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?
A
50J
B
100J
C
150J
D
200J
Verified Correct Answer: A.
Physical Principles & Numerical Solution
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
Source: WAEC / JAMB / NECO Classroom Archive (Record #60753)