1. Car Crash Video

2. Momentum, Impulse, and Collisions

3. Momentum (p) expresses the resistance of a body to slowing down.
related to inertia
p = mv
m = mass (kg)
v = velocity (m/s)
Unit is kg m/s

4. Example
A toy car has a mass of 5 kg. If it is moving at a velocity of 4 m/s, what is its momentum?
20 kg•m/s
Given
Unknown
Equation
Substitute
Solve

5. Your Turn
What is the momentum of a 25 kg boy moving at a speed of 1.2 m/s?
30 kg•m/s

6. Impulse = ∆ρ = m∆v = F(∆t)
Impulse (I)
measures a change in momentum
To calculate multiply the average force acting on a body by the length of time the force acts.
Impulse = ∆ρ = m∆v = F(∆t)
F = force (N or Newtons)
∆t = time (s)
Unit is N s
See SLO equation sheet.

7. Putting it all together…
Newton’s 2nd Law F = ma
a = Δv/t
F=mΔv / t
F(t) = m(vf-vi) Notice impulse, on the left side, is equal to the change in momentum of an object.

8. Impulse

9. Example
An 8 N force acts upon an object for 5 seconds. What impulse is given this object?
40 N•s
Given
Unknown
Equation
Substitute
Solve

10. (Hint: use Impulse=F(t))
Your Turn
A ball changes velocity from 20m/s to 30m/s. If it has a mass of 5kg, what impulse was necessary to cause this change?
50N·s
How long would it take a 5N force to change the velocity of the ball?
(Hint: use Impulse=F(t))
10s

11. The Law of Conservation of Momentum
In an isolated system, the total momentum does not change. Momentum is conserved in all closed systems!

12. There are 2 types of collisions
Elastic
m1v1i + m2v2i = m1v1f + m2v2f
Where 1 stands for object 1, 2 stands for object 2
Where i stands for initial & f stands for final
Use + and – velocities to show direction.

13. 2kg(2m/s)+3kg(-1m/s) = 2kg(v1f)+3kg(1.33m/s)
Example…
A 2 kg ball strikes another ball head on at an initial velocity of +2 m/s. If the second ball had a mass of 3 kg and was initially moving at -1 m/s, what is the final velocity of the 2 kg ball if the vf of the 3 kg ball is now +1.33m/s?
Define variables. Then:
m1v1i + m2v2i = m1v1f + m2v2f
2kg(2m/s)+3kg(-1m/s) = 2kg(v1f)+3kg(1.33m/s)
v1f = -1.5m/s

14. There are 2 types of collisions
Inelastic
m1v1i + m2v2i = (m1 + m2)v1&2f
Where v1&2f is the velocity of the two objects stuck together after the collision
Use + and – velocities to show direction.

15. Example…
A 500 kg car is driving at a velocity of 10 m/s. Another car (700 kg) hits it from behind at a velocity of 25 m/s and the two cars interlock bumpers. What is the resulting velocity of the two cars?
Define variables. Then:
m1v1i + m2v2i = (m1 + m2)v1&2f 500kg(10m/s)+700kg(25m/s) = (500kg+700kg)(v1&2f)
v1&2f = m/s