1. Conservation of Momentum
Contents:
New concept
Example
Whiteboards:

2. Conservation of Momentum - Definition
For a given interaction between objects, the net change in momentum is Zero. Or
Momentum is conserved.
(It don’t change)

3. pA = FBt pB = FAt FB = -FA t = t pA = -pB
Conservation of Momentum - Why is it conserved? A B
pA = FBt pB = FAt
FB = -FA
t = t
pA = -pB
Total p = pA+ pB = 0

4. What is the velocity of their embracing bodies?
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?

5. + Step 1: Draw a picture (decide +/-)
Conservation of Momentum - Using it.
Step 1: Draw a picture (decide +/-) +
4.5 m/s
60 kg
2.3 m/s
80 kg
v = ?
(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)
Step 3: Solve for the unknown

6. + Conservation of Momentum - Using it.
4.5 m/s
60 kg
2.3 m/s
80 kg
v = ?
(80 kg)(2.3 m/s) + (60 kg)(-4.5 m/s) = (140 kg)v
184 kgm/s kgm/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/s
They live happily ever after

7. Whiteboards

8. + 3.2 m/s 1.2 m/s v = ? 5.0 kg 4.0 kg (Stuck together)
Before
After
3.2 m/s
1.2 m/s
v = ?
5.0 kg
4.0 kg
(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

9. + 0 m/s 0 m/s v = ? 3.6 m/s 13 kg 17 kg 13 kg 17 kg
Before
After
0 m/s
0 m/s
v = ?
3.6 m/s
13 kg
17 kg
13 kg
17 kg
0 = (13kg)v + (17kg)(3.6m/s)
0 = (13kg)v kgm/s
-61.2kgm/s = (13kg)v
(-61.2kgm/s)/(13kg) = -4.7 m/s = v
-4.7 m/s

10. + 16.0 m/s 10.0 m/s v = ? 231 g 281 g (Stuck together)
Before
After
16.0 m/s
10.0 m/s
v = ?
231 g
281 g
(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

11. +
Before
After
6.20m/s
1.20 m/s
v = ?
13.0 kg
17.0 kg
13.0 kg
17.0 kg
(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

12. A 3452 kg truck going 23. 0 m/s collides with a 1253 kg car at rest
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

13. A 215 gram bullet going 542 m/s goes through a stationary 3
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

14. A 215 gram bullet going 542 m/s goes through a stationary 3
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 s, what force did the bullet exert on the block?
use mv = Ft:
(3.4 kg)(12m/s) = (F)(.025 s)
F = (3.4 kg)(12m/s)/(.025 s) = 1632 N = 1630 N
1630 N

15. A 2450. kg airplane (including ammunition) going 52. 40 m/s fires a 5
A kg airplane (including ammunition) going m/s fires a kg shell forward at m/s. What is the velocity of the plane afterward?
(2450kg)(52.40m/s) = ( kg)v+(5.12kg)890m/s
128,380kgm/s = ( kg)v kgm/s
123,823.2kgm/s = ( kg)v
(123,823.2kgm/s)/( kg) = m/s
50.65 m/s