aim how can we explain momentum and impulse n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Aim: How can we explain momentum and impulse? PowerPoint Presentation
Download Presentation
Aim: How can we explain momentum and impulse?

Loading in 2 Seconds...

play fullscreen
1 / 19

Aim: How can we explain momentum and impulse? - PowerPoint PPT Presentation


  • 79 Views
  • Uploaded on

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”

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Aim: How can we explain momentum and impulse?' - dot


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
aim how can we explain momentum and impulse

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?

slide2

Which is easier to do:

Stop a bullet fired from a gun or stop a bullet that is thrown at you?

Why?

momentum p
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
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

slide5

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

slide6

So how do we define a change in momentum?

  • Δp = pf – pi
      • Δp = mvf – mvi
      • Δp = m(vf – vi)
      • Δp = mΔv
slide7

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

slide8

Why do you follow through when:

  • Swinging a baseball bat
  • Swinging a tennis racket
  • Swinging a golf club
  • Kicking a football
impulse j
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

slide10

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

slide11

Egg Demo

Why doesn’t the egg break when it hits the bed sheet?

we know
We know:

Doesn’t J = Ft?

Doesn’t Δp = mΔv?

So J = Δp

Impulse is a change in momentum!!

slide13

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
Real-World Applications

Baseball and tennis player’s ‘following through’

Boxer’s ‘riding the punch’

Airbags

Padded dashboards

slide16

What if force and time are constant but the mass changes?

Astroblaster Demo

If mass decreases, velocity must increase!

slide17

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

slide18
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

slide19
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