Momentum and Collisions
Resource
http://www.physicsclassroom.com/
Class/momentum/momtoc.html
Define Inertia
• The property of any body to resist changes
in its state of motion.
• The measure of Inertia is:
• Mass (Kg)
• Which of Newton’s Laws is this associated
with?
• First Law
Why is a bullet that is thrown
not as dangerous as a bullet
that is fired from a rifle?
Think about this:
Which does more damage in striking a tree,
an F-150 (Ford Truck) 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?
• Which requires a greater stopping force?
Why?
Sooooo
• There is a relationship between
• Mass
• Velocity
• Force
Momentum is inertia “mass” in
motion• Momentum combines the first and second
law of 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.
the product ofmass and velocity
of an object
momentum = mass x velocity
ρ = mvWhere
p is momentum in kgm/s m is mass in kgv is velocity in m/s
SI units are kilogram x meters per second (kgm/s)
•Vector (direction matching that of the velocity)
Is momentum scalar or vector?
• impetus was the quality of an object that
was moving independent of an observed
force. Impetus comes from the Latin in- +
petere to go to, seek -- from Greek
petesthai to fly, piptein to fall, pteron wing.
Also, push and pull derive from the Latin
pellere.
Example 1
• An ostrich with a mass of 146kg is running to
the right with a velocity of 17m/s. Find the
momentum of the ostrich.
(146 kg)(17 m/s) = 2482 kg * m/s
Example 2
• What velocity would a 5.5g bullet have, if
its momentum was the same as the
ostrich in the previous problem?
v = p/m
5.5g = 0.0055 kg
v = 2482 kg * m/s/ 0.0055 kg
v = 451,273 m/s
3. A car has a momentum of 20 000 kg m/s . What would be the car's new momentum if ...
A. its velocity were doubled. B. its velocity were tripled. C. its mass were doubled (by adding more passengers and a greater load) D. both its velocity were doubled and its mass were doubled.
3. A car has a momentum of 20 000 kg m/s . What would be the car's new momentum if ...
A. its velocity were doubled. doubledB. its velocity were tripled. tripledC. its mass were doubled (by adding more passengers and a greater load) doubledD. both its velocity were doubled and its mass were doubled. quadrupled
• How can you change momentum of an object?
• Change the velocity. (Mass could change, but
then you are changing the object)
• 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.
The quantity of force applied during a time
interval is called Impulse
• Impulse is a change in momentum
• Impulse = FΔt
• Where F = Force and t = time
• What unit is impulse measured in?
• Ns (SI Unit)
• FYI: The symbol of Impulse is an I or a J.
We typically just write out Impulse
In your head…..
• Calculate the impulse when an average
force of 10N is exerted upon a cart for 2.5
seconds.
• 25N*s
• As impulse increases what happens to
momentum?
• It increases
• What happens to momentum if the impulse
decreases?
• It decreases
Impulse = change in momentumaka the Impulse–Momentum Theorem
FΔt = m(Vf-Vi)Ns = kgm/s
How can these be equal? This is and FYI
FΔt = mΔv
Impulse = change in momentum
FΔt = m(Vf-Vi)
So what does this mean?In simple terms, a __________ acting for a long time can produce the same change in momentum as a large force acting for a __________.
Small force
Short time
FΔt = mΔv
Increasing Momentum by
Increasing VelocityApplies to an object.
Therefore mass is usually constant.
If you increase momentum you get a greater
change in velocity.
Why would you want to increase
momentum? List some examples
NOTE: the time refers to how long the force is acting on the object
Example 4
• A hockey puck has a mass
of 0.12 kg and is at rest. A
hockey player makes a shot,
exerting a constant force of
30.0 N on the puck for 0.1 s.
With what speed does it
head toward the goal?
Example 4
• m = .12 kg
• vi = 0m/s
• t = 0.1 s
• F = 30N
• F∆t = m∆v
• Solve for ∆v (vf – vi)
• ∆v = F∆t/m
• ∆v = (30 N)(0.1sec)/0. 12kg
• ∆v = 25 m/s
Example 5
• A hockey puck has a mass
of 0.12 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 5
• F∆t = m∆v
• Solve for ∆v (vf – vi)
• ∆v = F∆t/m
• ∆v = (30 N)(0.16sec)/0. 12kg
• ∆v = 40 m/s
Example 6
• A hockey puck has a mass
of 0.12 kg and is at rest. A
hockey player makes a shot,
exerting a constant force of
36.0 N on the puck for 0.1 s.
With what speed does it
head toward the goal?
Example 6
• F∆t = m∆v
• Solve for ∆v (vf – vi)
• ∆v = F∆t/m
• ∆v = (36 N)(0.1 sec)/0. 12kg
• ∆v = 30 m/s
Example 7
• A hockey puck has a mass
of 0.12 kg and is at rest. A
hockey player makes a shot,
exerting a constant force of
36.0 N on the puck for 0.16
s. With what speed does it
head toward the goal?
Example 7
• F∆t = m∆v
• Solve for ∆v (vf – vi)
• ∆v = F∆t/m
• ∆v = (36 N)(0.16 sec)/0. 12kg
• ∆v = 48 m/s
The player should use
more force and follow
through!!!!
Apply the greatest force for as
long as possible
We just examined ways to
Increase Momentum by causing
an increase in velocity
Can you think of an example where mass
would change and therefore increase
momentum?
Infiniti
Decreasing Velocity
• You are driving at 50 mph and lose control of your car. You can hit a wall or a haystack. Which do you choose? Why?
• How is the momentum different?
• It isn’t
• Why ?
• Your momentum will be decreased by same impulse with either choice since momentum = impulse.
