An arrow of mass 0.1kg moving with a horizontal velocity of 15ms-1 is shot into a wooden block of mass 0.4kg lying at rest on a smooth horizontal surface. Their common velocity after impact is
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Correct Answer: Option D
Explanation:
M1U1 + M2U2
= (m1 + m2)v
= \(\frac{(0.1 {\times} 15) + (0.4 {\times} 0)}{(0.1 + 0.4)}\)
v = 3ms-1
M1U1 + M2U2
= (m1 + m2)v
= \(\frac{(0.1 {\times} 15) + (0.4 {\times} 0)}{(0.1 + 0.4)}\)
v = 3ms-1