Download Unit 8 Momentum 6 lessons - science-b

Do Now:
What makes the shuttle go UP?
Momentum
LESSON ONE
Objectives:
Utilize IMPULSE to calculate:
Force – time – change in velocity
Home work:
Page 233:
#’s 1 – 5 all
Page 250-52:
#’s 47, 51, 55, 60, 64
1
Momentum
ρ = mv
You must know how to do these actions:
1.
Calculate time needed to slow or accelerate an object to a certain velocity using applied Force with
known masses. IMPULSE (F)(Δt) = Δρ
2.
Find velocities of objects that collide in
1.
3.
(stick together, car crash, clay and balls etc.)
Find velocities of objects that collide in
1.
inelastic collisions
elastic collisions (
do NOT stick together, billiard balls)
4.
Using the conservation of momentum, find velocities of objects using just masses, velocities.
5.
Find velocities and directions of objects that collide in two dimensions.
1. IMPULSE
(F)(Δt) = Δρ =
mΔv  Δm Δv
3
Before the collision, the baseball moves toward the
bat. During the collision, the baseball is squashed
against the bat.
After the collision, however, the baseball moves at a
higher velocity away from the bat, and the bat
continues in its path, but at a slower velocity.
How are the velocities of the ball, before and after
the collision, related to the force acting on it?
Newton’s second law describes how the velocity of
an object is changed by a net force acting on it.
The change in velocity of the ball must have been
caused by the force exerted by the bat on the ball.
The force changes over time.
4
Newton’s second law of motion, F = ma, can be
rewritten by using the definition of acceleration as
the change in velocity divided by the time needed to
make that change.
It can be represented by the following equation:
v 
F = ma = m  
t 
Multiplying both sides of the equation by the time
interval, t, results in the following equation:

