A 8.39-g bullet is moving horizontally with a velocity of +350 m/s, where the si

Question

A 8.39-g bullet is moving horizontally with a velocity of +350 m/s, where the sign + indicates that it is frictionless surface. Air resistance is moving after the collision with the bullet. The mass of the first block is 1227 g, and its velocity is +0.597 m/s after the bullet the second mbeds itself drawing). The bullet is approaching two blocks resting on a horizontal completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated block after the bullet in ? (b) Find the ratio of the total kinetic energy after the collision to that before the collision. in part b. Note that both blocks are block is 1542 9. (a) What is the velocity of +350n's Block Block 2 (a) Before collision 0.597hvs mtixt 1-1222 bak 2 8.39 (D) After collision (a) vblock2 Number Units (b) KEafter/KEbefoNumber Units

Explanation / Answer

The problem can be solved using momentum conservation method

initial momentum of system = 0.00839*350=2.9365 kg-m/s

final momentum will be equal to initial momentum

velocity of block 1=0.597 m/s

mass of block 1=1.227 kg

momentum of block 1=0.597*1.227=0.7325 kg-m/s

momentum of bullet + block 2 = 2.9365-0.7325 = 2.204 kg-m/s

mass of bullet + block 2 = 1.542+0.00839 = 1.55039 kg

velocity of block 2 = 2.204/1.55039=1.42 m/s

initial kinetic energy = 0.5*0.00839*3502 = 513.8875 J

final kinetic energy = 0.5*1.227*0.5972 + 0.5*1.55039*1.422 = 1.78 J

kinetic energy after / kinetic energy before = 1.78/513.8875=0.0035

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