1. The force on an object may not be constant, but may vary over time. The force can be averaged over the time of application to find the impulse.

2. The impulse of a force is the product of the average force and the time interval during which the force acts. Impulse = F ave Δt The unit is the newtonsecond (Ns)

3. The linear momentum p of an object is the product of the object’s mass m and the velocity v. p = mv The unit is the kilogrammeter/second (kgm/s)

4. Remember F = ma, and a = (v f - v 0 )/ Δt. The final term can be substituted for a in F = ma. F = m v f - v 0 / Δt or F = mv f - mv 0 / Δt and finally: F Δt = mv f - mv 0

5. This is the impulse- momentum theorem, where impulse is equal to the change in momentum. F Δt = mv f - mv 0 impulse finalinitial momentum momentum

6. While it may be difficult to find average force, it is usually easy to find mv f and mv 0.

7. Ex. 1 - A baseball (m = 0.14 kg) has an initial velocity of v 0 = -38 m/s as it approaches a bat. The ball leaves the bat with a velocity of v f = +58 m/s. (a) Determine the impulse applied to the ball by the bat. (b) If the time of contact is Δt = 1.6 x 10 -3 s, find the average force exerted on the ball by the bat.

8. Ex. 2 - During a storm, rain comes straight down with a velocity of v 0 = -15 m/s and hits the roof of a car perpendicularly. the mass of the rain per second that strikes the roof of the car is 0.060 kg/s. Assuming that the rain comes to rest upon striking the car roof, find the average force exerted by the rain on the car roof.

9. In an isolated system no net external forces act on two colliding objects (The net external forces are zero). The sum of the internal forces (those caused by the collision itself) is also zero by Newton’s third law.

10. Therefore: (F 1 + F 2 )Δt = (mv f1 + mv f2 ) - (mv 01 + mv 02 ) becomes: 0 = (mv f1 + mv f2 ) - (mv 01 + mv 02 ) or: (mv f1 + mv f2 ) = (mv 01 + mv 02 )

11. This is the principle of conservation of linear momentum. The total linear momentum of an isolated system remains constant. (mv f1 + mv f2 ) = (mv 01 + mv 02 ) or: P f = P 0

12. Ex. 5 - A freight train is being assembled in a switching yard. Car 1 has a mass of m 1 = 65 x10 3 kg and moves with a velocity of v 01 = +0.80 m/s. Car 2, with a mass of m 2 = 92 x 10 3 kg and a velocity of v 02 = =1.2 m/s, overtakes car 1 and couples to it. Find the common velocity v f of the two cars after they become coupled.

13. Ex. 6 - Starting from rest, two skaters “push off” against each other on smooth level ice. One is a woman (m 1 = 54 kg), and one is a man (m 2 = 88 kg). The woman recoils with a velocity of v f1 = +2.5 m/s. Find the recoil velocity v f2 of the man.

14. Ex. 7 - When a gun fires a blank, is the recoil greater than, the same as, or less than when the gun fires a standard bullet?

15. The total linear momentum may be conserved even when the kinetic energies of the individual parts of a system do change. (e. g. the skaters)

16. Collisions are often classified by whether the total kinetic energy changes during the collision.

17. An elastic collision is one in which the total kinetic energy of the system after the collision is equal to the total kinetic energy before the collision. K.E. is conserved in the collision.

18. An inelastic collision is one in which the total kinetic energy of the system is not the same before and after the collision; if the objects stick together after colliding, the collision is said to be completely inelastic. Kinetic energy is not conserved. The coupling boxcars is an example of an inelastic collision.

19. In inelastic collisions, kinetic energy is lost in two main ways. First, it may become heat due to friction. Second, kinetic energy is lost when an object is permanently deformed by the collision.

20. Ex. 8 - A ball of mass m 1 = 0.250 kg and velocity v 01 = +5.00 m/s collides head-on with a ball of mass m 2 = 0.800 kg that is initially at rest. If the collision is elastic, what are the velocities of the balls after the collision?

21. Ex. 9 - A ballistic pendulum consists of a block of wood (mass m 2 = 2.50 kg) suspended by a wire. A bullet (mass m 1 = 0.0100 kg) is fired with a speed v 01. Just after the bullet collides with it, the block (now containing the bullet) has a speed v f and then swings to a maximum height of 0.650 m above the initial position. Find the speed of the bullet.