By utility we meanA 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?

By: |Published on: Mar 31|Categories: Physics JAMB| 0 comments

A. 50J
B. 100J
C. 150J
D. 200J

Correct Answer:

Option A = 50J

Explanation

Given

m1 = 0.05kg, u1 = 200ms−1, m2 = 0.95kg

K.E = 1/2 mv2
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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