Conservation of MomentumContents:

•New concept•Example•Whiteboards:•Brennen and the railcars

For a given interaction between objects, the net change in

momentum is Zero.

Or

Momentum is conserved.(It don’t change)

Conservation of Momentum - Definition

pA = FBt pB = FAtFB = -FA

t = t pA = -pB

Total p = pA+ pB = 0

Conservation of Momentum - Why is it conserved?

A B

Conservation of Momentum - Using it.

Example: 60. kg Sally going 4.5 m/s collides head on with 80. kg Bob who is going 2.3 m/s.

What is the velocity of their embracing bodies?

(80. kg)(2.3 m/s) + (60. kg)(-4.5 m/s) = (140. kg)v

Step 2: Write down the momentum of every object. (p = mv)

Conservation of Momentum - Using it.

Step 1: Draw a picture (decide +/-)

+4.5 m/s60 kg

2.3 m/s80 kg v = ?

Step 3: Solve for the unknown

Conservation of Momentum - Using it.

+4.5 m/s60 kg

2.3 m/s80 kg

(80 kg)(2.3 m/s) + (60 kg)(-4.5 m/s) = (140 kg)v184 kgm/s + -270kgm/s = (140 kg)v -86kgm/s = (140 kg)v (-86kgm/s)/(140 kg) = -.61 m/s

They are moving in Sally’s direction (to the left) at .61 m/sThey live happily ever after

v = ?

Whiteboards

Before After+3.2 m/s 1.2 m/s

5.0 kg 4.0 kg

v = ?

(Stuck together)

(5.0kg)(3.2m/s) + (4.0kg)(1.2m/s) = (9.0kg)v 20.8kgm/s = (9.0kg)v

v = (20.8kgm/s)/ (9.0kg) = 2.3 m/s

2.3 m/s

Before After+0 m/s 0 m/s

13 kg 17 kg

v = ?

0 = (13kg)v + (17kg)(3.6m/s) 0 = (13kg)v + 61.2kgm/s -61.2kgm/s = (13kg)v (-61.2kgm/s)/(13kg) = -4.7 m/s = v

-4.7 m/s

3.6 m/s

13 kg 17 kg

Before After+16.0 m/s 10.0 m/s

231 g 281 g

v = ?

(Stuck together)

(.231kg)(16m/s) + (.281kg)(-10m/s) = (.512kg)v .886 kgm/s = (.512kg) v (.886 kgm/s)/ (.512kg) = 1.73 m/s

1.73 m/s

Before After+6.20m/s

13.0 kg 17.0 kg

v = ?

(13kg+17kg)(6.2m/s) = (13kg)(-1.2m/s)+(17kg)v 186kgm/s = -15.6kgm/s+(17kg)v 201.6kgm/s = (17kg)v (201.6kgm/s)/(17kg) = 11.9 m/s = v

11.9 m/s

1.20 m/s

13.0 kg 17.0 kg

A 3452 kg truck going 23.0 m/s collides with a 1253 kg car at rest. What is the velocity of the wreckage just after the collision?

(3452kg)(23.0m/s)=(3452kg+1253kg)v 79396kgm/s = (4705kg)v(79396kgm/s)/(4705kg) = 16.9 m/s

16.9 m/s

A 215 gram bullet going 542 m/s goes through a stationary 3.40 kg block of wood. After this, the block is traveling at 12.0 m/s in the same direction, (a) what is the bullet’s velocity?

(. 215kg)(542 m/s) = (3.4kg)(12m/s) + (.215kg)v 116.53kgm/s = 40.8kgm/s + (.215kg)v 75.73kgm/s = (.215kg)v(75.73kgm/s)/(.215kg) = 352 m/s

352 m/s

A 215 gram bullet going 542 m/s goes through a stationary 3.40 kg block of wood. After this, the block is traveling at 12 m/s in the same direction, (b) if ripping through the block took .0250 s, what force did the bullet exert on the block?

use mv = Ft:(3.4 kg)(12m/s) = (F)(.025 s)F = (3.4 kg)(12m/s)/(.025 s) = 1632 N = 1630 N

