A .004 kg bullet is moving horizontally witha velocity of +355 m/s, where the +
ID: 1680790 • Letter: A
Question
A .004 kg bullet is moving horizontally witha velocity of +355 m/s, where the + sign indicatesthat is moving to the right. The bullet is approaching two blocksresting on a horizontal frictionless surface. Air resistance isnegligible. The bullet passes completely through the 1st block& embeds itself in the 2nd block. Note that both blocks aremoving after the collision with the bullet. The mass of the 1stblock is 1.15 kg and its velocity is +0.55 m/s after the bullet passes through it. The mass ofthe 2nd block is 1.53 kg. a) what is the velocity of the 2nd block after the bulletimbeds itself, v2? b) find the ration of the total kinetic energy after thecollision to the total kinetic energy before the collision. Thanks! A .004 kg bullet is moving horizontally witha velocity of +355 m/s, where the + sign indicatesthat is moving to the right. The bullet is approaching two blocksresting on a horizontal frictionless surface. Air resistance isnegligible. The bullet passes completely through the 1st block& embeds itself in the 2nd block. Note that both blocks aremoving after the collision with the bullet. The mass of the 1stblock is 1.15 kg and its velocity is +0.55 m/s after the bullet passes through it. The mass ofthe 2nd block is 1.53 kg. a) what is the velocity of the 2nd block after the bulletimbeds itself, v2? b) find the ration of the total kinetic energy after thecollision to the total kinetic energy before the collision. Thanks! A .004 kg bullet is moving horizontally witha velocity of +355 m/s, where the + sign indicatesthat is moving to the right. The bullet is approaching two blocksresting on a horizontal frictionless surface. Air resistance isnegligible. The bullet passes completely through the 1st block& embeds itself in the 2nd block. Note that both blocks aremoving after the collision with the bullet. The mass of the 1stblock is 1.15 kg and its velocity is +0.55 m/s after the bullet passes through it. The mass ofthe 2nd block is 1.53 kg. a) what is the velocity of the 2nd block after the bulletimbeds itself, v2? b) find the ration of the total kinetic energy after thecollision to the total kinetic energy before the collision. Thanks!Explanation / Answer
m = 0.004 kg, v = 355 m/s,m1 = 1.15 kg, v1 =0.55 m/s, m2 = 1.53 kg.a) what is the velocity of the 2nd block after the bulletimbeds itself, v2?
momentum conservation:
mv = m1v1 + (m +m2)v2
v2 = (mv - m1v1)/(m +m2) = 0.51 m/s
b) find the ratio of the total kinetic energy after thecollision to the total kinetic energy before the collision.
total kinetic energy after the collision =m1v12/2 + (m +m2)v22/2
total kinetic energy before the collision = mv2/2
the ratio = [m1v12 + (m +m2)v22]/(mv2) =0.0015
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