Transcript Momentum and Collisions - PHYSICS I PRE-AP

Momentum and Collisions
Chapter 6
International regulations specify
the mass of official soccer balls.
 How does the mass of a ball affect the way
it behaves when kicked?
 How does the velocity of the player’s foot
affect the final velocity of the ball?
Knowledge to Expect
 Energy cannot be created or destroyed, but
only changed from one form to another.
 An object that is not being subjected to a
force will continue to move at constant
speed in a straight line.
Knowledge to Review
 A force on an object is a push or pull that tends to
cause a change in motion. Forces can be field or
contact forces.
 Newton's laws of motion describe the effects of
forces on objects and the idea that forces always
exist in pairs.
 Energy of motion, called kinetic energy, depends
on the mass and speed: KE=1/2mv2
 Energy is neither created nor destroyed, but it can
be converted from one form to another.
Momentum
A vector quantity defined as the
product of an object's mass and
velocity.
Equation
 “p” is the letter used to denote
momentum. It comes from Leibniz’s
use of the term progress, defined as the
“the quantity of motion which which a
body proceeds in a certain direction.”
 Momentum = mass x velocity
 p=mv
Who was Leibniz?
 Leibniz was one of the great polymaths of the
modern world.
 As an engineer he worked on calculating
machines, clocks and even mining machinery.
 As a librarian he more or less invented the modern
idea of cataloguing.
 As a mathematician he came up with the calculus
independently of (though a few years later) than
Newton, and his notation has become the
standard.
 As a physicist he made advances in mechanics,
specifically the theory of momentum.
Momentum Unit
kg·m/s
Momentum is a vector
quantity.
It has magnitude and
direction
Practice
 A 2250 kg pickup truck has a velocity of 25
m/s to the east. What is the momentum of
the truck?
 M=2250 kg v=25 m/s p=?
 P=2250 kg x 25 m/s
 P= 5.6 x 104 kg·m/s
Momentum is closely related to
force.
A change in momentum takes force
and time.
A change in momentum is caused
by an impulse.
What is Impulse?
 Impulse is the product of the force and the
time over which it acts on an object for a
constant force.
 Impulse (J) is force x the change in time.
 J=Ft
Impulse –Momentum Theorem
Impulse is equal to the change
in momentum.
J= p
p= mvf - mvi
Ft= mv
v= Vf - Vi
Practice
 A 1400 kg car moving westward with a velocity of
15 m/s collides with a utility pole and is brought to
rest in 0.30 s. Find the magnitude of the force
exerted on the car during the collision.
 M= 1400 kg
t= 0.30 s vf= 0 m/s
 vi= 15 m/s
F=?
 Ft= mvf - mvi
 F= 1400 kg(0m/s) – 1400 kg (15)
0.30s
F=-7.0 x 104 N to the east
Impulse-Momentum Theory
 A change in momentum over a longer time
requires less force.
 This relationship is used to design safety
equipment
– Trampolines
– Nets
– Safety belts
Impulse-Momentum Theory
Ft Large force with small impact
time….much damage
f
TSmall force with large impact
time….less damage
Ft
= P
Ft = M1Vf - M1Vi
Ft = M2Vf – M2Vi
The Law of Conservation of
Momentum
 When two objects interact the sum of each
of their individual momentums BEFORE
the interaction will EQUAL the sum of their
individual momentums AFTER the
interaction.
When is momentum conserved?
 Elastic Collisions
 Objects pushing away from each other.
 PA + PB = PA’ + PB’
 TOTAL INITIAL MOMENTUM =
TOTAL FINAL MOMENTUM
PRACTICE
 A 76 KG BOATER, INITIALLY AT REST
IN A STATIONARY 45 KG BOAT, STEPS
OUT OF THE BOAT AND ONTO THE
DOCK. IF THE BOATER MOVES OUT
OF THE BOAT WITH A VELEOCTIY OF
2.5 M/S TO THE RIGHT, WHAT IS THE
FINAL VELOCITY OF THE BOAT?
 PA + PB = PA’ + PB’
SOLUTION
 M1 = 76 KG
M2 = 45 KG
 V1 = 0
V2 = 0
 V1’ = 2.5 M/S
V 2’ = ?
 PA + PB = PA’ + PB’
 M1V1 + M2V2 = M1’V1’ + M2’V2’
 76 X 0 + 45 X 0 = 76 X 2.5 + 45 X V2’
 0 = 190 +45V2’
 -4.2 = V2’
Two Types of Collisions
 Elastic – mechanical energy is conserved
 M1V1 + M2V2 = M1’V1’ + M2’V2’
 Inelastic – some mechanical energy is
transformed into other forms of energy there
fore mechanical energy is not conserved.
