A lead bullet of mass 0.05 kg is fired with a speed of 200 m/s into a lead block of mass 0.95 kg at rest. Given that the lead block moves after the impact, determine its kinetic energy.
A.
150J
B.
200J
C.
50J
D.
100J
Correct answer is C
\(M_{bullet}\) = 0.05kg, \(V_{bullet}\) = 200m/s, \(M_{block}\) = 0.95kg and \(V_{block}\) = 0
for an inelastic collision
\(M_{bullet} \times V_{bullet} + M_{block} \times V_{block} = (M_{bullet} + M_{block})\)v
0.05 x 200 + 0.95 x 0 = ( 0.05 + 0.95)v
10 + 0 = v
v = 10m/s
K.E of the block = \(\frac{1}{2}mv^2\)
K . E = \(\frac{1}{2} \times(0.05 + 0.95) \times10^2\) ( bullet now part of the block)
K . E = \(\frac{1}{2} \times1 \times100\)
= 50J
