A 4.00-g bullet is moving horizontally with a velocity = 365 m/s, where the sign

Question

A 4.00-g bullet is moving horizontally with a velocity = 365 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 1150 g, and its velocity is +0.477 m/s after the bullet passes through it. The mass of the second block is 1530 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. I have already found Part (a), which is 0.59 m/s. please find part (b)

Explanation / Answer

Mb (mass of bullet) = .004 kg M1 (mass of block 1) =1.15 kg M2 (mass of block 2) =1.53 kg. MV = P (momentum) Mb (Vob- Velocity of Bullet) = P .004 (355) = 1.42 kgm/s Final momentum of the bullet = Mbvf MbVf = 1.42 - M1 Vfb1 (final velocity of block 1) 1.42-.632=.004 (v) Vfb= 197m/s Now to find the final velocity of the Block 2 .7875= 1.534 (v) Vfb2= .514 m/s Thats the first part, My second part is this .5(m)(v^2) = KE KE is not conserved, therefore KEf/KEo= the ratio .002 (355^2) =252 J =KEo .5(1.15)(.55^2) + .5(1.534)(.514^2) ,375 J .375/252=.0014 .14:1 is the ratio

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