1 / 21

# Momentum and Impulse - PowerPoint PPT Presentation

Momentum and Impulse. Principles of Physics. Momentum. - property of an object related to its mass and velocity. - “mass in motion” or “inertia in motion” p = momentum (vector) p = mv m = mass (kg) v = velocity (m/s) (vector) units for momentum are kg m/s

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

## PowerPoint Slideshow about ' Momentum and Impulse' - chandler-wiley

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

### Momentum and Impulse

Principles of Physics

- property of an object related to its mass and velocity.

- “mass in motion” or “inertia in motion”

p = momentum (vector)

p = mv m = mass (kg)

v = velocity (m/s) (vector)

units for momentum are kg m/s

**If an object is not moving, it has no momentum

p = mv

Example: Calculate the momentum of a 250 kg cart with a velocity of 25 m/s.

p = mv

p = 250 kg (25 m/s)

p = 6250 kg m/s

p = m1v1 + m2v2 + m3v3 + …

Example: A 300 kg car is travelling to east at 45 m/s and a 500 kg truck is travelling west at 30 m/s. Calculate the total momentum of the system.

p = m1v1 + m2v2

p = 300 kg(45 m/s) + 500 kg(-30 m/s)

p = -1500 kg m/s

• Conservation – to keep constant even though changes occur

• whatever you had before an event you will still have after the event

The total momentum of a system is the same before and after a collision

ptotal before = ptotal after

ptotal before = ptotal after

ptotal before= m1v1i + m2v2i + m3v3i + …

**remember v has direction

ptotal after = m1v1f + m2v2f + m3v3f + …

Elastic - bounce

• objects hit and bounce off from each other

Inelastic – stick

• multiple objects hit and stick together

or

• one objects separates into 2 or more (explosion)

A cue ball with mass 0.050 kg moves at a velocity of 0.20 m/s. It collides with the 8 ball (0.050 kg) moving in the same direction with velocity 0.10 m/s. After the collision, the cue ball is moving with a velocity of 0.08 m/s in the same direction. What is the 8 ball’s final velocity?

Write down given information

Givens:

m1 = 0.050 kg

m2 = 0.050 kg

v1i= 0.20 m/s

v2i= 0.10 m/s

v1f = 0.08 m/s

v2f = ?

A cue ball with mass 0.050 kg moves at a velocity of 0.20 m/s. It collides with the 8 ball (0.050 kg) moving in the same direction with velocity 0.10 m/s. After the collision, the cue ball is moving with a velocity of 0.08 m/s in the same direction. What is the 8 ball’s final velocity?

Determine the momentum of the cue ball before and after the collision

Givens:

m1 = 0.050 kg

m2 = 0.050 kg

v1i= 0.20 m/s

v2i= 0.10 m/s

v1f = 0.08 m/s

v2f = ?

p1i = m1v1i

= 0.050 kg(0.20 m/s)

= 0.01 kg m/s

p1f = m1v1f

= 0.050 kg(0.08 m/s)

= 0.004 kg m/s

A cue ball with mass 0.050 kg moves at a velocity of 0.20 m/s. It collides with the 8 ball (0.050 kg) moving in the same direction with velocity 0.10 m/s. After the collision, the cue ball is moving with a velocity of 0.08 m/s in the same direction. What is the 8 ball’s final velocity?

Determine the momentum of the 8 ball before the collision

Givens:

m1 = 0.050 kg

m2 = 0.050 kg

v1i= 0.20 m/s

v2i= 0.10 m/s

v1f = 0.08 m/s

v2f = ?

p2i = m2v2i

= 0.050 kg(0.10 m/s)

= 0.005 kg m/s

A cue ball with mass 0.050 kg moves at a velocity of 0.20 m/s. It collides with the 8 ball (0.050 kg) moving in the same direction with velocity 0.10 m/s. After the collision, the cue ball is moving with a velocity of 0.08 m/s in the same direction. What is the 8 ball’s final velocity?

Set up a conservation equation

pbefore = pafter

p1i + p2i = p1f+ p2f

A cue ball with mass 0.050 kg moves at a velocity of 0.20 m/s. It collides with the 8 ball (0.050 kg) moving in the same direction with velocity 0.10 m/s. After the collision, the cue ball is moving with a velocity of 0.08 m/s in the same direction. What is the 8 ball’s final velocity?

Substitute momentum values into the conservation equation and solve for the momentum of the 8 ball after the collision

pbefore = pafter

p1i + p2i = p1f+ p2f

0.01 kg m/s + 0.005 kg m/s = 0.004 kg m/s + p2f

p2f = 0.011 kg m/s

A cue ball with mass 0.050 kg moves at a velocity of 0.20 m/s. It collides with the 8 ball (0.050 kg) moving in the same direction with velocity 0.10 m/s. After the collision, the cue ball is moving with a velocity of 0.08 m/s in the same direction. What is the 8 ball’s final velocity?

Determine the final velocity of the 8 ball

p2f = m2v2f

0.011 kg m/s = 0.050 kg (v2f)

v2f = 0.22 m/s

A 0.015 kg bullet is shot into a 5.085 kg wood block at rest on a frictionless surface. The block and the bullet acquire a speed of 1 m/s as a result of the collision. Calculate the initial velocity of the bullet.

Write down given information

Givens:

mb = 0.015 kg

mbb = 5.085 kg

vbi= ?

vbbi= 0 m/s

vbf = 1 m/s

vbbf = 1 m/s

A 0.015 kg bullet is shot into a 5.085 kg wood block at rest on a frictionless surface. The block and the bullet acquire a speed of 1 m/s as a result of the collision. Calculate the initial velocity of the bullet.

Determine the momentum of the block before and after the collision

p2i= m2v2i

= 5.085 kg(0 m/s)

= 0 kg m/s

Givens:

m1= 0.015 kg

m2= 5.085 kg

v1i= ?

v2i= 0 m/s

v1f= 1 m/s

v2f= 1 m/s

p2f= m2 v2f

= (5.085 kg) (1 m/s)

= 5.085 kg m/s

A 0.015 kg bullet is shot into a 5.085 kg wood block at rest on a frictionless surface. The block and the bullet acquire a speed of 1 m/s as a result of the collision. Calculate the initial velocity of the bullet.

Determine the momentum of the bullet after the collision

Givens:

m1= 0.015 kg

m2= 5.085 kg

v1i= ?

v2i= 0 m/s

v1f= 1 m/s

v2f = 1 m/s

p1f= (m1)v1f

= (0.015 kg) (1 m/s)

= 0.015 kg m/s

A 0.015 kg bullet is shot into a 5.085 kg wood block at rest on a frictionless surface. The block and the bullet acquire a speed of 1 m/s as a result of the collision. Calculate the initial velocity of the bullet.

Set up a conservation equation

pbefore = pafter

p1i+ p2i = p1f + p2f

A 0.015 kg bullet is shot into a 5.085 kg wood block at rest on a frictionless surface. The block and the bullet acquire a speed of 1 m/s as a result of the collision. Calculate the initial velocity of the bullet.

Substitute the momentum values and determine the momentum of the bullet before the collision

pbefore = pafter

p1i+ p2i = p1f + p2f

p1i+ 0 = 0.015 kg m/s + 5.085 kg m/s

p1i = 5.1 kg m/s

A 0.015 kg bullet is shot into a 5.085 kg wood block at rest on a frictionless surface. The block and the bullet acquire a speed of 1 m/s as a result of the collision. Calculate the initial velocity of the bullet.

Determine the initial velocity of the bullet

p1i= m1v1i

5.1 kg m/s = 0.015 kg(v1i)

v1i = 340 m/s