Momentum

“The Big Mo”

“Momentum’s on our side!”

“Feel the Momentum shift!”

(insert any other sports cliché here)

How is velocity affected by force?

Newton’s 1st law says if no net force acts on an object its velocity will not change

How much the velocity changes depends on two things: The magnitude of the force, and how long the force acts on it

So then what?

Newton’s 2nd law can help us with the math:

F = (m)(a)

Can be rewritten as:

F = (m)(∆v/∆t)

Which can be rearranged to show:

(F)(∆t) = (m)(∆v)

(F)(∆t) = (m)(∆v)

The left side of the equation is called impulse (the product of the magnitude of the force and the time it is applied)

The right side of the equation is called momentum (the product of an object’s mass and velocity)

Taking it one step further…

The equation:(F)(∆t) = (m)(∆v)

Can be rewritten again:

(F)(∆t) = (m2v2) - (m1v1) Momentum is symbolized by “p”, so the equation is

written as:

(F)(∆t) = p2 - p1

This equation is called the impulse-momentum theorem (impulse causes a change in momentum)

Example #1

An 2200 kg SUV traveling at 60 mph (26 m/s) stops in .22 s when it hits a concrete wall. What is the force applied by the wall to stop the SUV?

(F)(∆t) = p2 - p1

F(.22s) = 2200(0) - 2200(26)

F = -260000 N

(the answer makes sense: the force is negative to show it is acting opposite the motion, and it is pretty big)

Oh yeah, one more thing:

Momentum is a conserved quantity What does is mean to be conserved?

The momentum of a system is neither lost nor gained

Meaning, the total momentum of a system remains the same

In math terms, it is written as:

pbefore = pafter

To Use the Conservation of Momentum

Keep in mind that since momentum is dependent on velocity, it is considered a vector quantity

This means direction is important Use your signs correctly in a situation; one

direction is “+”, the opposite direction is “-”

Example #2

A .105 kg hockey puck, moving at 24 m/s, is caught by a 75 kg goalie at rest. What is the speed of the goalie after the catch?

pbefore = pafter

(m)(v)puck + (m)(v)goalie = (m)(v)puck+goalie

(.105)(24) + (75)(0) = (75.105)(x)X = .0336 m/s

(notice the answer makes sense; it is really small, the sign is positive, meaning the goalie will move in the same

direction the puck was traveling)

Example #3:

A 4.00 kg model rocket is launched, shooting 50.0 g of fuel from its exhaust at 625 m/s. What is the velocity of the rocket after the fuel is burned?

pbefore = pafter

(m)(v)fuel + rocket = (m)(v)rocket +(m)(v)fuel

(.050 + 4.00)(0) = (.050)(625) + (4.00)(x)X = -7.81 m/s

(the answer makes sense; the velocity is negative, meaning the rocket moves in the opposite direction of the fuel)

Homework

Chapter 9 Pages 217-221 #15, 24, 25, 31, 36, 40, 42