1. Physics 111

2. Equations:  Momentum = (p) = m(v)  Is in kg* m/s  Net Force*(t) = m* delta v = delta p  Impulse = (J) = Delta F* t = N* seconds  If no external forces, Pf=Pi So mv i = mv f  If elastic, KE i = KE f Can use V 1i – v 2i = -(v 1f – v 2f ) Also use: v 1f = (m 1 – m 2 / m 1 + m 2 )*v 1i And v 2f = (2m / m 1 +m 2 )*v 1i  If inelastic, M1v + m2v = (m1 +m2)v

3. Concept Check  Consider these situations  (i) a ball moving at speed v is brought to rest  (ii) the same ball is projected from rest so that it moves at speed v  (iii) the same ball moving at speed v is brought to rest and then projected backward to its original speed. In which case(s) does the ball undergo the largest change in momentum?

4. Answer: Situation 3  Why? Let's say the ball has inertial mass m and velocity v. The decrease in momentum in case (i) is 0 - mv = -mv (final momentum minus initial momentum). In case (ii), we find mv - 0 = +mv. In case (iii), we have m(-v) - mv = -2mv because the ball's velocity is now in the opposite direction. So the magnitude of the change is greatest in the third case.

5. Warmup:  A bicycle has a momentum of 24 kgm/s. What momentum would the bicycle have if it had … twice the mass and was moving at the same speed? one-half the mass and was moving with twice the speed? three times the mass and was moving with twice the speed?

6. Answer:  1) 48 kg*m/s P=mv P=(1)(24) initially ○ You are given twice the weight so ○ P=(2)(24) = 28  2) 24 kg*m/s P=(.5)(48) = 24  3)144 kg*m/s P= (3)(48) = 144

7. Practice Problem:  Jerome plays middle linebacker for South's varsity football team. In a game against cross- town rival North, he delivered a hit to North's 82- kg running back, changing his eastward velocity of 5.6 m/s into a westward velocity of 2.5 m/s. a. Determine the initial momentum of the running back. b. Determine the final momentum of the running back. c. Determine the momentum change of the running back. d. Determine the impulse delivered to the running back.

8. Solution:  a. 460 kgm/s, east (rounded from 459 kgm/s) P=mv = 82* +5.6  b. - 205 kgm/s or 205 west P=mv=82 * -2.5 = -205  c. -665 kgm/s, or 665 west Delta p = P final – P initial = -205 – 460 = -665  d. -660 Ns Impulse (J) = delta p

9. Problem:  An 82-kg male and a 48-kg female pair figure skating team are gliding across the ice at 7.4 m/s, preparing for a throw jump maneuver. The male skater tosses the female skater forward with a speed of 8.6 m/s. Determine the speed of the male skater immediately after the throw.

10. Solution: 6.7 m/s  M1 = 82  M2 = 48  Vi=7.4  V1f = 8.6  V2f= ??  M1v + m2v = (m1 +m2)v  (48)(8.6) + (82)(v2f) = (48 +82)(7.4) Solve for V2f