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?
Given
m\(_{1}\) = 0.05kg, u\(_{1}\) = 200ms\(^{-1}\), m\(_{2}\) = 0.95kg
K.E = \(\frac{1}{2}\)m\(_{r}\)v\(^{2}\)
m\(_{1}\)u\(_{1}\) = v(m\(_{1}\) + m\(_{2}\)) [law of conversation of momentum]
v = \(\frac{0.05 \times 200}{0.05 + 95}\) = 10ms\(^{-1}\)
K.E = \(\frac{1}{2}\)(1)10\(^{2}\) = 50J
There is an explanation video available below.
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