1. Abelardo M. Zerda III Michael O. Suarez Jm Dawn C. Rivas Leslie Kate Diane Berte

2. Momentum and Impulse Conservation of Momentum Collisions in One Dimensions Collisions in Two Dimensions

3. Measure of one’s motion, equivalent to the product of one’s mass and velocity. Momentum involves motion and mass. Expressing the definition mathematically, p = mv where: p = momentum in kg-m/s units m = mass in kg units v = velocity of the object in m/s units

4. Calculate the momentum of a 100-kg missile traveling at 100 m/s eastward. A 100 kg car strikes a tree at 50 km/h. Find its momentum.

5. Impulse – force (F) applied by one object to another object within a given time interval Δt The formula to get impulse is: F Δt = Δp where: F Δt = impulse in N-s (newton-second) units Δp = change in momentum in kg-m/s units

6. A person in a sled with a total mass of 125 kg, slides down a grassy hill and reach a speed of 8 m/s at the bottom. If a pile of grass can exert a constant force of 250 newtons, how fast will the sled stop? A 100 kg car strikes a tree at 20 km/h and comes to a stop in 0.5 seconds. Find its initial momentum and the force on the car while it is being stopped.

7. In a collision, energy is not always conserved but all collisions have to conserve momentum if there is no net applied force. ∑ p before collision = ∑ p after collision

8. There are three types of collisions: 1.Elastic collisions – conserve kinetic energy 2.Inelastic collisions – do not conserve kinetic energy 3.Completely inelastic collision

9. Let us assume that for the two balls, one is running on a straight track toward the second one, which is stationary. Assume the following items are given: m 1 – mass of object 1 m 2 – mass of object 2 v 1 – velocity of object 1 before collision v 2 – velocity of object 2 before collision v 1 ’ – velocity of object 1 after collision v 2 ’ – velocity of object 2 after collision

10. Solving for the velocity of mass 1 after collision v 1 ’: v 1 ’ = [(m 1 – m 2 ) / (m 1 + m 2 )] v 1 Solving for the velocity of mass 2 after collision v 2 ’: v 2 ’ = [2m 1 / (m 1 + m 2 )] v 1

11. v 1 ’ = [(m 1 – m 2 ) / (m 1 + m 2 )] v 1 = [(m-m) / (m+m)] v 1 = 0 After colliding with m 2, m 1 stops. v 2 ’ = [2m 1 / (m 1 + m 2 )] v 1 = (2m / 2m) v 1 = v 1 Total energy transfer occurred between m 1 and m 2.

12. Since m 1 > m 2, the term (m 1 – m 2 ) is positive. v 1 ’ = [(m 1 – m 2 ) / (m 1 + m 2 )] v 1 m 1 still moves and v 1 ’ has the same direction as v 1. Since all the terms are nonzero then, v 2 ’ = [2m 1 / (m 1 + m 2 )] v 1 m 2 also moves and v 2 ’ has the same direction as v1

13. Since m 1 < m 2, the term (m 1 – m 2 ) is negative. v 1 ’ = [(m 1 – m 2 ) / (m 1 + m 2 )] v 1 m 1 still moves but in the opposite direction. v 2 ’ = [2m 1 / (m 1 + m 2 )] v 1 Since all the terms are nonzero, then m 2 also moves and v 2 has the same direction as v 1.

14. Inelastic collisions produce a combined mass after collision. The equation for this type of collision is: m 1 v 1 + m 2 v 2 = ( m 1 + m 2) v f where: m 1 + m 2 is the combined mass of the object V f is the final velocity value for the combined mass If m 2 v 2 = 0, v f = v 1 m 1/ ( m 1 + m 2)

15. Balanced momentum in the x-direction and y-direction Angles are arbitrary

16. Along the x-axis:  p x before collision =  p x after collision p 1 cos  1 + p 2 cos  2 = p 1 cos  1 + p 2 cos  2 Along the y-axis:  p y before collision =  p y after collision p 1 sin  1 + p 2 sin  2 = p 1 sin  1 + p 2 sin  2

17. A pool ball weighing 2 kg is traveling at 30  at 0.8 m/s hits another ball moving at 0.5 m/s at 180 . If the second ball leaves the collision at 0  and the first moves away at 150 , find the final velocity vectors of the balls.