A lead bullet of mass 0.05kg is fired with a velocity of ...
A lead bullet of mass 0.05kg is fired with a velocity of 200ms−1 into a block of mass 0.95kg. Given that the lead block can move freely, the final kinetic energy after impact is?
50J
100J
150J
200J
Correct answer is A
Given
m1 = 0.05kg, u1 = 200ms−1, m2 = 0.95kg
K.E = 12mrv2
m1u1 = v(m1 + m2) [law of conversation of momentum]
v = 0.05×2000.05+95 = 10ms−1
K.E = 12(1)102 = 50J
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