Download Unit 5 Notes

Physics
Unit 5 Notes
This unit corresponds to Chapter 6 of our textbook.
Momentum
Momentum is a quantity separate from kinetic energy. Momentum is defined as the
product of mass and velocity:
P = mv
P = momentum
m = mass
v = velocity
The rate of change of momentum is equal to the net force applied to it. As an example,
let’s look at a 1.00kg object that has a 2.0N force applied to it for 3.0s:
P = [2.0(kg)(m)/s2][3.0s]
P = 6.0kgm/s
How fast is the object traveling?
P = mv
6kgm/s = 1kg(v)
v = 6m/s
Is that what we would have found using our previous physics knowledge? We know that
if a 2N force is applied to a 1kg object, that object will accelerate at 2m/s2. If that object
accelerates at that rate for 3.0s, then its final velocity will be 6m/s.
Let’s look at another example:
A tennis player hits the ball and the ball achieves a velocity of 55m/s. The ball was in
contact with the racket for .004s and the ball has a mass of .06kg. Find the force exerted
on the ball by the racket.
We can use the following equation:
P = Ft
F = P/t
F = (.06kg)(55m/s)/.004s
F = 800N
Momentum in an isolated system is conserved. That is, it is not lost or gained when a
process occurs. An isolated system is one in which there are no outside forces. The most
common application of this is in collisions. There are two main types of collisions,
elastic, and inelastic. This has nothing to do with our everyday connotation of the word
elastic. Elastic collisions are collisions in which both momentum and kinetic energy are
conserved. In an elastic collision, we don’t see deformation of the objects that collide. A
good example of an elastic collision is when billiard balls collide. Inelastic collisions are
collisions where momentum is conserved but kinetic energy is not conserved. Car
collisions are good examples of this because the crumpling of the cars structures absorbs
some kinetic energy. A perfectly inelastic collision is one where the two objects that
collide stick together after the collision. Let’s look at an example of two railroad cars
colliding:
A 10,000kg railroad car traveling at 20.0m/s strikes another identical car. The two cars
couple together and continue moving along the rails. What is their velocity?
We will use conservation of momentum to solve this problem:
P = P0
P0 = (10,000kg)(20.0m/s)
P = (20,000kg)(v)
v = 10m/s
Let’s look at another example:
A 5kg rifle fires a .02kg bullet at 620m/s. Find the recoil velocity of the rifle.
Prifle = Pbullet
5kg(v) = .02kg(620m/s)
v = 2.5m/s
Let’s look at another application of momentum, the ballistic pendulum:
Ballistic Pendulum
In a ballistic pendulum a projectile is launched and strikes a target. The projectile
embeds itself in the target. The target then moves upwards. The velocity of the projectile
can then be calculated.
We have to make an important assumption about this type of problem. We assume that
the projectile hits the target, becomes embedded and stops. This is an inelastic collision.
Then, the target moves upwards, after the projectile has stopped. This approximation is
acceptable.
Let’s work this example:
Calculate the velocity of a .002kg bullet that strikes a 2kg ballistic pendulum that lifts the
pendulum 1.00m.
We will assume that all of the kinetic energy possessed by the pendulum with the bullet
stuck in it is converted to gravitational potential energy. Therefore:
ke = mgh
ke = (2.02kg)(9.8m/s2)(1m)
ke = 1/2mv2
1/2mv2 = 19.8J
Remember, m = 2.02kg so
v = 4.47m/s
Let’s use that velocity in the momentum formula:
Pbullet = Ppendulum+bullet
.02kg(v) = 2.02kg(4.47m/s)
v = 447m/s
Billiard Balls and Collisions in 2-D
Let’s assume that a billiard ball has a mass of .15kg. When the player hits the cue ball,
we will then consider the cue ball and the target ball as the system. A system is defined
as arbitrarily chosen mass and energy that is of interest to us as scientists. The cue ball
and target ball are not isolated before the cue ball is hit, they are subject to the outside
force of the player and the cue stick. We will start our problem after the cue ball is hit so
that we can say that we have an isolated system.
The cue ball is moving at a velocity of 3.00m/s in the x direction. It strikes another ball
and the two balls move away at angles of 45 degrees with the x direction. Calculate the
velocity of the balls in the x and y directions.
Let’s think about this problem. What was the momentum in the y-direction before the
collision? It was zero. That means that after the collision, the net momentum in the y
direction must be 0. We can use vectors to solve the problem:
In the y direction the momentum of the cue ball and the target ball are equal in magnitude
and opposite in direction. The velocities equal
sin45(vcue) and sin45(vtarget)
In the x direction the velocities are:
cos45(vcue) and cos45(vtarget)
The momentum in the x direction is conserved so:
.15kg(3.00m/s) = .15kg(cos45)vcue + .15kg(cos45)vtarget
We know that vcue = vtarget from the y direction.
Therefore vcue = vtarget = 2.1m/s
Impulse and Collisions
Impulse is defined as force multiplied by time. If we look at the units for impulse, what
do we find?
Impulse = F(t)
Force is in Newtons. Remember that a Newton is a kilogram meter per second squared:
1N = 1kgm/s2
If we multiply the unit Newton by the unit seconds, we get a kilogram meter per second.
This is the same as the unit for momentum. Therefore, the impulse delivered to an object
is equal to the change in momentum of the object. Let’s use this to solve a problem.
What is the velocity of a 2 kilogram object if it has a 10N force applied to it for 5s?
We could solve this problem by finding the acceleration of the object, and using a in a
kinematics equation. Or we could just use the fact that the impulse equals the change in
momentum:
10N(5s) = 2kg(v)
v = 25m/s
Center of Mass
The center of mass of an object is the part of an object that would move in the same way
as a particle would if subjected to the same net force. In other words, if we had a uniform
2m long pipe, the center of mass would be at the 1m mark. If we had a 1m diameter
uniform circle made from plywood, the center of mass would be in the center. The
important point we need to remember is that the center of mass of an object cannot move
unless the object is acted upon by an outside force. Let’s try an example problem:
A 75kg angler is standing at the back of his 3m long 150kg boat. His boat is up against
the dock and the angler’s arms can reach 1m. If he walks to the front of the boat, will he
be able to reach the rope on the edge of the dock to tie up his boat?
Let’s not calculate the center of mass of the boat-angler system. Instead, let’s say the
center of mass stays wherever it is (this is true). We know if a 75kg angler moves 3m,
then the 150kg boat must move______.
3m(75kg) = xm(150kg)
X = 1.5m
The boat moves back 1.5m and the angler can’t reach the edge of the dock to grab the
rope.