A large block of wood of mass m1 =1.25 kg hangs from a long mass less cord. A bu

Question

A large block of wood of mass m1 =1.25 kg hangs from a long mass less cord. A bullet of mass m2 = 0.0625 kg is fired into the block, coming quick to rest. The block plus bullet system ion swing upward and rise a vertical distance h = 0.135 m before the system comes momentarily in rest at the end of its arc. Assume the air resistance is negligible. The speed V of the block and bullet just after the impact; what is the speed v01, of the bullet just prior to the collision? What is the kinetic energy KEi of the bullet and block system immediate before the collision? What is the kinetic energy KEr of the bullet and block system right after the collision? Find the kinetic energy loss due to the collision.

Explanation / Answer

(a) Let the velocity of the block and bullet system just after the collision = V

Applying energy conservation between the state of the system just after the collision and the state of maximum height

1/2*(m1+m2)*v2 = (m1+m2)*g*h or v = (2*g*h)0.5 = (2*9.8*0.135)0.5 = 1.627 m/s

(b) Now applying momentum conservation before and after the collision

m2*V01 = (m1+m2)*V or V01 = (1.25+0.0625)*1.627/0.0625 = 34.167 m/s

(c) Kinetic energy of system just before collision (KEi) = 1/2*m2*V012 = 1/2*0.0625*(34.167)2 = 36.48 J

(d) Kinetic energy of system just after collision (KEf) = 1/2*(m1+m2)*V2 = 1/2*1.3125*(1.627)2 = 1.74 J

(e) Loss in Kinetic energy = KEi - KEf = 36.48 - 1.74 = 34.74 J

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