.B: Collsions in 20 A bullet of mass 0.116 kg traveling horizontally at a speed
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Question
.B: Collsions in 20 A bullet of mass 0.116 kg traveling horizontally at a speed of 100 m/s embeds itself in a block of mass 3.5 kg that is sitting at rest on a nearly frictionless surface. of the block after the bullet embeds itself in the bl m/s (b) Calculate the kinetic energy of the bullet plus the block before the collision (c) Calculiate the knetic energy of the bullet plus the block after the collision: d) Was this collision elastic or inelastic? O Inelastic D not enough information to tell O elastic (e) Calculate the rise in thermal energy of the bullet plus block as a result of the collision: + Ethermal.block- (n What was the transfer of energy Q (microscopic work) from the surroundings into the block +bullet system during the collision? (Remember that Q represer energy transfer due to a temperature difference between a system and its Additional Materials D) Section 10.6Explanation / Answer
(a)By the law of momentum conservation:-
=>m1u1+m2u2=(m1+m2)v
=>0.116 x 100 = (0.116+3.5) x v
=>v = 3.2 m/s
(b) KE(before) = 1/2 x m1 x u1^2 = 1/2 x 0.116 x (100)^2 = 580 J
(c) KE(after) = 1/2 x (m1+m2) x v^2 = 1/2 x 3.616 x (3.2)^2 = 18.5 J
(d)Rise in thermal energy = Loss in KE = KE(before) - KE(after) = 580 - 18.5 = 561.49 J
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