Question

Forensics Test 1 2 3 4 As part of a forensics test, a 0.005 kg bullet is fired into a 0.52 kg stationary wooden block. The bullet becomes embedded in the block. The initial speed of the bullet is 390 m/s. 1) What is the initial velocity of the center of mass of the bullet-block system before the collision? m/sec Submit 2) What is the velocity of the block (with the embedded bullet) after the collision? m/sec Submit Help 3) How much mechanical energy was lost during the collision? (This mechanical energy was transformed into sound and friction generated heat.) J Submit Help 4) Now lets say that instead of being embedded in the block, the bullet comes out the other side of the block with a velocity of 284 m/sec. Assuming the block remains intact, what is its velocity after the collision? m/sec Submit Help

Answer 1

Blues Band 1092 11 -525 7> V, .525 mitma 0.005 ☺ ucom = (0.005) (390) + (52)(0) Sm, uitm U23 - 3 mitra 9 (0.005 +.520) com = 3.71 mis ") Velocity of block Affece embeded Pe=pf For block (390) (0.005) to = (0.525) v, bullet system so v = 3.71 mis () Koce = (+) (0.005)(390)2 = 380.25 kile = (5) (.525) (3.7103 = 3.613 so lass= kiceg - kocê = 376.637 Joule. | using momentum conservation Pe=pf (Block - Bullet system? (390) (0.005) +0 = (284) (0.005)+ \ (:52) TV2 = 10ls