Transcript Momentum (p) - Shenendehowa Central Schools

Aim: How can we
explain momentum and
impulse?
Do Now:
Which is easier to do:
Stop a skateboard traveling
at 5 m/s or stop a car
traveling at 5 m/s?
Why?
Which is easier to do:
Stop a bullet fired
from a gun or stop a
bullet that is thrown
at you?
Why?
Momentum (p)
• “Mass in motion”
• Momentum = mass * velocity
• Or p = mv
• Vector quantity
What about the units?
• Units are kg·m/s
What is the momentum of
a 60 kg halfback moving
9 m/s, eastward?
p = mv
p = (60 kg)(9 m/s)
p = 540 kg·m/s
eastward
If a 1 kg ball bounces off of a wall with the
same velocity as shown below, is there a
change in momentum?
v = 10 m/s
v = 10 m/s
Yes because there has been a change in
direction
Momentum is a vector quantity
So how do we define a
change in momentum?
Δp = pf – pi
Δp = mvf – mvi
Δp = m(vf – vi)
Δp = mΔv
Calculate the change in momentum of
the bouncing ball
Δp = mΔv
Δp = (1 kg)(-10 m/s – 10 m/s)
Δp = (1 kg)(-20 m/s)
Δp = -20 kg·m/s
The negative sign indicates a change in
direction
Why do you follow through when:
•Swinging a baseball bat
•Swinging a tennis racket
•Swinging a golf club
•Kicking a football
Impulse (J)
• A force must act on an object
for a time in order to change its
velocity
• Impulse (J) = Force * time
• Or J = Ft
• Vector quantity
What about the units?
• The units are N·s
Calculate the impulse on a
baseball being hit by a baseball
bat with a force of 1200 N over
0.02 s
J = Ft
J = (1200 N)(0.02 s)
J = 24 N·s
Egg Demo
Why doesn’t the egg break
when it hits the bed sheet?

F We
mknow:
*a 
m
*

v v
t
F  ma m
F  m*a  m*
tt

v
or
F

m
or
or
t
F * t  m * v
Ft
v
Ftmm
v
Doesn’t J = Ft?
Doesn’t Δp = mΔv?
So J = Δp
Impulse is a change in momentum!!
Ft  mv
mΔv is a constant
•mass has not changed
•initial and final velocities
have not changed
Time to slow down increased
Therefore force has to decrease
Hence, the egg does not break!
Real-World Applications
•Baseball and tennis player’s
‘following through’
•Boxer’s ‘riding the punch’
•Airbags
•Padded dashboards
F

m
*
a

m
What if force and time are
constant but the mass changes?
Astroblaster Demo
or
Ft  mv
If mass decreases, velocity must
increase!
A car with m=725 kg is moving at 32 m/s to the east.
The driver applies the brakes for 2 s. An average force
of 5.0 x 103 N is exerted on the car. What is the change
in momentum?
Δp = Ft
Δp = (-5 x 103 N)(2 s)
Δp = -1 x 104 N
What is the impulse on the car?
J = Δp
J = -1 x 104 N
What is the car’s final velocity?
Δp = mΔv
-1 x 104 N = (725 kg)(vf – 32 m/s)
-1 x 104 = 725vf – 23,200
vf = 18.2 m/s
An impulse of 30.0 N·s is applied to
a 5.00 kg mass. If the mass had a
speed of 100 m/s before the
impulse, what is its speed after
the impulse?
J = mΔv
30 N·s = (5 kg)(vf – 100 m/s)
30 = 5vf – 500
vf = 106 m/s
A car with a mass of 1.0 x 103 kg is
moving with a speed of 1.4 x 102 m/s.
What is the impulse required to bring
the car to rest?
J = mΔv
J = (1 x 103 kg)(0 m/s – 1.4 x 102 m/s)
J = -1.4 x 105 N·s