Final Exam Review

Final Exam Review

Conservation of EnergyImpulse & Momentum

Conservation of Momentum

Conservation of Energy

• Def: Total energy in a system remains constant. Energy is not created or destroyed.

• Key Point: Energy can change it’s form, but it never really disappears. (Rabbit into the hat)

How to find the total Energy

• 2 Ways to determine the total energy in a system:

• 1) Calculate energy when it’s ALL Potential

• 2) Calculate energy when it’s ALL Kinetic

Example

• A 1000kg roller coaster ride begins at the top of a 50m hill and has a drop to ground level before continuing onto the rest of the ride.

• A) What is the Potential Energy at the beginning?

• B) What is the Kinetic Energy at the bottom of the first hill?

• C) What is the total energy in this system?

Example

50m

1000 kgPE = mgh

Example1000 kgPE = mgh = 490,000 J

50m

KE = ?

Example1000 kgPE = mgh = 490,000 J

50m

KE = 490,000J

ExamplePE = mgh = 490,000 J

50m

KE = 490,000J

ALL energy is PE here.

All energy is KE here.

Total energy in the system is 490,000 J

ExamplePE = mgh = 490,000 J

KE = 490,000J

Total energy in the system is 490,000 J

50m

25m

What is the PE here?What is the KE?

ExamplePE = mgh = 490,000 J

KE = 490,000J

Total energy in the system is 490,000 J

50m

25m

What is the PE here? PE=mgh=(1000)(9.8)(25)=245,000JWhat is the KE?

ExamplePE = mgh = 490,000 J

KE = 490,000J

Total energy in the system is 490,000 J

50m

25m

What is the PE here? PE=mgh=(1000)(9.8)(25)=245,000JWhat is the KE? KE=Total-PE= 490,000-245,000=245,000J

Practice

• Pg. 177 Practice E• Do 1,2 &4 ONLY.

Momentum

• Formula: p = mv momentum(kg*m/s) = mass(kg) * velocity(m/s)

Impulse

• Def: The applied force needed to change the momentum of an object.

• Formula: FΔt = m(vf-vi) Impulse=Δp

Practice

• Pg 199- Practice A• #1 ONLY• Pg 201- Practice B• #1 & 2 ONLY

Conservation of Momentum

• Key Point: The total momentum between two objects always remains constant.

Example

• A 3 kg ball rolling to the right at 5m/s hits a 2kg ball at rest. The 3kg ball comes to rest after the collision. What is the velocity of the ball initially at rest?

3kg 2kg

5m/s 0m/s

Example

• A 3 kg ball rolling to the right at 5m/s hits a 2kg ball at rest. The 3kg ball comes to rest after the collision. What is the velocity of the ball initially at rest?

3kg 2kg

5m/s 0m/s

Total momentum:(3)(5) + (2)(0) = 15kgm/s

Example

• After collision:

3kg 2kg

0m/s ? m/s

Total momentum: 15kgm/s

Total momentum: 15 = (3)(0) + (2)(v) 15 = 2v 7.5 m/s = v

Practice

• Pg. 211 – Section Review• #3 ONLY.