1. Momentum and Collisions
Chapter 6

2. Chapter Organization Momentum of a single object
Momentum between objects or collisions

3. Single Object Momentum
Momentum (r) = mv
Momentum (r) – kg m/s or N s
Mass (m) – kg
Velocity (v) – m/s

4. Change in Momentum Change in momentum = D r r = m D v r = m(vf –vi)
r = mvf – mvi
r = rf - ri
D r = Impulse = F t

5. Single Object Momentum-EX
A car weighing 15,680 N and moving at 20.0 m /s is acted upon by 640 N force until it is brought to a halt.
What is the mass and weight of the car?
Weight = 15,680 N m = W/g = 1568 kg
What is the change in momentum?
Dr = m D v = kg (0 – 20.0 m /s)
= -31,360 kg m /s (momentum is decreasing as it slows down so it is negative)

6. Single Object Momentum-EX
What is the initial momentum ?
ri = m vi = 1568 kg (20.0 m /s)
ri =31,360 kg m/s
What is the final momentum?
rf = m vf = 1568 kg (0 m /s)
rf = 0 kg m/s

7. Single Object Momentum-EX
What is the impulse of the car?
Dr = rf - ri = 0 kg m /s – 31,360 kg m /s
Dr = impulse = -31,360 kg m /s
How long does the braking force act on the car to bring it to a halt?
Dr = F t t = Dr / F
t= -31,360 kg m /s / N = 49 s

8. Conservation of Momentum
m1v1 + m2 v2 = m1 v1’ + m2v2’
Subscript 1 is object one
Subscript 2 is object two
‘ is called prime
Prime is used to indicate after the collision

9. Three types of collisions
“Nonsticky”
“Sticky”
“Newton’s Third”

10. Nonsticky Same direction
EX: A plastic ball of mass 0.200kg moves with a velocity of 0.30 m/s. This plastic ball collides with a second plastic ball of mass 0.100kg that is moving along the same line at a velocity of 0.10 m/s. After the collision, the velocity of the 0.100kg ball is 0.26 m/s. What is the velocity of the first ball?
0.2 ( 0.3) (0.1) = 0.2 (v1’) (0.26 )
= 0.2 (v1’)
0.07 – = 0.2 (v1’) – 0.026
/ 0.2 = 0.2 (v1’) /0.2
0.22 m/s = v1’

11. Nonsticky
Opposite direction Sign rule: 1st object is right &positive and left is negative
EX: A 0.50kg ball traveling at 6.0 m/s collides head-on with a 1.00kg moving in the opposite direction at a velocity of m/s. The 0.50kg moves away at -14 m/s after the collision. Find the velocity of the 1.00 kg ball after the collision.
.5(6) + 1(-12) = .5(-14 ) + 1(v2’)
-9 = (v2’)
-2 = 1(v2’)
-2 m/s = v2’

13. Sticky
Sticky means that the two objects stick together when they collide and fly off together at the same speed.
You still use
m1v1 + m2 v2 = m1 v1’ + m2v2’
but v1’ = v2’

14. Example of a Sticky m1v1 + m2 v2 = (m1 +m2) v’
Moving at 20.0 m/s, a car of mass 700Kg collides with a stationary truck of mass 1400 kg. If the two vehicles interlock as a result of the collision, what is the velocity of the car-truck system?
700kg(20m/s) kg(0m/s)= ( ) v’
14000 kg m/s = 2100kg ( v’)
14000 kg m/s/ 2100kg = 2100kg ( v’) / 2100 kg
6.67 m/s = v’

17. Newton’s 3rd Law Problems
m1v1 + m2 v2 = m1 v1’ + m2v2’
Where v1 = v2 = 0
Ex: A 40.0 kg projectile leaves a 2000 kg launcher with a velocity of 800 m/s. What is the recoil velocity of the launcher?
2000(0) +40 (0) = 2000 (v1’)+ 40 (800)
0 = 2000 (v1’)+ 32,000
-32,000 = 2000 (v1’)
-32,000 /2000 = 2000 (v1’) / 2000
-16 m/s = v1’