Question

A 0.005 kg bullet is travelling horizontally at a velocity of 250 m/s when it strikes a wooden block of mass 12 kg on a table at rest. the bullet sticks in the block of wood. assuming momentum is conserved what is the velocity of the block of wood after the collision?

Answer 1

Use the conservation of momentum formula: m1v1 + m2v2 = v(m1 + m2)

(0.005kg)(250m/s) + (12kg)(0m/s) = v(0.005kg + 12kg)

v = ((0.005kg)(250m/s)) / (0.005kg + 12kg)

v = 0.104 m/s