a lead bullet of mass 0.05kg is fired with a velocity of 200ms-1 into a lead block of mass 0.95kg. Given that the lead block can move freely, the final kine...
Question 1 Report
a lead bullet of mass 0.05kg is fired with a velocity of 200ms-1 into a lead block of mass 0.95kg. Given that the lead block can move freely, the final kinetic energy after impact is
Answer Details
From principle of conservation of linear momentum,
(0.05 x 200) - (0.95 x 0) = (0.05 + 0.95) x V (since collision is inelastic).
(0.05 x 200) - (0.95 x 0) = (0.05 + 0.95) x V (since collision is inelastic).
Thus V = 10m/s.
K.E = 1/2 (0.05 + 0.95) x 102
K.E = 1/2 (1 x 100) = 50 J
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