1. Agenda
Aim: SWBAT recognize kinetic energy is related to momentum by the end of today’s lesson.
Tasks:
Kinetic Energy and Momentum PPT notes
Example Problems
Calculate the momentum and kinetic energy of a moving object given the mass and velocity.
Example Problem: Solve a one-dimensional elastic collision problem using momentum conservation.
Describe the relationship between linear momentum and kinetic energy.

2. 12.1 Kinetic Energy and Momentum
Kinetic energy and momentum are different quantities, even though both depend on mass and speed.
Kinetic energy is a scalar quantity.
Momentum is a vector, so it always depends on direction.
Two balls with the same mass and speed have the same kinetic energy but opposite momentum.

3. Task 1: Kinetic Energy and Momentum
Given: 𝐾𝐸= 1 2 𝑚 𝑣 2 𝑎𝑛𝑑 𝑝=𝑚𝑣
Show that 𝐾𝐸= 1 2 𝑚 𝑣 2 = 𝑝 2 2𝑚

4. 12.1 Calculating Momentum p = m v
The momentum of a moving object is its mass multiplied by its velocity.
That means momentum increases with both mass and velocity.
Momentum
(kg m/sec)
p = m v
Velocity (m/sec)
Mass (kg)

5. Comparing momentum You are asked for momentum.
A car is traveling at a velocity of 13.5 m/sec (30 mph) north on a straight road. The mass of the car is 1,300 kg. A motorcycle passes the car at a speed of 30 m/sec (67 mph). The motorcycle (with rider) has a mass of 350 kg. Calculate and compare the momentum of the car and motorcycle.
You are asked for momentum.
You are given masses and velocities.
Use: p = m v
Solve for car: p = (1,300 kg) (13.5 m/s) = 17,550 kg m/s
Solve for cycle: p = (350 kg) (30 m/s) = 10,500 kg m/s
The car has more momentum even though it is going much slower.

6. Find the Kinetic Energy of this system using 𝐾𝐸= 𝑝 2 2𝑚
A car is traveling at a velocity of 13.5 m/sec (30 mph) north on a straight road. The mass of the car is 1,300 kg. A motorcycle passes the car at a speed of 30 m/sec (67 mph). The motorcycle (with rider) has a mass of 350 kg. Calculate and compare the momentum of the car and motorcycle.
You are asked for Kinetic Energy.
You previously found the momentum
Use: 𝐾𝐸= 𝑝 2 2𝑚
Solve for car: p = 17,550 kg m/s
KEcar = ?
Solve for cycle: p = 10,500 kg m/s
KEcycle = ?

7. 12.1 Conservation of Momentum
The law of conservation of momentum states when a system of interacting objects is not influenced by outside forces (like friction), the total momentum of the system cannot change.
If you throw a rock forward from a skateboard, you will move backward in response.

8. 12.1 Conservation of Momentum
To see the relationship, consider two balls
connected by a spring. The balls are motionless and therefore have no momentum.
When you compress the spring, the third law says the balls exert equal forces
(through the springs) in opposite directions on one another, -F1 = F2
Remember from Newton’s second law that the equal and opposite forces create
opposite accelerations, which create opposite velocities.
The accelerations are inversely proportional to the masses, so the velocities are also inversely
proportional to the masses. Heavy objects end up with less velocity and light
objects with more velocity. The velocities caused by the original equal and
opposite forces are exactly as predicted by the law of momentum conservation.

9. 12.1 Collisions in One Dimension
A collision occurs when two or more objects hit each other.
During a collision, momentum is transferred from one object to another.
Collisions can be elastic
or inelastic.

10. 12.1 Collisions

11. Elastic collisions
Two kg billiard balls roll toward each other and collide head-on.
Initially, the 5-ball has a velocity of 0.5 m/s.
The 10-ball has an initial velocity of -0.7 m/s.
The collision is elastic and the 10-ball rebounds with a velocity of 0.4 m/s, reversing its direction.
What is the velocity of the 5-ball after the collision?

12. Elastic collisions
You are asked for 10-ball’s velocity after collision.
You are given mass, initial velocities, 5-ball’s final velocity.
Diagram the motion, use m1v1 + m2v2 = m1v3 + m2v4
Solve for V3 : (0.165 kg)(0.5 m/s) + (0.165 kg) (-0.7 kg)=
(0.165 kg) v3 + (0.165 kg) (0.4 m/s)
v3 = -0.6 m/s

13. Inelastic collisions You are asked for the final velocity.
A train car moving to the right at 10 m/s collides with a parked train car.
They stick together and roll along the track.
If the moving car has a mass of 8,000 kg and the parked car has a mass of 2,000 kg, what is their combined velocity after the collision?
You are asked for the final velocity.
You are given masses, and initial velocity of moving train car.

14. Inelastic collisions
Diagram the problem, use m1v1 + m2v2 = (m1v1 +m2v2) v3
Solve for v3= (8,000 kg)(10 m/s) + (2,000 kg)(0 m/s)
(8, ,000 kg)
v3= 8 m/s
The train cars moving together to right at 8 m/s.

15. 12.2 Force is the Rate of Change of Momentum
Essential Question: How are force and momentum related?

16. 12.2 Force is the Rate of Change of Momentum
Momentum changes when a net force is applied.
The inverse is also true:
If momentum changes, forces are created.
If momentum changes quickly, large forces are involved.

17. 12.2 Force and Momentum Change
The relationship between force and motion follows directly from Newton's second law.
Force (N)
F = D p
D t
Change in momentum
(kg m/sec)
Change in time (sec)

18. Calculating force You are asked for force exerted on rocket.
Starting at rest, an 1,800 kg rocket takes off, ejecting 100 kg of fuel per second out of its nozzle at a speed of 2,500 m/sec. Calculate the force on the rocket from the change in momentum of the fuel.
You are asked for force exerted on rocket.
You are given rate of fuel ejection and speed of rocket
Use F = Δ ÷Δt
Solve: Δ = (100 kg) (-25,000 kg m/s) ÷ (1s) = - 25,000 N
The fuel exerts and equal and opposite force on rocket of +25,000 N.

19. 12.2 Impulse
The product of a force and the time the force acts is called the impulse.
Impulse is a way to measure a change in momentum because it is not always possible to calculate force and time individually since collisions happen so fast.

20. 12.2 Force and Momentum Change
To find the impulse, you rearrange the momentum form of the second law.
Impulse (N•sec)
F D t = D p
Change in
momentum
(kg•m/sec)
Impulse can be expressed in kg•m/sec (momentum units) or in N•sec.

21. Jet Engines
Nearly all modern airplanes use jet propulsion to fly. Jet engines and rockets work because of conservation of linear momentum.
A rocket engine uses the same principles as a jet, except that in space, there is no oxygen.
Most rockets have to carry so much oxygen and fuel that the payload of people or satellites is usually less than 5 percent of the total mass of the rocket at launch.