Download Section 9.1 Impulse and Momentum

Mr. Borosky
Physics Section 9.1 Notes
Page 1 of 3
Chapter 9 Momentum and Its Conservation
In this chapter you will:
Describe momentum and impulse and apply them to the
interactions between objects.
Relate Newton’s third law of motion to conservation of
momentum.
Explore the momentum of rotating objects.
Sections
Section 9.1: Impulse and Momentum
Section 9.2: Conservation of Momentum
Section 9.1 Impulse and Momentum
Objectives
Define the momentum of an object.
Determine the impulse given to an object.
Define the angular momentum of an object.
Read intro paragraph p. 229
IMPULSE AND MOMENTUM
I suggest you read chapter 9 from the old book.
Read section.
F = ma
a = Δv / Δt
F = m(Δv / Δt)
Multiply both sides by Δt
FΔt = mΔv
Impulse – is the product of force and time interval over which it
acts. It is a vector quantity in the direction of the force. It is
measured in units Newton seconds (N*s). If force varies with time
it is found by determining the Area under the curve of a Force Time
Graph as in Figure 9-1.
Physics Principals and Problems © 2005 Started 2006-2007 School Year
Mr. Borosky
Physics Section 9.1 Notes
Page 2 of 3
Momentum - product of an object’s mass and velocity. It is a vector
quantity that has the same direction as the velocity of the object.
It is denoted by “p” (lower case). It is mass in motion. The unit
for Momentum is kg m/s. To find Momentum you use
p = m*v
Δp = mΔv
FΔt = mΔv = mvF – mvI = Δp = pF – pI
1 N*s = 1 kg*m/s
Impulse Momentum Theorem – states impulse given to an object is
equal to its change in momentum. Or the impulse of an object is
equal to the object’s Final Momentum Minus its Initial Momentum.
Thus Ft = p or FΔt = pF – pI
If the force is constant the Impulse is the Product of the Force
multiplied by the time interval over which it acts.
If Force is not constant then the Impulse is found using the Average
Force multiplied by the time interval or by finding the area under
the curve of a Force Time Graph.
USING THE IMPULSE MOMENTUM THEOREM
Read Section.
Go over baseball example p. 231
USING THE IMPULSE MOMENTUM THEOREM TO SAVE LIVES
Read Section.
A large impulse results from a large force over a short period of
time or a small force over a long period of time.
An airbag reduces the force by increasing the time interval during
which it acts.
Example Problem 1 p. 232
a) FΔt = mΔv
b) FΔt = mΔv
F(21) = 2200(-26)
F(3.8) = 2200(-26)
21F = -57200
3.8F = -57200
F = -2,723.81 N
F = -15,052.63 N
c) FΔt
F(.22)
.22F =
F =
= mΔv
= 2200(-26)
-57200
-260,000 N
Physics Principals and Problems © 2005 Started 2006-2007 School Year
Mr. Borosky
Physics Section 9.1 Notes
Page 3 of 3
Do Practice Problems p. 233 # 1-5
ANGULAR MOMENTUM
Read Section.
τ = IΔω / Δt
τΔt = IΔω
Angular Impulse – is the product of the Torque and the time
interval. It equals τΔt.
Angular Momentum – is the product of the Moment of Inertia and the
Angular Velocity. It equals IΔω. It is denoted by “L”.
Angular Impulse Angular Momentum Theorem – states the angular
impulse on an object is equal to the object’s final angular momentum
minus the object’s initial angular momentum.
If there are no torques acting on an object then its angular
momentum is constant.
Do 9.1 Section Review p. 235 # 6-9 (Skip 10-12)
Physics Principals and Problems © 2005 Started 2006-2007 School Year