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Momentum Problem of a Projectile
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[QUOTE="Libohove90, post: 2972442, member: 188210"] [h2]Homework Statement [/h2] A 224 kg projectile is fired at 116 m/s at a 60 degree angle. At its highest point, it explodes into three fragments. Two fragments have the same speed and move vertically downward. The other fragment moves horizontally. a) what is the velocity of the third fragment right after the explosion? [h2]Homework Equations[/h2] Initial Momentum = Final Momentum Momentum equation [h2]The Attempt at a Solution[/h2] I need to find the x component of the velocity, thus I do 116*cos 60 = 58 m/s The initial momentum is 224 kg * 58 m/s = 12,992 kg m/s. Since the initial momentum = final momentum, I look at the x component of the momentum. 12,992 kg m/s = 1/3 (224 kg) V ^Only one fragment has a horizontal velocity. I get confused because everywhere else I look, the way to solve this is 12,992 kg m/s = 1/3 (224 kg)(58 m/s) + 1/3 (224 kg) V I got lost at that point. Why does the initial momentum = 1/3 (224 kg)(58 m/s) + 1/3 (224 kg) V? [/QUOTE]
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Momentum Problem of a Projectile
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