Transcript Momentum - Conroe High School

“The quality of Motion”
Momentum
•A vector quantity
defined as the product
of an objects mass and
velocity.
• In every day speech, momentum is
often misused- in science,
momentum has a very specific
meaning
• Momentum = mass x velocity
• p = mv
• units are kgm/s
• momentum is a vector quantity that
acts in the same direction as the
velocity
p.209 Practice 6A
Problems 1-3
Delta V
V
• p=m v
• Watch the change in velocity when
there is a change in direction.
•
V = Vf – Vi
• Example; in a car crash, a car going
50mph west then 50mph east has a
of 100mph.
V
Impulse= force x time
• Impulse is what is needed to bring an
object to a stop.
• Impulse is equivalent to momentum
• Look at Newton's 2nd Law
– F = ma F= m (ΔV/ ΔT)
– F Δt = m ΔV
– Impulse= momentum
• The units for impulse are Newton Seconds
– Shouldn’t impulse & momentum share the
same units…?
They do!
• N•S = Kg • m/s
• Kg m/s2 • sec
• Kg • m/s = kg • m/s
• F Δt = Δp
• This is the impulse momentum
theorem. Often the force applied is
NOT constant. If so we will use the
average force over the time interval
Which is worse, 50mph car
crash in a single crash with a
wall, or 2 cars(of the same mass) in
opposite directions going
50mph?
• http://www.youtube.com/watch?v=r8E5
dUnLmh4
• 6min clip
• Car Crash
– Inertia
– Momentum
– Impulse
– air bag vs. windshield
http://www.youtube.com/watch?
v=d7iYZPp2zYY&feature=relat
ed
Airbag Deployment
• Real time
http://www.youtube.com/watch?v=_Av0
WGrlTGY
• Slow motion
http://www.youtube.com/watch?v=A2fAg
W_1nD0
Example Problem
•
A 144g baseball is pitched
horizontally at +38m/s. After it is hit
by a bat it moves horizontally at –
38m/s
a. what impulse did the bat deliver to the
ball?
b. If the bat and ball were in contact .8
milliseconds, what was the average
force the bat exerted on the ball?
c. Find the average acceleration of the
ball during the contact with the bat
A
Given
Formula
m= .144 kg
F Δt = m ΔV
Vi = +38m/s
Ft = m(Vf - Vi)
Vf = -38m/s
t = .0008sec
Solution
Ft = .144 (-38 – (+38)
= .144(-76m/s)
= -10.9NS (in
direction of hit ball)
B
Given
Formula
Δp = -10.9kgm/s
Δp = FΔt
F= Δp / Δt
Solution
F = -10.9/ .0008
F = -13,625N
C
Given
F= -13, 625N
Formula
F= ma
a= F/m
Solution
a= -13,625/ .144
a = -94,618 m/s2
Bungee Jump
Bungee Jump
• Take a look at how stopping time
affects force.
p. 211 Practice 6B
Problems 1-3
p.213 probs 1 & 2
6C
Momentum is a way to describe
collisions
• In all collisions we must define our
system (often this is naming the
objects involved)
• Example: batter hits ball
– we would likely define our system as
the bat and ball
– However, hands, batter body, shoes,
dirt, & air can all affect momentum.
– When we study momentum changes, we
will use a closed, isolated system
Two Kinds of Collisions
• Elastic: objects bounce off each other
after the collision or pass through.
• Inelastic: objects stick together or couple
after they collide
In BOTH collisions momentum is conserved
Another type of interaction is an explosion.
Two or more objects are together and
then they move apart.
Law of conservation of
momentum
• The momentum of any
closed, isolated system, does
NOT change
• P before collision = P after
collision
• Ptot = P’tot
V = 2.2m/s
1000kg
V=0
Before
m = 2000kg
After
V = 1.1m/s
1000kg
Ptotal = P’ total
’ means after collision
PA + PB = P’A + P’B
MA V’A + MB V’B (elastic)
MA VA + MB VB =
(MA+MB ) V’AB (inelastic)
• Look at previous box cars…
MA
VA
MB
VB =
MA
MB
V’AB
1000kg (2.2) + 1000kg (0) = (1000 + 1000) V’
1000(2.2) + 1000 (0)
1000 + 1000
2200 + 0
2000
= V’AB
= V’AB
V’AB = 1.1m/s
Explosion Example
A 3.8kg gun is fired. The 42
gram bullet leaves the muzzle
at 470m/s. Find the guns recoil
velocity.
