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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 200ms1 into a block of mass 0.95kg. Given that the lead block can move freely, the final kinetic energy after impact is?

A.

50J

B.

100J

C.

150J

D.

200J

Correct answer is A

Given 
m1 = 0.05kg, u1 = 200ms1, m2 = 0.95kg

K.E = 12mrv2

m1u1 = v(m1 + m2) [law of conversation of momentum]

v = 0.05×2000.05+95 = 10ms1

K.E = 12(1)102 = 50J