Momentum and Collisions

1. Momentum and Collisions

2. Momentum and Impulse • The momentum of an object is the product of its mass and velocity: p=mv • Units of momentum: kg·m/s

3. Momentum and Impulse • If the resultant force F is zero, the momentum of the object does not change.

4. Impulse-Momentum Theory I = FΔt = Δp = mvf – mvi (I = Impulse) • *If we exert a force on an object for a time interval Δt, the effect of this force is to change the momentum of the object from some initial value mvi to some final value mvf.

5. Example: A golf ball of mass 50. g is struck with a club. Assume that the ball leaves the club face with a velocity of +44m/s.A) Estimate the impulse due to the collision.B) Estimate the length of time of the collision and the average force on the ball. pi = mvi = 0 (ball starts at rest) pf = mvf = (0.05 kg)(44 m/s) pf = 2.2 kg·m/s Impulse = Δp = pf – pi Δp = 2.2 kg·m/s - 0 kg·m/s Δp = 2.2 kg·m/s

6. Example: A golf ball of mass 50. g is struck with a club. Assume that the ball leaves the club face with a velocity of +44m/s.A) Estimate the impulse due to the collision.B) Estimate the length of time of the collision and the average force on the ball. • Estimate distance ball travels while in contact with club = displacement of ball while in contact ≈ 2.0 cm

7. Applications of impulse momentum theory: FDt = mvf - mvi • Follow through in golf swing, batting, tennis. • Catching a water balloon. • Moving with the punch in boxing. • Padding boxing gloves, goalie gloves in hockey, baseball mitts, inside of helmets…

8. Example: In a crash test, a car of mass 1500 kg collides with a wall. The initial velocity of the car is 15.0 m/s east and the final velocity of the car is 2.6m/s west. The collision lasts for 0.150 s. find (A) How can the car’s initial velocity be to the east and its final velocity to the west? Calculate (B) the impulse due to the collision and (C) the average force exerted on the car.

9. Example:In a crash test, a car of mass 1500 kg collides with a wall. The initial velocity of the car is 15.0 m/s east and the final velocity of the car is 2.6m/s west. The collision lasts for 0.150 s. find (A) How can the car’s initial velocity be to the east and its final velocity to the west? Calculate (B) the impulse due to the collision and (C) the average force exerted on the car. pi = mvi pi = (1500kg)(-15m/s) = -2.25x104kg·m/s pf = mvf pf = (1500kg)(2.6m/s) = +3.90x103kg·m/s I = Δp = pf – pi I = 3900 – (-22500) I = 2.6x104kg·m/s

10. Example: In a crash test, a car of mass 1500 kg collides with a wall. The initial velocity of the car is 15.0 m/s east and the final velocity of the car is 2.6m/s west. The collision lasts for 0.150 s. find (A) How can the car’s initial velocity be to the east and its final velocity to the west? Calculate (B) the impulse due to the collision and (C) the average force exerted on the car.

11. Rebounding FDt = mvf - mvi • Is it better to bounce or to hit and stick/stop?

12. Conservation of Momentum pi = pf • Law of conservation of momentum: when no external forces act on a system consisting of 2 objects, the total momentum of the system before the collision is equal to the total momentum of the system after the collision.

13. Example: A baseball player uses a pitching machine to help him improve his batting average. He places the 50.-kg machine on a frozen pond. The machine fires a 0.15-kg baseball with a speed of 36 m/s in the horizontal direction. What is the recoil velocity of the machine?

14. Example: A baseball player uses a pitching machine to help him improve his batting average. He places the 50.-kg machine on a frozen pond. The machine fires a 0.15-kg baseball with a speed of 36 m/s in the horizontal direction. What is the recoil velocity of the machine? pbefore = pafter m1v1i + m2v2i = m1v1f + m2v2f (+50.kg 0.15kg)(0) = (50.kg)(v1f) + (.15kg)(36m/s) 0 = (50.kg)(v1f) + (.15kg)(36m/s) (50.kg)(v1f) = -5.4kg·m/s v1f = -0.11 m/s (moves .11 m/s opposite to the motion of the ball)

15. Collisions • For any type of collision, the total momentum is conserved. • However, the total kinetic energy is generally NOT conserved.

16. Elastic Collisions • Elastic collisions: both momentum and kinetic energy are conserved. • Example: two balls bouncing perfectly off each other. m1v1i + m2v2i = m1v1f + m2v2f and ½ m1v1i2 + ½ m2v2i2 = ½ m1v1f2 + ½ m2v2f2

17. Inelastic Collisions • Inelastic collisions: momentum is conserved, but kinetic energy is not. • When 2 objects collide and stick together, the collision is PERFECTLY INELASTIC; in this case, their final velocities are the same. For perfectly inelastic collisions, m1v1i + m2v2i = (m1 + m2)v2f

18. Example: A large luxury car with a mass of 1800. kg stopped at a traffic light is struck from the rear by a compact car with a mass of 900. kg. The two cars become entangled as a result of the collision. If the compact car was moving at 20.0 m/s before the collision, what is the velocity of the entangled mass after the collision?

19. Example: A large luxury car with a mass of 1800. kg stopped at a traffic light is struck from the rear by a compact car with a mass of 900. kg. The two cars become entangled as a result of the collision. If the compact car was moving at 20.0 m/s before the collision, what is the velocity of the entangled mass after the collision? pbefore = pafter M1v1i + m2v2i = (m1+m2)vf (1800 kg)(0) + (900.kg)(20.0m/s) = (1800+900)vf 1.80 x 104 kg·m/s = (2700)vf 1.80x104 = (2700)vf v1f = 6.67 m/s

20. Example: Two kids are playing marbles. The shooter rolls toward a stationary marble at 5.0 cm/s. They two collide and the smaller marble speeds away at 8.0 cm/s. What is the final velocity of the shooter? Assume the shooter is three times as massive as the smaller marble. (have faith with the masses…) pi = pf 3mvsi + mvmi = 3mvsf + mvmf 3m(5.0cm/s) + 0 = 3mvsf + m(8.0cm/s) 15 cm/s + 0 = 3vsf + 8.0cm/s 7cm/s = 3vsf Vsf = 2.3 cm/s