Recoil – Explosion solution.
Mg x vg = Mb x vb
Vg = .042kg x 470m/s
3.8kg
Vg = 5.2 m/s
Elastic Example problem
• Lab cart A of mass .355kg
moves along with a velocity of
.095m/s. Cart A collides with
cart B of mass .71kg moving in
the same direction with a
velocity of .045m/s. After
collision cart A has velocity of
.035m/s. What is the velocity of
cart B after they collide?
Given
Formula
MA = .355kg
PA + PB = P’A + P’B
VA= .095m/s
MAVA + MBVB = MAV’A + MBV’B
VB = .045m/s
MAVA + MBVB - MAV’A = MBV’B
MB= .71kg
MAVA + MBVB - MAV’A = V’B
V’A= .035m/s
MB
V’B = ?
Solution
elastic
collision
.355(.095) + .71(.045) - .355(.035) = V’B
.71
.0337 + .032 - .0124 = V’B
.71
= .075m/s
Prac Prob #5
• In a hockey game, a 75kg goalie
catches a 105g puck in his leather
glove. The puck was traveling at 48m/s
when he caught it. The goalie was at
rest when he caught it. Calculate his
velocity after the catch.
5
Given
MP =.105 kg
VP= 48m/s
MG= 75kg
VG= 0
Formula
MPVP + MGVG = (MP + MG) VPG
MPVP + MGVG = VPG
MPG
V’P+G = ?
Solution
Inelastic
.105(48) + 0
75.105kg
VPG = .067m/s
Prac. Prob #6
• A bullet was fired into a block of wood.
The bullet embedded into the wood and
moved off with a velocity of 8.6m/s to
the right. Find the bullets velocity
before striking the wood. The bullets
mass is 35 grams and the blocks mass
is 5kg.
Given
Formula
MB= .035kg
MBVB + MWVW = (MB + MW)V’BW
MW= 5kg
VB = (MB + MW)VBW – MWVW
V’BW= 8.6m/s
MB
6
VW= 0
VB=?
Inelastic
Solution
VB = 5.035(8.6) – 5(0)
.035
VB = 1237.2m/s
Prac Prob #7
• Another gun fires a bullet at a wooden
block. This time the bullet goes
through the block. Calculate the
velocity of the wooden block after the
bullet passes through. You know the
bullets mass 35g, blocks mass 2.5kg,
bullets initial speed 475m/s, and bullets
final speed 275m/s.
7
Given
Formula
MB = .035kg
MBVB + MWVW = MBV’B + MWV’W
VB = 475 m/s
MBVB + MWVW – MBV’B = V’W
MW= 2.5kg
MW
V’B= 275 m/s
Solution
VW= 0m/s
.035(475) + 2.5(0) - .035kg(275m/s)
V’W= ?
2.5
elastic
16.63 + 0 – 9.63
2.5
2.8 m/s
Practice Problems p. 219,224,
226
Head-On Collision
Car “Rear Ends” Truck
Truck “Rear-Ends” Car
The Cart and the Brick
Overview
• Know
– Momentum
– Impulse
– Collisions
• Law of Conservation of momentum
• Difference between elastic and inelastic
collisions
• How mass, velocity, force, and time are interrelated.
• Remember old formulas from past chapters.
Momentum =
Impulse
Force – time graph
Force
(N)
Time (sec)
Dr. Weatherly jumps off the high dive (7m) and
splashes into the swimming pool. It was
measured that it took the water .7 seconds to
bring him to a stop underwater. His mass is
90kg.
A- What is his velocity as he hits the water?
B- What is his momentum as he strikes the
water?
C- What is the average force the water puts on
our Principal to bring him to rest underwater?