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. T...

Answer Details

We can solve this problem using the law of conservation of momentum, which states that the total momentum of a closed system remains constant before and after an interaction. In this case, the arrow and the wooden block form a closed system. Let the velocity of the arrow after impact be v. We can set up the following equation: (mass of arrow * initial velocity of arrow) = (total mass * final velocity) 0.1 kg * 15 m/s = (0.1 kg + 0.4 kg) * v 1.5 kg m/s = 0.5 kg * v v = 3 m/s Therefore, the common velocity of the arrow and the wooden block after impact is 3 m/s. None of the answer options provided matches this result.