Conservation of Momentum - Mrs. Gamzon's Course Website
Transcript of Conservation of Momentum - Mrs. Gamzon's Course Website
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Conservation of Momentum
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Where To Next?
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Meet BennuAsteroid Bennu is classified as a potentially hazardous asteroid.
Large enough to reach Earth’s surface
Predicted orbit brings it within 5 million miles of Earth’s orbit
In 2135, Bennu will pass Earth within the Moon’s orbit.Close interaction with the Earth will nudge the asteroid
into a new orbit.
Why does that happen??
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Colliding In Space
A collision is an interaction between two objects during which a force is exerted for a
relatively short amount of time.
The forces can be either a contact or non-contact force!
At the microscopic and cosmic levels, non-contact
forces dominate.
Microscopic - Electrostatic Cosmic - Gravity
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Non-Contact CollisionAs objects come closer to each other, they will
experience a strong force for a given amount of time.We used these types of interactions to direct the
Voyager Missions.
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What’s the Impact?
We are not certain, but it is possible that the new orbit can lead to an impact between 2175 and 2199.
What will happen if Bennu hits Earth?
What can we do to prevent it from hitting Earth?
1 in 2800 chance that there will be a direct hit!
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Modeling CollisionsWe can model collisions in space, using
contact collisions in the lab.
Let’s use our dynamic carts to see what happens to both objects before, during, and after the collision.
What happens when object’s collide?
How do forces change after an impact?
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Back To Newton Again!When there is a collision between two objects, they
exert a force on each other for a given amount of time.
FCB = -FBC
Δt • FCB = -FBC • Δt
JCB = -JBC
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What Happens?If these two objects collide, what will happen?
FCB = -FBC
Why is the result different if the impulses are the same?
Ma = maAll organisms respond to rapid changes in velocity!
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What Does This Mean?The two objects experience the same impulse
Impulse is equal to the change in momentum!
JCB = -JBC
ΔpC = -ΔpB mCvCf - mCvCo = -(mBvBf - mBvBo)mCvCf - mCvCo = -mBvBf + mBvBo
mCvCo + mBvBo = mCvCf + mBvBf
Total po = Total pf
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Types of Collisions
Inelastic Elastic
Objects stick together
Energy is lost
Objects bounce
Energy is conserved
Same final velocity Different final velocities
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EnergyThere are different types of energy.
Elastic
Radiant
Chemical
Nuclear
Gravitational
Sound
Motion
Thermal
Electrical
Before a collision, the two objects have motion energy.
In Physics, motion energy is referred to as kinetic energy.
KE = 1/2mv2
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Energy
During an inelastic collision, kinetic energy is transformed into other forms of energy.
Elastic
Radiant
Chemical
Nuclear
Gravitational
Sound
Motion
Thermal
Electrical
Energy is never destroyed, but it is lost by the system!
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Nothing Is Perfect
Elastic
The ball bounces off the player’s head.
Inelastic
Most collisions are somewhere in between!
Energy is lost to deformation.
Elastic
The marbles bounce.
Inelastic
Energy is lost to sound and heat.
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Let’s Try One!On a greasy, essentially frictionless lunch counter, a 0.500 kg submarine sandwich, which is moving 3.00 m/s to the right, collides with a 0.250 kg grilled cheese sandwich moving to the left with a speed of 3.50 m/s. If the two sandwiches stick together and continue sliding across the counter, what is the final velocity of the resulting grilled cheese/submarine combo?
Inelastic collision!
ms = 0.500 kg
vs = +3.00 m/s
mg = 0.250 kg
vg = -3.50 m/s
msg = 0.750 kg
vsg = ? m/s
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Let’s Try One!
po = pf
ms = 0.500 kg
vs = +3.00 m/s
mg = 0.250 kg
vg = -3.50 m/s
msg = 0.750 kg
vsg = ? m/s
msvso + mgvgo = (ms+mg)vsg
vsg= (msvso + mgvgo) /(ms+mg)
vsg= +0.83 m/s
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Let’s Try Another One!Jeanne rolls a 7.0-kg bowling ball down the alley for the league championship. One pin is still standing, and Jeanne hits it head-on with a velocity of 9.0 m/s. The 2.0-kg pin acquires a forward velocity of 14.0 m/s. What is the new velocity of the bowling ball?
Elastic collision!
mB = 7.0 kg
vB = +9.00 m/s
mP = 2.0 kg
vP = 0 m/svp = +14.0 m/s
vB = ? m/s
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po = pf
mBvBo + mPvPo = mBvBf + mpvpf
vBf= (mBvBo + mPvPo - mPvPf)/(mB)
vsg= +5.0 m/s
mB = 7.0 kg
vB = +9.00 m/s
mP = 2.0 kg
vP = 0 m/svp = +14.0 m/s
vB = ? m/s
Let’s Try Another One!
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Reality Isn’t So CertainOften we know the initial information about objects colliding,
but we have to figure out the final information.Not a problem with inelastic problems!
When there is an elastic collision, things become more challenging.
po = pf
mAvAo + mBvBo = mAvAF + mBvBF
30 kg•m/s + 12 kg•m/s = 2vAF + 4vBF
42 kg•m/s = 2vAF + 4vBF
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How Do You Get Unstuck?In an elastic collision, energy is also conserved.
KEo = KEf
12
12
12
12mAvAo2 + mBvBo2 = mAvAF2 + mBvBF2
It’s time to solve a system of equation!!
42 = 2vAF + 4vBF
243 = 1vAF2 + 2vBF2
Maybe there is something else we
can do?