HW_9_sol - HW 9 solutions P112 S07 1) Solving the rocket...

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Unformatted text preview: HW 9 solutions P112 S07 1) Solving the rocket equation, v- v = v ex ln m m , for the ratio m m , with v = 0, m m = exp v v ex = exp 8.00 km s 2.10 km s = 45.1. 2) The trick here is to realize that the center of mass will continue to move in the original parabolic trajectory, “landing” at the position of the original range of the projectile. Since the explosion takes place at the highest point of the trajectory, and one fragment is given to have zero speed after the explosion, neither fragment has a vertical component of velocity immediately after the explosion, and the second fragment has twice the velocity the projectile had before the explosion. a) The fragments land at positions symmetric about the original target point. Since one lands at 1 2 R , the other lands at 3 2 R = 3 2 v 2 g sin2 a = 3 2 (80 m / s) 2 (9.80 m / s 2 ) sin120 o = 848 m. b) In terms of the mass m of the original fragment and the speed v before the explosion, K 1 = 1 2 mv 2 and K 2 = 1 2 m 2 (2v) 2 = mv 2 ,so DK = mv 2- 1 2 mv 2 = 1 2 mv 2 ....
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This note was uploaded on 03/27/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Spring '07 term at Cornell.

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HW_9_sol - HW 9 solutions P112 S07 1) Solving the rocket...

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