Transcript Momentum and Collisions

1/3/11 Welcome Back!!!!
Today: Review concept
of work and power
through lab activity.
1/4/11
Today: Review concept of work, energy,
and momentum through lab activity.
If you did not turn in power lab, place it in
blue sorter now!
Those of you who missed the test on
momentum can take it in class today or
tomorrow.
1/4/11
Today: Finish review of concept of work,
energy, and momentum through lab
activity.
When completed, place it in blue sorter.
Those of you who missed the test on
momentum can take it in class today or
tomorrow.
Momentum and
Collisions
http://www.physicsclassroom.c
om/Class/momentum/momtoc.
html
Think about this:
Which does more damage in striking a
tree, an F-150 or a Mini-Cooper?
 Is this always true?
 What other information do you need
to determine your response?
 What term do you think describes
this?
Define Inertia
 The
property of any body to resist
changes in its state of motion.
 The measure of Inertia is:
 Mass
 Which of Newton’s Laws is this
associated with?
 First Law
Momentum is inertia in
motion
 The
linear momentum of an object of
mass m moving with velocity v is
defined as the product of the mass
and the velocity. Momentum is
represented by the symbol p.
Momentum p = mv

Where
p
is momentum in kgm/s
 m is mass in kg
 v is velocity in m/s
Is momentum scalar or vector?
 Vector (direction matching that of the
velocity)
 SI units are kilogram x meters per second
(kgm/s)

p = mv
 Determine
the
momentum of a
60-kg halfback
moving eastward
at 9 m/s.
540 kg m/s
Example 1
 Determine
the momentum of a F 150
truck moving northward at 45 mph.
Assume a weight 5000 pounds.
 Remember the SI unit for momentum
in kg * m/s
45400 kg * m/s
 1 mile = 1.6 km
 1 kg = 2.2 pounds
Example 2
 How
fast (in mph) would a Mini
Cooper (2500 pounds) need to be
traveling to have the same
momentum as the truck in example
1? 1 mile = 1.6 km
 1 kg = 2.2 pounds
90 mph you could do this
in your head, just look at
Example 3
 Calculate
the
momentum of
the Titanic, of
mass 4.2 x 107
kg, moving at 14
knots (1 knot =
1.852 km/h).
3.02 x 108 kg * m/s
12/13 Almost done…..
 Goal:
Discuss Impulse Momentum
Theory and Conservation of
Momentum
 Tomorrow you will be working on
problem solving
 Wednesday we will consider special
situations
 Thursday we will have a test.






How can you change momentum of an object?
Change the velocity.
What term describes a change in velocity?
Acceleration
How do you change the velocity, ie cause
acceleration?
Apply a Net force.






As force increases, what happens to
momentum?
It increases.
Will the change be instantaneous?
No. It takes time.
As time increases, what happens to
momentum?
It increases.
The quantity of force applied during
a time interval is called Impulse
• Impulse = F∆t
• As impulse increases what happens to the
change in momentum?
• It increases
• What happens to change in momentum if
the impulse decreases?
• It decreases
Impulse-Momentum Theorem

The expression FΔt = Δp is called the impulsemomentum theorem.
Impulse = Change in Momentum

F∆t = m∆v = ∆p = mvf – mvi







What part of the equation describes the impulse?
FΔt
Is impulse a scalar or vector quantity?
Vector quantity in the direction of the force.
What units is Impulse is measured in?
Ns
In other words
A
net force, F, applied to an object for a
certain time interval, t, will cause a
change in the object’s momentum equal
to the product of the force and time
interval.
 In simple terms, a small force acting for a
long time can produce the same change
in momentum as a large force acting for a
short time.
Increasing Momentum
 Teeing
Off a golf ball or Swinging at a
baseball
 How would you increase the
momentum?
 Apply the greatest force for as long as
possible
 In other words: FOLLOW THROUGH!!!
Impulse = F∆t
 Calculate
the impulse when an
average force of 10N is exerted upon
a cart for 2.5 seconds.
 25N*s
Answer example 4 and 5
Example 4
A
hockey puck has a
mass of 0.115 kg and is
at rest. A hockey player
makes a shot, exerting a
constant force of 30.0 N
on the puck for 0.16 s.
With what speed does it
head toward the goal?
Example 4
F∆t = m∆v
 Solve for ∆v (vf – vi)
 ∆v = F∆t/m
 ∆v = (30 N)(0.16sec)/0.0115kg
 ∆v = 41.7 m/s
 How would the velocity of the puck
change if the hockey player didn’t follow
through with the shot?
 Velocity would be less giving goalie more
time to react and block the shot .

