 – 8)11/10 Rotation11/15 Gravity11/17 Waves & Sound11/22 Temperature11/29 Finish up & Review12/01 Review (last day of classes 12/02)Final 12/06 @ 2:00 – 4:30 for section 2

Racing Balls1

2

When the two balls are launched from one end of the track with the same initial velocity, what will happen: 1) Ball #1 on the straight track arrives at the other end first 2) Ball #2 on the track with the dip arrives at the other end first 3) the race is a tie - both balls reach the other end at the same time?

Chapter 9Linear Momentum and Collisions

Linear momentum is defined as:

p = mv

Momentum is given by mass times velocity.

Momentum is a vector.

The units of momentum are (no special unit):

[p] = kg·m/s

Momentum = p = mv (kg.m/s)Since p is a vector, we can also consider the

components of momentum:

px = mvx

py = mvy

Note: momentum is “large” if m and/or v is large. • Name an object with large momentum but small

velocity.• Name an object with large momentum but small mass

Momentum

Is there a relationship between F and P?

Another way of writing Newton’s Second Law is

F = p/t= rate of change of momentum

This form is valid even if the mass is changing.

This form is valid even in Relativity and Quantum Mechanics.

Impulse

We can rewrite F = p/t as:

Ft = p

I = Ft is known as the impulse.

The impulse of the force acting on an object equals the change in the momentum of that object.

Exercise: Show that impulse and momentum have the same units.

Example

A 0.3-kg hockey puck moves on frictionless ice at 8 m/s toward the wall. It bounces back away from the wall at 5 m/s. The puck is in contact with the wall for 0.2 s.

(a) What is the change in momentum of the hockey puck during the bounce? [3.9 kg.m/s]

(b) What is the impulse on the hockey puck during the bounce? [3.9 kg.m/s]

(c) What is the average force of the wall on the hockey puck during the bounce? [19.5 N]

If there are no external forces on a system, then the total momentum of that system is constant. This is known as:

The Principle of

Conservation

of

Momentum

In that case, pi = pf.

Internal vs. External Forces

system

Here the system is just the box and table. Any forces between those two objects are internal. Example: The normal forces between the table and the box are internal forces. Internal forces on the system sum to zero.

External forces do not necessarily sum to zero. Something outside the circle is pushing or pulling something inside the circle.

ExampleTwo skaters are standing on frictionless ice. Skater A has a mass of 50

kg and skater B has a mass of 80 kg. Skater A pushes Skater B for 0.25 s, causing Skater B to move away at 10 m/s.

(a) What is the velocity of Skater A after he pushes Skater B? [16 m/s]

(b) What is the change in momentum of Skater B? [800 kg.m/s]

(c) What is the average force exerted on Skater B by Skater A during the 0.2 s push? [4000 N]

(d) What is the change in momentum of Skater A? [800 kg.m/s]

(e) What is the average force exerted on Skater A by Skater B during the 0.2 s? [4000 N]

A honeybee with a mass of 0.175 g lands on one end of a popsicle stick. After sitting at rest for a moment, the bee runs toward the other end with a velocity 1.55 cm/s relative to the still water. What is the speed of the 4.65 g stick relative to the water? (Assume the bee's motion is in the negative direction.)

Problem #4 (HW 10)

[0.583e-3] m/s = 0.583 mm/s

Collisions

In general, a “collision” is an interaction in which

• two objects strike one another

• the net external impulse is zero or negligibly small (momentum is conserved)

Examples: car crash; billiard balls

Collisions can involve more than 2 objects

From the conservation of momentum:

pi = pf

m1v1,i + m2v2,i = m1v1,f + m2v2,f

v2,f

v2,i

v1,f

v1,i

What about conservation of energy?

We said earlier that the total energy of an isolated system is conserved, but the total kinetic energy may change.

• elastic collisions: K is conserved• inelastic collisions: K is not conserved

• perfectly inelastic: objects stick together after colliding

Perfectly Inelastic Collisions

After a perfectly inelastic collision the two objects stick together and move with the same final velocity:

pi = pf

m1 v1,i + m2 v2,i = (m1+ m2)vf

This gives the maximum possible loss of kinetic energy.In non-relativistic collisions, the total mass is conserved

Problem #3 (HW 10)A car with a mass of 950 kg and an initial speed of v1 = 17.8 m/s approaches an intersection, as shown in the figure. A 1300 kg minivan traveling northward is heading for the same intersection. The car and minivan collide and stick together. If the direction of the wreckage after the collision is 37.0° above the x axis, what

is the initial speed of the minivan and the final speed of the wreckage?[9.8] m/s , [9.41] m/s

Elastic Collisions

Kinetic energy is conserved in addition to momentum:

pi = pf

Ki = Kf

2,22

2,11

2,22

2,11

,22,11,22,11

21

21

21

21

ffii

ffii

vmvmvmvm

mmmm

vvvv

Lots of variables, keep track!

Elastic Collisions in 1-dimension

Kinetic energy is conserved in addition to momentum:

pi = pf

Ki = Kf

f2i2f1i1

f1f2i2i1

i2f2f1i1

i2f2i2f22f1i1f1i11

2i2

2f22

2f1

2i11

2f22

2f11

2i22

2i11

i2f22f1i11

f22f11i22i11

vvvv

vvvv

vvvv

vvvvmvvvvm

vvmvvm

vm2

1vm

2

1vm

2

1vm

2

1

vvmvvm

vmvmvmvm

,,,,

,,,,

,,,,

,,,,,,,,

,,,,

,,,,

,,,,

,,,,

:Divide

Relative velocity of approach before collision = relative velocity of separation after collision

Center of MassThe center of mass (CM) of an object or a group of objects (system) is the “average” location of the mass in the system. The system behaves as if all of its mass were concentrated at the center of mass.

The center of mass is not always located on the object.

Where is the CM for this object?

Calculating the Center of Mass

For two objects:

Mxmxm

mmxmxm

X cm2211

21

2211

In general, X coordinate of center of mass:

Mmx

mmxmxm

X cm

...

...

21

2211

In general, Y coordinate of center of mass:

Mmy

mmymym

Ycm

...

...

21

2211

Exercise

Find the x coordinate of the center of mass of the bricks shown in the Figure.x = 0.92 L