Decreasing Velocity
• So what is different?
• Remember impulse is Force x time
• Hitting the haystack increases the time, thus decreasing the Force
• The change in momentum is the same regardless!
• Actually affecting Force or Time
Example 8
• A 1400kg car is travelling eastward at a
velocity of 15m/s, when it veers off the
road and collides with a pole and is
brought to rest in 0.30s. How much force
is exerted on the car during the collision?
Example 8
• F∆t = m∆v
• Solve for F
• F = (m∆v)/ ∆t
• F = [(1400 kg)(0-15 m/s)]/0.3 s
• F = -70000 N
• Why is F negative?
• It is a stopping force
Example 9
• A 1400kg car is travelling eastward at a
velocity of 15m/s, when it veers off the road
and instead of colliding with a pole it collides
with a barrier of sand and is brought to rest in
0.70s. How much force is exerted on the car
during the collision?
Example 9
• F∆t = m∆v
• Solve for F
• F = (m∆v)/ ∆t
• F = [(1400 kg)(0-15 m/s)]/0.7 s
• F = -30000 N
• How does this force compare to previous example?
• Smaller force: hopefully less damage to the car!
Can you think of other
examples where you would
want to decrease force?
Effect of Collision Time Upon the
Force….or why a boxer “rides”
the punch
Spreading impulse out over a longer time means that the
force will be less; either way, the change in momentum of
the boxing glove, fist, and arm will be the same.
What about vertical situations?
• Think of an object falling: its vi is 0 m/s, it
accelerates and reaches a vf max just
before impact. We are considering the
time frame of the stopping force.
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 PART II
and Multiple Objects
• This section we will observe more than
one object, and how they interact.
Law of Conservation of
Momentum• The momentum of any closed, isolated
system does not change.
• individual parts of the system 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.
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.
• To solve conservation of momentum
problems, use the formula:
• The sum of the momenta before the
collision equals the sum of the momenta
after the collision.
afterbefore pp
p1 + p2 = p’1 + p’2
'
22
'
112211 vmvmvmvm 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)
p1 + p2 = p’1 + p’2
Can be further extended to :
Solve for v2’
m1v1 + m2 v2 = m1v1’ + m2v2’
[m1v1 + m2 v2 - m1v1’ ] /m2 = v2’
There are 2 main types of
collisions• Inelastic collisions: two objects stick or join after the
collision. They each have the same velocities after the event.
• Elastic collision: two objects "bounce" apart when they collide. They each have different velocities after the event. They may or may not go in the same direction
• Explosions: special type. One object splits into multiple objects after explosion. Momentum before is zero. Sum of momentum after is zero.
The animation below portrays the inelastic collision between a 1000-kg car and a 3000-kg truck. The before- and after-collision velocities and momentum are shown in the data tables.
The animation below portrays the elastic collision between a 3000-kg truck and a 1000-kg car. The before- and after-collision velocities and momentum are shown in the data tables.
When objects collide and
“STICK”
• Two objects collide and continue jointly in
same direction
• This is called Inelastic
• Train hits Car - Train crashes into Car -
Train Car Accident Shocking video –
YouTube
Example A
Example A:
• A grandma is roller skating at 6 m/s and collides with a little boy who is stationary. If grandma has a mass of 80 kg and the boy has a mass of 40 kg, what is their velocity after impact?
• m1v1 + m2v2 = m1v1’ + m2v2’
• Modify formula to show they stick and that you are solving for final velocity
Example A
Grandma is m1 Child is m2
m1v1 + m2v2 = (m1 + m2)v3
(m1v1 + m2v2 )/(m1 + m2) = v3
(80 kg)(6 m/s) + (40 kg)(0 m/s)/(80 kg + 40 kg) = v3
v3 = 4 m/s
A grandma is roller skating at 6 m/s and collides with a little boy
who is stationary. If grandma has a mass of 80 kg and the boy
has a mass of 40 kg, what is their velocity after impact?
When objects go different
directions or “BOUNCE”This is called an elastic collision
Exercise Ball Fails Compilation! – YouTube
Determine the formula: Indicate Opposing Directions
In these ex I am assuming v1 starts right, v2starts left)
Then Rearrange Formula to Solve for v2’
EX B
• 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 velocity of the blue
stone after the collision.
• Determine the formula: Red in m1, blue is m2
• Indicate Directions
• Then Rearrange Formula to Solve for v ’
m1v1 + m2v2 = m1v1’ + m2v2’
[m1v1 + m2v2 – m2v2’]/m1 = v1’
(17 kg x -5 m/s) + 0 – (17 kg x 1.25 m/s) /20
kg = -5.31 m/s
Explosions
• One object explodes (seperates) into two
objects moving in opposite direction
• Canon Recoil
• Recoil Music Video - YouTube
Ex C 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?
• Determine the formula: Astronaut is m1, tank is m2
• Then Rearrange Formula to Solve for v1’
Ex C
m1v1 + m2v2 = m1v1’ + m2v2’
0 = m1v1’ + m2v2’
– m2v2’/m1 = v1’
– (10 kg x 12 m/s)/63 kg = -1.90 m/s
More Practice
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?
• m1v1 + m2v2 = m1v1’ + m2v2’
• [m1v1 + m2v2 - m1v1’]/m2 = v2’
0 + 0 – (76 kg x 2.5 m/s)/45 kg = -4.22 m/s
Sample B
• 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?
Sample B
(5kg)(2.4m/s) + (6kg)(4.6m/s) = (6kg)(4.1m/s) + (5kg)(? m/s)
(? m/s) = (39.6 kg*m/s – 24.6 kg*m/s)/(5kg)
3m/s