Example 5
 If
the momentum of the NASA space
shuttle as it leaves the atmosphere is
3.75 x 108 kg m/s, and its mass is
75,000 kg, what is its speed?
 (3.75 x 108 kg m/s) / 75,000 kg
 5000 m/s
Decreasing “Momentum”
You lose control of your car. You can hit
a wall or a haystack. Which do you
choose? Why?
 Your momentum will be decreased by
same impulse with either choice since
momentum = impulse.
 So what is different?
 Remember impulse is Force x time
 Hitting the haystack increases the time,
thus decreasing the Force

Effect of Collision Time Upon the
Force….or why a boxer “rides”
the punch
Example 6
A
2200-kg sport utility vehicle
traveling at 26 m/s can be stopped in
21 s by gently applying the brakes, in
5.5 s in a panic stop, or in 0.22 s if it
hits a concrete wall. What is the
momentum in all of these situations?
What average force is exerted on the
SUV in each of these stops?
Example 6
2200 kg at 26m/s
F∆t = m∆v
Determine momentum
 mv = (2200kg)(0m/s - 26m/s) = -57200 kg*m/s
 Why is momentum negative?
 We are stopping.
 Set momentum equal to impulse and solve for
Force
 At 21sec: F(21s) = -57200 kg*m/s
 F = (-57200 kg*m/s)/21s = -2720 N

Example 6
•
•
•
•
At 5.5sec: F(5.5s) = -57200 kg*m/s
F = (-57200 kg*m/s)/5.5s = -10400 N
At .22sec: F(.22) = -57200 kg*m/
F = (-57200 kg*m/s)/.22s = -260000 N
Think…..
 When
a dish falls, will the impulse be
less if it lands on a carpet than if it
lands on a hard floor?
 No. The impulse will be the same for
either surface because the same
momentum change occurs for each.
Force is less on carpet because of
greater time for momentum change.
Momentum and Multiple Objects
 In
the last section, we looked at the
momentum of objects one at a time.
This section we will observe more
than one object, and how they
interact.
 Take for example a game of
billiards…
What happens…
 …when
one billiard ball is knocked
into another?
Several Outcomes
 …but
 Either
all have something in common.
the cue ball stops completely
or slows down, and,
 The ball it hits goes from being still to
moving at some speed.
If…
If the two balls were BOTH moving, there
would be even more possible outcomes,
such as…
 The cue ball would bounce back while the
other ball went forward,
 The cue ball knocks the other forward,
and it keeps going forwards, too,
 The cue ball and the other ball roll
backwards.

AND, what if…
 If
you imagine playing billiards with
different sized balls, you get even
more possible outcomes!
 What if the ball you were hitting with
the cue ball was a bowling ball?
What would happen?
Conservation of Momentum
In all of those examples, momentum is
conserved.
 Conservation of momentum means that in
any collision, the total amount of
momentum remains constant.
 In other words, when two objects collide,
the momentum of object A plus the
momentum of object B initially is equal to
the momentum of object A plus the
momentum of object B at the end.

Most Generally:
 The
total momentum of all objects
interacting with one another remains
constant regardless of the nature of
forces between the objects.
How to calculate?


Compare the total momentum of
two objects before and after they
interact.
The momentum of each object
changes before and after an
interaction, but the total momentum
of the two objects together remains
constant.
Law of Conservation of
Momentum
The law of conservation of momentum
states: The momentum of any closed,
isolated system does not change.
 During any event, such as an explosion or
a collision, individual parts of the system
of objects involved may experience
changes in momentum.
 However, the total momentum of the
system before the event must equal the
total momentum of the system after the
event.


To solve conservation of momentum
problems, use the formula:
pbefore  pafter

The sum of the momenta before the
collision equals the sum of the momenta
after the collision.
p1 + p2 = p’1 + p’2
p1 + p2 = p’1 + p’2
Can be further extended to :
m1v1  m2v2  m v  m v
'
1 1
p = momentum before collision (kg m/s)
p’ = momentum after collision (kg m/s)
m1 = mass of object 1 (kg)
v1 = velocity of object 1 before the collision (m/s)
m2 = mass of object 2 (kg)
v2 = velocity of object 2 before the collision (m/s)
v1’ = velocity of object 1 after the collision (m/s)
v2’ = velocity of object 2 after the collision (m/s)
'
2 2

