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
The correct answer is: C
Explanation
\(M_{bullet}\) = 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