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 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 momentum is zero p = mv

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

How is velocity affected by force? Impulse causes change in momentum

F = ma F = m ( ∆v/∆t) F∆t = m∆v N2L

F ∆t = p 2 – p 1 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 time or 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 = v 1 = v 2 = t= Let’s solve a problem 1400 kg 15 m/s west 0 m/s west 0.3 s F = (mv 2 – mv 1 )/∆t 7.0 X10 4 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 m 1 v 1 +m 2 v 2 = m 1 v 1 ’ + m 2 v 2 ’ 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 default.html#momentum default.html#momentum Elastic Collision 3_cY UF8

Inelastic Collisions In any case where objects stick together they will both have the same velocity and will end up sharing the momentum of the first object. m 1 v 1 +m 2 v 2 = (m 1 +m 2 )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 kg6 kg4 kg6 kg V 1 = 10 m/s V 2 = 0 m/s V 1 ‘ = ? m/sV 2 ‘ = 8 m/s

1 st let’s start with the collision equation m 1 v 1 +m 2 v 2 = m 1 v 1 ’ + m 2 v 2 ’ m 1 = 4 kg m 2 = 6 kg v 1 = 10 m/s v 2 = 0 m/s v 1 ’ = ? v 2 ’ = 8 m/s 4(10) + 6(0) = 4(v 1 ’) + 6(8) = 4(v 1 ’) = 4(v 1 ’) -2 = v 1 ’

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 kg6 kg4 kg6 kg V 1 = 10 m/s V 2 = 0 m/s V‘ = ? m/s

Again start with the collision equation m 1 v 1 +m 2 v 2 = (m 1 + m2)v’ m 1 = 4 kg m 2 = 6 kg v 1 = 10 m/s v 2 = 0 m/s v’ = ? 4(10) + 6(0) = 4 + 6(v’) = 10(v’) 4 = v’