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?
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option A
Explanation:
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
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