Block A weighs 19.9 lb and is moving toward the left with a velocity of 4 ft/s j
ID: 1896353 • Letter: B
Question
Block A weighs 19.9 lb and is moving toward the left with a velocity of 4 ft/s just before it is struck by the bullet B from a caliber 50 machine gun. The coefficient of kinetic friction between A and the plane is 0.25. The bullet weighs 0.10 lb and is fired into the center of block A with a velocity of 2900 ft/s to the right.a) What is the final velocity of the block-bullet combination?
b) how long will block A continue to move after the embedment of the bullet?
c) What percentage of system energy is lost during impact?
Explanation / Answer
1 lb = 0.45359237 Kg
1 ft = 0.3048 m
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a)
conservation of momentum:
-ma va + mb vb = (ma + mb) v
(-19.9)*4 + 0.10*2900 = (19.9 + 0.10) * v
>>>> v = 10.52 ft/s = 10.52*0.3048 = 3.206496 m/s
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b)
m = ma+mb = 19.9+0.10 = 20 lb = 20*0.45359237 Kg = 9.0718474 Kg
acceleration = a = fk/m = k g = 0.25 * 9.8 = 2.45 m/s2
t = v/a = 3.206496/2.45 = 1.31 s
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c)
K = 0.5*ma*(va)2 + 0.5*mb*(vb)2 - 0.5*(ma + mb)*v2
= 0.5*19.9*0.45359237*(4*0.3048*4*0.3048)+0.5*0.10*0.45359237*(2900*0.3048*2900*0.3048) - 0.5*9.0718474*3.206496*3.206496
= 17680 J
K/K * 100 = (17680/17727)*100 = 99.7%