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
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’