 M1V1 + M2V2 = M1’V1’ + M2’V2’
 M1V1 + M2V2 = (M1’ + M2’)Vf ’
COLLISION FORCES ARE NOT
CONSTANT
 FORCES VARY WITH TIME
THROUGHOUT THE COLLISION.
 FORCES ARE ALWAYS EQUAL AND
OPPOSITE DURING THE COLLISION.
PERFECTLY INELASTIC
COLLISION
 WHEN TWO OBJECTS COLLIDE AND
MOVE TOGETHER AS ONE MASS
 A NEARLY PERFECT INELASTIC
COLLISION IS WHEN ONE OBJECT
COLLIDES WITH ANOTHER OBJECT
AND STOPS WITHIN THE OBJECT IT
STRUCK.
 M1V1 + M2V2 = (M1’ + M2’)Vf ’
PRACTICE
 A 1850 kg LUXURY SEDAN STOPPED AT A
TRAFFIC LIGHT IS STRUCK FROM THE
REAR BY A COMPACT CAR WITH A MASS
OF 975 KG. THE TWO CARS BECOME
ENTANGLED AS A RESULT OF THE
COLLISION. IF THE COMPACT CAR WAS
MOVING AT A VELOCITY OF 22.0 M/S TO
THE NORTH BEFORE THE COLLISION,
WHAT IS THE VELOCITY OF THE
ENTANGLED MASS AFTER THE
COLLISION?
SOLUTION
 M1 = 1850 KG
 V1 = 0
M2 = 975 KG
V2 = 22 M/S
 V’ = ?
 M1V1 + M2V2 = (M1’ + M2’)Vf ’
 1850 X 0 + 975 X 22 = (1850 + 975) V’
 7.59 M/S TO THE NORTH = V’
ENERGY
 KINETIC ENERGY IS NOT CONSTANT
IN INELASTIC COLLISIONS
 KE = ½ MV2
 KEi = KE1i + KE2i
 KEf = KE1f + KE2f
 ∆KE = KEf – KEi
PRACTICE
 TWO CLAY BALLS COLLIDE HEAD-ON IN A
PERFECTLY INELASTIC COLLISION. THE
FIRST BALL HAS A MASS OF .500 KG AND
AN INTIAL VELCITY OF 4.00 M.S TO THE
RIGHT. THE MASS OF THE SECOND BALL
IS .250 KG, AND IT HAS AN INITIAL
VELCOTIY OF 3.00 M/S TO THE LEFT. WHAT
IS THE FINAL VELOCITY OF THE
COMPOSITE BALL OF CLAY AFTER THE
COLLISISION?
SOLUTION
 M1 = .500KG
M2 = .250 KG
 V1 = +4.00 M/S
V2 = -3.00 M/S
 V’ = ?
∆KE = ?
 M1V1 + M2V2 = (M1’ + M2’)Vf ’
 .500 X 4.00 + .250 X -3.00 = (.500 + .250) Vf’
 1.67 M/S = Vf ’
WHAT IS THE DECREASE IN
KINETIC ENERGY DURING THE
COLLISION?








KEi = KE1i + KE2i
KEi = 1/2(.500)(4.OO)2 + 1/2(.250)(-3.00)2
KEi = 5.12J
KEf = KE1f + KE2f
KEf = 1/2(.500 + .250)(1.67)2
KEf = 1.05 J
∆KE = 1.05J – 5.12J
∆KE = -4.07
KINETIC ENERGY IS
CONSERVED IN ELASTIC
COLLISIONS
 MOMENTUM AND KINETIC ENERGY
REMAINS CONSTANT IN AN ELASTIC
COLLISION.
 M1V1i + M2V2i = M1V1’ + M2V2’
 ½ M1V1i2 + ½ M2V2i2 = ½ M1V1’2 + ½
M2V2’2
PRACTICE
 A .015 KG MARBLE MOVING TO THE RIGHT AT .225
M/S MAKES AN ELASTIC HEAD ON COLLION WITH
A .030 KG SHOOTER MARBLE MOVING TO THE
LEFT AT .180 M/S. AFTER THE COLLISION, THE
SMALLER MARBLE MOVES TO THE LEFT AT .315
M/S. ASSUME THAT NEITHER MARBLE ROTATES
BEFORE OR AFTER THE COLLISION AND THAT
BOTH MARBLES ARE MOVING ON A
FRICITONLESS SURFACE. WHAT IS THE VELOCITY
OF THE .030 KG MARBLE AFTER THE COLLISION?
SOLUTION
 M1V1 + M2V2 = M1’V1’ + M2’V2’
 MAKE SURE YOU GIVE DIRECTIONS
BY POSITIVE AND NEGATIVE.
 V2; = .09 M/S TO THE RIGHT