Ft = mv
Impulse, or Ft, is the product of the average force
on an object and the time interval over which it acts.
Impulse is measured in newton-seconds.
5
IMPULSE
What does
an air bag
do from an
“IMPULSE”
point of
view?
LESSON ONE SUMMARY
6
Using the Impulse-Momentum Theorem
to Save Lives
A large change in momentum occurs only when there is a
large impulse.
A large impulse can result either from a large force acting over a short
period of time or from a smaller force acting over a long period of time.
What happens to the driver when a crash suddenly stops a car?
An impulse is needed to bring the driver’s momentum to zero.
An air bag reduces the force by increasing the time interval during
which it acts.
It also exerts the force over a larger area of the person’s body, thereby
reducing the likelihood of injuries.
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Do Now:
What is the impulse needed
to accelerate
a 5.00 Kg mass
from rest to 10.0 m/s?
Momentum
LESSON TWO
Jan 19
Elastic events
Objectives:
Utilize ELASTIC to calculate:
Force – time – change in velocity
Home work:
Page 240:
#’s 19 – 21 all
Page 250-52:
#’s 50, 70, 72, 73, 74
8
Momentum
Two-Particle Collisions
LESSON TWO
Although it would be simple to consider the bat as a
single object, the ball, the hand of the player, and the
ground on which the player is standing are all objects that
interact when the baseball player hits the ball.
A much easier system is the collision of two balls of
unequal mass and velocity.
During the collision of the two balls, each briefly exerts a
force on the other. Despite the differences in sizes and
velocities of the balls, the forces they exert on each other
are equal and opposite, according to Newton’s third law
of motion.
These forces are represented by the following equation.
FB on A = -FA on B
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Elastic Collisions
Momentum
two masses stay separate LESSON TWO
Kinetic Energy is conserved Jan 19
10
Use the result of Newton’s third law of motion
–FA on B = FB on A.
Momentum
LESSON TWO
Jan 19
PA2 = -FA on B Δt + PA1
Add the momenta of the two balls.
THIS IS IT!!!!
'
v
m1v1 + m2v2 = m1
1
+ m2
'
v
2
This shows that the sum of the momenta of the
balls is the same before and after the collision.
That is, the momentum gained by ball 2 is
equal to the momentum lost by ball 1.
If the system is defined as the two balls, the
momentum of the system is constant.
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What is happening??
Newton’s Cradle: Dominique Toussaint on wikipedia
List ELASTIC “Events”
1.
2.
3.
4.
5.
6.
7.
13
A 4.00 Kg rocket launches from rest.
0.05 Kg of fuel is exhausted @ – 625 m/s for 1.00 sec.
How far up does the rocket travel?
F Δt = Δm Δv
F = Δm Δv / Δt = 3.95 (7.91) / 1.00 = 31.2445 N
Wt = mg = 3.95 * 9.8 = 38.71 N …..
It won’t get off the ground!
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IN-Elastic Event
Momentum
Do Now:
LESSON THREE
Jan 20
Two balls collide.
The 1st ball has a mass of 3.00 Kg and an initial velocity of 2.00 m/s @ 0.00º
The 2nd ball has a mass of 9.00 Kg and an initial velocity of 0.00 m/s
After the collision:
The 1st ball has a velocity of 0.5 m/s @ 0.00º
What is the velocity of the second ball if it too moves off @ 0.00º ?
Objectives:
Utilize IN-ELASTIC to calculate:
Force – time – change in velocity
Home work Page 238:
#’s 13 – 18 all
Page 252-53:
#’s: 76, 77 (add part “b” the bullet imbeds itself in the block)
#’s: 86, 90
15
List IN - ELASTIC “Events”
1.
2.
3.
4.
5.
6.
7.
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In Elastic
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
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THIS IS the “other” IT!!!! Momentum
LESSON THREE
m1v1 + m2v2 = (m1+ m2)v‘3
Jan 23
Homework
Page 255: # 2- 5, 7
Page 243 #22
#1 Two cars collide…and they stick together.
The 1st car has a mass of 1875 Kg and an initial velocity of 23.00 m/s @ 0.00º
The 2nd car has a mass of 1025 Kg and an initial velocity of 17.00 m/s @ 0.00º
After the collision:
What is the velocity of the two cars if they both move off @ 0.00º ?
#2 Two cars collide…and they stick together.
The 1st car has a mass of 1875 Kg and an initial velocity of 23.00 m/s @ 0.00º
The 2nd car has a mass of 1025 Kg and an initial velocity of 17.00 m/s @ 180.0º
After the collision:
What is the velocity (speed and direction) of the two cars?
#3 If the cars in #2 took 1.00 seconds to collide and stick together,
What was the force on the SECOND Car?
What was the acceleration felt by a 60.00 Kg passenger in car number 2?
17
Momentum
LESSON Four
Jan 23
Two-Dimensional Collisions
Up until now, you have looked at momentum in one
dimension only. The law of conservation of
momentum holds for all closed systems with no
external forces. It is valid regardless of the
directions of the particles before or after they
interact.
Now you will look at momentum in two dimensions.
For example, billiard ball A strikes stationary ball
B. Consider the two balls to be the system.
Home work
Page 242:
Example Problem # 4
Page 243:
#’s 22– 25 all
The original momentum of the moving ball is pA1;
the momentum of the stationary ball is zero.
Therefore, the momentum of the system before the
collision is equal to pA1.
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After Collision
After the collision, both balls are moving and have
momenta. According to the law of conservation of
momentum, the initial momentum equals the vector
sum of the final momenta, so pA1 = pA2 + pB2.
The equality of the momentum before and after the
collision also means that the sum of the components
of the vectors before and after the collision must be
equal.
v1'
m1
Therefore, the sum of the final
y-components must be zero.
pA2y + pB2y = 0
m2
v2'
They are equal in magnitude but have opposite
signs. The sum of the horizontal components is also
equal.
pA1 = pA2x + pB2x
19
Example:
A 2.00-kg ball, A, is moving at a speed of 5.00
m/s. It collides with a stationary ball, B, of the same mass.
After the collision, A moves off in a direction 30.0º to the
left of its original direction. Ball B moves off in a
direction 90.0º to the right of ball A's final direction. How
fast are they moving after the collision?
PHET Momentum Billiard Balls
Solution: 1. Sketch the before and after states
After
Before
A
vA1
B
vB1=0
A
vA2
B
vB2
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2. Draw a momentum vector diagram. Note that
pA2 and pB2 form a 90º angle.
pA2
pA1
•
pB1
90º
pB2
30º
p2
3. Perform calculations:
Determine initial momenta:
pA1 = mava1 = (2.00 kg)(5.00m/s) = 10.0 kg•m/s
pB1 = 0
Find p2:
p2 = p1 = 10.0 kg•m/s
21
Find pA2 and pB2.
pA2 = p2 cos 30.0°
pB2 = p2 sin 30.0°
= (10.0 kg•m/s)(0.866)
= (10.0 kg•m/s)(0.500)
= 8.66 kg•m/s
= 5.00 kg•m/s
Find final speeds.
pA2
vA2 = m
A
8.66 kg•m/s
=
2.00 kg
pB2
vB2 =
mB
5.00 kg•m/s
=
2.00 kg
= 4.33 m/s
= 2.50 m/s
22
Two Dimensional IN CLASS

“Initial”ρx


“Initial”ρy



= M1 (V1)(Cos{Θ }) + M2 (V2)(Cos{Θ })
= M1 (V1)(Sin{Θ }) + M2 (V2)(Sin{Θ })
Two Equations two Unknowns
Page 253 # 79
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