A Two ice skaters—one guy and
one girl—are standing, facing each
other on a frozen pond. They push
off of each other, and the guy, who
has a mass of 64.2kg, travels
backwards at a velocity of 5.6m/s.
If the girl has a mass of 49.3 kg,
what will be her velocity in the
opposite direction?
 Insert
one of these to solve for mass
Example 7

A 5.0 kg bowling ball with a
velocity of 6.0 m/s strikes a
1.5 kg standing pin
squarely. If the ball
continues on at a velocity of
3.0 m/s what will be the
velocity of the pin after the
collision?
Example 8 REWRITE
ANSWER IN long equation set
up

Before the collision:




(5kg)(6m/s) = 30 kg*m/s
(1.5kg)(0m/s) = 0 kg*m/s
Total p = 30 kg*m/s + 0 kg*m/s = 30 kg*m/s
After the collision:




Total p = 30 kg*m/s (From before collision)
(5kg)(3m/s) + (1.5kg)(? m/s) = 30 kg*m/s
(? m/s) = 30 kg*m/s – [(5kg)(3m/s)]/ (1.5kg)
10 m/s
Example 9
A
5 kg bowling ball is rolling in the gutter
towards the pins at 2.4 m/s. A second
bowling ball with a mass of 6 kg is thrown
in the gutter and rolls at 4.6 m/s. It
eventually hits the smaller ball and the 6
kg ball slows to 4.1 m/s. What is the
resulting velocity of the 5 kg ball?
Example 9

Before the collision:




(5kg)(2.4m/s) = 12 kg*m/s
(6kg)(4.6m/s) = 27.6 kg*m/s
Total p = 12 kg*m/s + 27.6 kg*m/s = 39.6 kg*m/s
After the collision:




Total p = 39.6 kg*m/s (From before collision)
(6kg)(4.1m/s) + (5kg)(? m/s) = 39.6 kg*m/s
(? m/s) = 39.6 kg*m/s – [(6kg)(4.1m/s)]/(5kg)
3m/s
Example 10
 Two
people are practicing curling. The red
stone is sliding on the ice towards the
west at 5.0 m/s and has a mass of 17.0
kg. The blue stone has a mass of 20.0 kg
and is stationary.
After the collision,
the red stone moves east at 1.25 m/s.
Calculate the momentum and velocity of
the blue stone after the collision.
Example 10

Before the collision:




(17kg)(-5m/s) = -85.0 kg*m/s
(20kg)(0m/s) = 0 kg*m/s
Total p = kg*m/s + 0 kg*m/s = -85.0 kg*m/s
After the collision:





Total p = - 85.0 kg*m/s (From before collision)
(17kg)(1.25m/s) + (20kg)(? m/s) = - 85kg*m/s
(? m/s) = -85.0 kg*m/s – [(17kg)(1.25m/s)]/ (20kg)
-5.31m/s blue stone
p = -106 kg*m/s
Sample Problem 9
4 m/s
The Astronaut’s Game
Momentum Conserved…
 Momentum
is conserved in collisions,
like the billiard balls we talked about.
 Momentum is also conserved in two
objects pushing away from each
other.
 What happens if two ice skaters
facing each other push?
The Ice Skaters
 They
begin at rest (total momentum
of zero), and they end going in
opposite directions (the total
momentum still adds to zero).
12/15
 Please
pick up your work from
yesterday
 Today’s Goal: Discuss different
types of conservation of momentum
problems.


