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A 5.00-g bullet is shot through a 1.00-kg wood block suspended on a string 2.00 m long. The center of mass of the block rises a distance of 0.38 cm.
Find the speed of the bullet as it emerges from the block if its initial speed is 450 m/s.
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Added Thu, 23 Jul '15
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m=.005 kg= mass of bullet
M=1 kg = mass of wooden block

v_i = initial velocity of bullet = 450 m/s
v_f = final velocity of bullet
V = velocity of wooden block

mv_i = mv_f + MV
V = sqrt(2gh) = sqrt(2*9.81*.0038) = 0.273

v_f is what youre looking for, so
v_f = (mv_i - MV)/m = (0.005*450 - 1*0.273)/0.005 = 395.4 m/s
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