1630 N

A 2450. kg airplane (including ammunition) going 52.40 m/s fires a 5.120 kg shell forward at 890.0 m/s. What is the velocity of the plane afterward?(2450kg)(52.40m/s) = (2444.88 kg)v+(5.12kg)890m/s 128,380kgm/s = (2444.88 kg)v+4556.8kgm/s 123,823.2kgm/s = (2444.88 kg)v(123,823.2kgm/s)/(2444.88 kg) = 50.65 m/s

50.65 m/s

A 153 gram bullet going 452 m/s goes through the first of two stationary 3.50 kg blocks of wood, and sticks in the second. After this, the first block is traveling at 6.50 m/s in the same direction, what speed are the second block and bullet going?(.153kg)(452m/s) = (3.5kg)(6.5m/s)+(.153kg+3.5kg)v 69.156kgm/s = 22.75kgm/s + (3.653kg)v 47.406kgm/s = (3.653kg)v(47.406kgm/s)/(3.653kg) = 12.7 m/s12.7 m/s

A 153 gram bullet going 452 m/s goes through the first of two stationary 3.50 kg blocks of wood, and sticks in the second. After this, the first block is traveling at 6.50 m/s in the same direction, What is the bullet’s velocity between the blocks?(.153kg)(452m/s) = (3.5kg)(6.5m/s)+(.153kg)v 69.156kgm/s = 22.75kgm/s + (.153kg)v 46.406kgm/s = (.153kg)v(46.406kgm/s)/(.153kg) = 303.31 = 303 m/s

303 m/s

A 153 gram bullet going 452 m/s goes through two stationary 3.50 kg blocks of wood. After this, the first block is traveling at 6.50 m/s and the second is going 8.30 m/s in the same direction. What is the bullet’s velocity after this?(.153kg)(452m/s)= (3.5kg)(6.5m/s)+(3.5kg)(8.3m/s)+(.153kg)v

69.156kgm/s = 22.75kgm/s+29.05kgm/s+(.153kg)v17.356kgm/s = (.153kg)v(17.356kgm/s)/(.153kg) = 113.44 = 113 m/s

113 m/s

A 153 gram bullet going 452 m/s is heading toward two stationary 3.50 kg blocks of wood. What is the velocity if the bullet sticks in the first block, and the two blocks then slide and stick together?(.153kg)(452m/s)= (3.5kg+3.5kg+.153kg)v69.156kgm/s = (7.153kg)v(69.156kgm/s)/(.153kg) = 9.668 = 9.67 m/s

9.67 m/s

60. kg Brennen is playing on two flatbed rail cars initially at rest. Car A has a mass of 560 kg and B 780 kg. He reaches a velocity of +5.2 m/s on A, before jumping to B where he slows to +3.4 m/s before jumping off the other end. The cars are uncoupled, and rest on a frictionless track1. What is the velocity of car A when he is in midair?

0 = (60kg)(+5.2 m/s) + (560 kg)vv = -.557 m/s

-.56 m/s

A B

60. kg Brennen is playing on two flatbed rail cars initially at rest. Car A has a mass of 560 kg and B 780 kg. He reaches a velocity of +5.2 m/s on A, before jumping to B where he slows to +3.4 m/s before jumping off the other end. The cars are uncoupled, and rest on a frictionless track2. What is the velocity of car B when he leaves it?

(60 kg)(5.2 m/s) + 0 = (60 kg)(3.4 m/s) + (780 kg)vv = +.138 m/s

+.14 m/s

A B

60. kg Brennen is playing on two flatbed rail cars initially at rest. Car A has a mass of 560 kg and B 780 kg. He reaches a velocity of +5.2 m/s on A, before jumping to B where he slows to +3.4 m/s before jumping off the other end. The cars are uncoupled, and rest on a frictionless track3. What would have been the velocity of car B had he remained there, and not jumped off?

+.37 m/s

A B

(60 kg)(5.2 m/s) + 0 = (60 kg+ 780 kg)vv = +.371 m/s

60. kg Brennen is playing on two flatbed rail cars initially at rest. Car A has a mass of 560 kg and B 780 kg. He reaches a velocity of +5.2 m/s on A, before jumping to B where he slows to +3.4 m/s before jumping off the other end. The cars are uncoupled, and rest on a frictionless track4. What would the velocity of car B have been had he jumped off the back of it to give himself a velocity of zero?

(60 kg)(5.2 m/s) + 0 = (60 kg)(0 m/s)+ (780 kg)vv = +.40 m/s

+.40 m/s

A B