A billiard ball approaches a
cushioned edge of a billiard table
with momentum, p. After the
collision with the cushion, it
bounces straight back with the
same amount of momentum in the
opposite direction. What is the
impulse on the ball?
 = -p –p = -2p
Sample Problem A
A
76kg man is standing at rest in a
45kg boat. When he gets out of the
boat, he steps out with a velocity of
2.5m/s to the right (onto the dock).
What is the velocity of the boat?
+ m2v2 = m1v1’ + m2v2’
 [m1v1 + m2v2 - m1v1’]/m2 = v2’
0 + 0 – (76 kg x 2.5 m/s)/45 kg = -4.22 m/s
 m1v1
Sample Problem B
A
63.0kg astronaut is on a spacewalk
when his tether to the shuttle breaks.
He is able to throw a 10.0kg oxygen
tank away from the shuttle with a
velocity of 12.0m/s. Assuming he
started from rest, what is his
velocity?
+ m2v2 = m1v1’ + m2v2’
 [m1v1 + m2v2 – m2v2’]/m1 = v1’
0 + 0 – (10 kg x 12 m/s)/63 kg = -1.90 m/s
 m1v1
Sample Problem C
A
6.50kg bowling ball is moving at a
velocity of 3.5m/s when it hits a
0.30kg billiard ball going 1.3m/s in
the opposite direction. If the bowling
ball stops moving after the collision,
what is the velocity of the billiard
ball?
+ m2v2 = m1v1’ + m2v2’
 [m1v1 + m2v2 - m1v1’]/m2 = v2’
 m1v1
[(6.5 kg x 3.5 m/s) + (.3kg x -1.3m/s) -0]/.3 kg = 74.5 m/s
Sample Problem D
A
65-kg ice skater traveling at 6.0
m/s runs head on into an 85-kg
skater traveling straight forward at
4.5 m/s. At what speed and in what
direction are the ice skaters traveling
if they move joined together after the
collision?
 m1v1
+ m2v2 = (m1 + m2)v3
 (m1v1 + m2v2 )/(m1 + m2) = v3
Sample Problem E
A
4.0-kg object traveling westward at
25 m/s hits a 15-kg object at rest.
The 4.0-kg object bounces eastward
at 8.0 m/s. What is the speed and
direction of the 15-kg object?
Does this seem familiar?
 The
idea of momentum being
equal—especially when two
objects are pushing away from
each other—may seem familiar.
 The conservation of momentum
equation is derived from Newton’s
3rd Law.
 Newton’s 3rd law says…
Newton and Momentum
 From
last section:
 Ft = p, or (as we’ll use it) Ft = mvf
- mvi
 If we consider two forces, we’ll call
them F1 and F2, we have that:
 F1t = m1v1,f – m1v1,i …and…
 F2t = m2v2,f – m2v2,i
 Since these are the only two forces
Newton and Momentum
 Since
they are in the same collision,
the time is exactly the same, so:
 F1t = -F2t
 Which
means (substituting from
momentum):
 m1v1,f – m1v1,i = -(m2v2,f – m2v2,i)
In real collisions…



In this chapter (and this book altogether) we
treat force as constant in collisions.
The reality is that force is not constant, but
varies throughout a collision.
Think of two rubber balls hitting each other…at
first, the balls apply force to each other, but the
more contact, the more they are compressed
and the more force they push with, until their
velocities are reversed—then the force
decreases until they lose contact and the force
exerted is zero.
Average Force
 Because
of that, when we calculate
impulse, we generally use average
force over the collision.
Collisions
 One
type of collision we have not
considered is one in which two
objects stick together and move with
a common velocity after colliding.
 This is called a perfectly inelastic
collision.
 In
a perfectly inelastic collision,
Elastic or Inelastic?!?
What does the word elastic mean in
physics?
 When we think of something as elastic it
tends to keep its shape—when bent or
stretched it returns to normal.
 In a collision, elastic or inelastic refers to
whether or not an object is deformed
during the collision.

Inelastic Collisions
 In
an inelastic collision, kinetic
energy is not constant.
 Some energy is converted to sound,
some is converted to internal energy
as the objects deform.
Sample Problem
 Two
clay balls collide head on in a
perfectly inelastic collision. The first
ball has a mass of 0.500kg and an
initial velocity of 4.00m/s to the right.
The second ball is 0.250kg, and has
an initial velocity of 3.00m/s to the
left. What is the final velocity of the
two balls after they collide and stick
together, and how much kinetic
Practice Problem #3
A
0.25kg arrow with a velocity of
12m/s west strikes a 6.8kg target
(standing still). What is the final
velocity of the target and arrow?
What is the decrease in kinetic
energy?
Elastic Collisions
 In
elastic collisions, the objects
collide and return back to their
original shape.
 The overall kinetic energy in an
elastic collision is conserved.
 This means that:

KE1,i + KE2,i = KE1,f + KE2,f
Practice Problem #4
A 0.015kg marble moving to the right at
0.225m/s makes an elastic head on
collision with a 0.030 marble moving left
at 0.180m/s. After the collision, the
smaller marble moves to the left at a
velocity of 0.315m/s.

A) Find the final velocity of the
bigger marble.

B) Verify by comparing kinetic
energy before and after collision.

In reality…
In real life, there are very, very few
collisions that are perfectly inelastic or
elastic.
 Most collisions lose energy in the form of
sound, heat, etc.
 The third category of collisions is called
inelastic collisions; collisions in which
objects bounce off of each other, but
kinetic energy is not conserved.

Inelastic Collisions
 For
the problems we will solve, we
will consider all collisions either
perfectly inelastic or elastic.