A lead bullet of mass 0.05 kg is fired with a velocity of 200 m/s into a lead block of mass 0.95 kg. Given that the lead block can move freely, the final kinetic energy after impact is
A
100J
B
150J
C
50J
D
200J
Verified Correct Answer: C.
Physical Principles & Numerical Solution
From the law of conservation of linear momentum,
\(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\)
Since the collision is inelastic, we have
\((0.05 \times 200) - (0.95 \times 0) = (0.05 + 0.95)V\)
\(10 = V\)
\(V = 10ms^{-1}\)
Hence the Kinetic energy = \(\frac{1}{2} (0.05 + 0.95) \times 10^2\)
= \(\frac{1}{2} \times 100\)
= 50J
Source: Myschool Past Questions Archive (Record #47400)