Solution: Individual Exercise #1 (Part a)

1. m= 1 ton vf = ? Dt=9s final initial F Solution: Individual Exercise #1 (Part a) Burning fuel neglected, so mass is constant

2. Solution: Individual Exercise #1 (Part b)

3. Initial Final Dt=0.020 s Fave vi=0 km/h vf=100 km/h Solution: Pairs Exercise #1(Part a)

4. Solution: Pairs Exercise #1(Part b) By Newton’s 3rd law, the force on the hockey stick is the same as that on the puck.

5. Helicopter w/ ammo mi = 1 ton vi = 40 mph Helicopter w/ less ammo mf = ? vf = ? Solution to Pairs Exercise #2 bullets Initial Final i.e., the helicopter is going backwards!!

6. Solution to Pairs Exercise #3: No, it would not change i.e., the answer stays the same

7. block mw = 500 g block mw = 500 g bullet mb=5g vb,f=100 m/s bullet mb=5g vb,i=300 m/s vw,i = 0 m/s vw,f = ? m/s Initial Final Solution to Individual Exercise #2

8. Solution: Pairs Exercise #4 White Ball: Dpwhite = mwhite(vwhite,final- vwhite,initial) = 2 lbm · (1.5 - 2) ft/s = -1 lbm·ft/s Black Ball: Dpblack = mblack(vblack,final- vblack,initial) = 1 lbm · (1 - 0) ft/s = 1.0 lbm·ft/s More…

9. Solution: Pairs Exercise #4 (con’t) Initial Kinetic Energy: KEinitial = ½ mwhite v2white, initial + ½ mblack v2black, initial = ½ (2 lbm) (2 ft/s)2+ ½ (1 lbm) (0 ft/s)2 = 4 lbm ft2/s2 Final Kinetic Energy: KEfinal = ½ mwhite v2white, final + ½ mblack v2black, final = ½ (2 lbm) (1.5 ft/s)2+ ½ (1 lbm) (1 ft/s)2 = 2.75 lbm ft2/s2 Energy is conserved, but kinetic energy is not.

10. m1 = 1 kg v1 = 1 m/s m3 = ? v2 = ? m2 = 1 kg v2 = 10 m/s Sol’n 5.1 a. b. c. Total = 50.5 J d.

11. 2 1 Throat Diffuser Conventional Jet Ejector 3

12. Sol’n 5.2 m1 = 1 kg v1 = 1 m/s m3 = ? v2 = ? m2 = 1 kg v2 = 2 m/s a. b. c. Total = 2.5 J d.

13. 1 1.5 1 2.8 1 1 Advanced Jet Ejector Numbers are Mach numbers.