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?

Question

Raymond Ward

Physics

30 December, 10:47

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?

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Jaylin Conner · 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