A rifle of mass 10 kg fires a 10 g bullet into a 3 kg block of woodwhich is susp

Question

A rifle of mass 10 kg fires a 10 g bullet into a 3 kg block of woodwhich is suspended by a thin wire, forming a ballistic pendulum.The bullet remains embedded in the block of wood, which is observedto swing so that it reaches a vertical height of 20 cm above itsstarting point. What was the speed of the bullet, and with whatspeed did the rifle recoil?

Ignore air resistance, friction in the suspension, and the verticalmotion of the bullet, and assume that the block of wood is smallrelative to the length of the wire.

Explanation / Answer

m1 = 10 kg m2 = 10 g = 0.01 kg m3 = 3 kg h = 20 cm = 0.2 m From conservation of energy, you can find the speed of thebullet-block after the collision: PE = KE (m2 + m3)gh = (1/2)(m2 +m3)v2 v = (2gh) This is the speed acquired after the (inelastic) collision, inwhich momentum is conserved. Let v2 be the speedof the bullet before the collision. pi = pf m2v2 + 0 = (m2 +m3)v v2 = [(m2 + m3)/m2]v The initial firing was itself a collision, in which the initialmomentum was zero pi = pf 0 = m1v1 + m2v2 v1 = -(m2/m1)v2 The answer is negative because the rifle is moving in a directionopposite that of the bullet. Since the question asks for thespeed, just take the absolute value of the answer.

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