Transcript momentum

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