A 4.00 g bullet is moving horizontally with a velocity of +355 m/s (moving to th

Question

A 4.00 g bullet is moving horizontally with a velocity of +355 m/s (moving to the right). The bullet is approaching two blocks resting on a horizontal frictionless surface (air resistance is negligible). The bullet passes completely through the rst block (mass 1150 g) (an inelastic collision) and embeds itself in the second one (mass 1530 g), the velocity of the rst block after the bullet passes through is +0.550m/s. (a) What is the velocity of the second block after the bullet embeds itself? (b) Find the ratio of the total kinetic energy after the collisions to that before the collisions.

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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