conservation of momentum - mrs. gamzon's course website

Conservation of Momentum

Where To Next?

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??

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

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.

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!

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?

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

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!

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

Types of Collisions

Inelastic Elastic

Objects stick together

Energy is lost

Objects bounce

Energy is conserved

Same final velocity Different final velocities

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

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!

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.

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

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

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

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!

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

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?