Momentum Physics 2015

Physics Definition : Linear momentum of an object of mass (m) moving with a velocity (v) is defined as the product of the mass and the velocity. Momentum is represented by the symbol (p).

Linear Momentum

Important Info.

Momentum – • The product of the mass and velocity of an

object• Is a vector quantity: has magnitude and

direction• Is “moving mass” or “inertia in motion”• Depends on both mass and velocity

The Equation:

p = mv

The faster an object is moving, we often express the object as :

“its picking up speed” or

“its gaining momentum”

Relationship

Relationship

http://www.youtube.com/watch?v=y2Gb4NIv0Xg

A 2250 kg pickup truck has a velocity of 25 m/s to the east. What is the momentum?

p = m = v =

Let’s solve a problem

2250 kg

25 m/s east p = mv

56250 kg * m/s east

Relationships• If you double the mass• If you double the

velocity• If at rest (velocity is 0)

• Then double momentum• Then double momentum• Then momentum is zero

p = mvhttp://www.youtube.com/watch?v=hTZI-kpppuw

Physics Definition of: The product of the average net force exerted on an object and the time interval over which the force acts.

Impulse

http://www.youtube.com/watch?v=cuWTZDne918

(I ) I=Ft

How hard is it to stop a moving object?

To stop an object, we have to apply a force over a period of time.

This is called Impulse

Impulse (I) = FΔt Units: N s∙

F = force (N)Δt = time elapsed (s)

http://www.youtube.com/watch?v=DU7HDgv8Zzghttp://www.youtube.com/watch?v=bERoaOSoGK8

How is velocity affected by force?

Impulse causes change in momentum

F = maF = m ( ∆v/∆t)

F∆t = m∆v

N2L

F ∆t = p2 – p1

Impulse – Momentum Theorem

Large change in momentum occurs only when there is a large impulse.

Large impulses can occur when:1. A large force over a short period of

timeor

2. A small force over a long period of time

Saves lives

A 1400 kg car moving westward with a velocity of 15 m/s collides with a utility pole and is brought to rest in 0.3 seconds. Find the force exerted on the car during the collision.

F = m = v1 =

v2 =

t=

Let’s solve a problem

1400 kg15 m/s west0 m/s west0.3 s

F = (mv2 – mv1)/∆t7.0 X104 N east

Physics Definition: Law of conservation of momentum- the total p of all objects interacting with one another remains constant regardless of the nature of the forces between the object.

Momentum is Conserved

Collisions

• Collisions are when two or more objects run into each other.– They can stick together – Spring back apart

p before a collision = p after a collision

Momentum is conserved for objects pushing away from each other.

Momentum is conversed in collisions.

1. Elastic Collisions - Bounces off

2. Inelastic Collisions - Sticks together

Collisions

Elastic Collisions

• No permanent deformation • Energy and momentum is conserved• Ex: Billiard ball colliding with another

Inelastic Collision

• Objects sticks together• Momentum is conserved • Energy is not conserved• Ex: a ball of putty hitting and sticking to

another ball• Ex: Two railroad cars colliding and coupling

together

Partially Elastic Collision

• There is some deformation• They do not stick together• Automobile collision is a good example

(mathematically), a partially elastic collision is handled the same way as an elastic collision except that energy is not conserved.

Conservation of Momentum Equation

m1v1 +m2v2 = m1v1’ + m2v2’

• Note: regardless of the type of collision, you will most likely use momentum to solve it!

• Each object has a velocity before and after the collision but their masses remain the same.

Let’s do some word problems first…

1. Ping-pong ball striking a pool ball

ELASTIC COLLISION

#2

• Collision between two air-hockey pucks on a frictionless surface

• http://physics.doane.edu/physicsvideolibrary/default.html#momentum

Elastic Collision

https://www.youtube.com/watch?v=qNou0xg3_cYhttps://www.youtube.com/watch?v=4IYDb6K5UF8

Inelastic CollisionsIn any case where objects stick together they will both have the

same velocity and will end up sharing the momentum of the

first object. m1v1+m2v2 = (m1+m2)v’

Let’s Calculate Now

• A 4kg block with an initial velocity of 10 m/s that is colliding with an 6 kg block that is stationary. After the collision, the 6 kg block is seen to be moving 8 m/s. Your job is to determine the velocity of the 4 kg block after the collision.

4 kg 6 kg 4 kg 6 kg

V1 = 10 m/s V2 = 0 m/s V1 ‘ = ? m/s V2 ‘ = 8 m/s

1st let’s start with the collision equation

• m1v1 +m2v2 = m1v1’ + m2v2’

• m1= 4 kg

• m2 = 6 kg

• v1 = 10 m/s

• v2 = 0 m/s

• v1’ = ?

• v2’ = 8 m/s

4(10) + 6(0) = 4(v1’) + 6(8)40 + 0 = 4(v1’) + 48

-8 = 4(v1’) -2 = v1’

Another one….

• A 4 kg block with an initial velocity of 10 m/s is colliding with an 6 kg block that is stationary. After the collision, both blocks are stuck together and are moving together. Your job is to determine the velocity of the linked blocks after the collision.

4 kg 6 kg 4 kg 6 kg

V1 = 10 m/s V2 = 0 m/sV‘ = ? m/s

Again start with the collision equation

• m1v1 +m2v2 = (m1+ m2)v’

• m1= 4 kg

• m2 = 6 kg

• v1 = 10 m/s

• v2 = 0 m/s• v’ = ?

4(10) + 6(0) = 4 + 6(v’) 40 + 0 = 10(v’)

4 = v’