Question

Please let me know the answer and how to find it. Thank you!

A 8.57-g bullet is moving horizontally with a velocity of +355 m/s, where the sign + indicates that it is moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1176 g, and its velocity is +0.681 m/s after the bullet passes through it. The mass of the second block is 1618 g. (a) What is the velocity of the second block after the bullet imbeds itself? (b) Find the ratio of the total kinetic energy after the collision to that before the collision. +355m/s Block 1 Block 2 (a) Before collision +0.68 lm/s block 2 m block 1 = 1176 mblock 2 = 1618 mbullet = 8.578 (b) After collision (a) Vblock2 = Number Units m/s (b) KE after/KEbefore = Number Units

Answer 1

=) 8657(355) = 1176(0-681) + (1618 +857) Vz :0-681mbs u=355 m/s m2 mi m2 tmb mi ny 8.57 g 11769 16189 Conservation of linear Momentum (molu + mi (0) +m2(0) = m; v; +(matme) Yz Dlv; = 10348 mis? kEofler - m, 0,2+ + ₂ tmp) V₂² £m4²4 oto KE before /m2 + men hel + - (355 +2) ma mb 1176 (0.681 2 1618 +1161.37812 8.57 355 8.57 = 5.0497X16** + 0.28598 NIO 5.336 X10 all the Best!!