A lead bullet of mass 0.05 kg is fired with a speed of 20...
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.
150J
200J
50J
100J
Correct answer is C
Mbullet = 0.05kg, Vbullet = 200m/s, Mblock = 0.95kg and Vblock = 0
for an inelastic collision
Mbullet×Vbullet+Mblock×Vblock=(Mbullet+Mblock)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 = 12mv2
K . E = 12×(0.05+0.95)×102 ( bullet now part of the block)
K . E = 12×1×100
= 50J
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