A 4.00-g bullet is moving horizontally with a velocity of +355 m/s, where the +

Question

A 4.00-g bullet is moving horizontally with a velocity of +355 m/s, where the + sign indicates that it is moving to the right.  The bullet is approching two blocks resting  on a horzizonta frictionless surface.  Air resistance is negligible.  The bullet passes completely through the first block (an inelastic collision)  and embeds itself in the second block.  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 +0.550 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

Explanation / Answer

Given :    From the conservation of the momentum we have                         0.004 kg * 355 m/s = 1.15kg *0.55 m/s + (1.53kg +0.004kg )*v where v is the velocity of the second block so solving this we get   v = +0.5133 m/s the ratio is = [0.5(1.15kg )(0.55) 2+(1.534kg)(0.5133)2 ] / [ 0.5 (0.004kg)(355)2 ]                   = --------------- Solve it I hope it helps you                         0.004 kg * 355 m/s = 1.15kg *0.55 m/s + (1.53kg +0.004kg )*v where v is the velocity of the second block so solving this we get   v = +0.5133 m/s the ratio is = [0.5(1.15kg )(0.55) 2+(1.534kg)(0.5133)2 ] / [ 0.5 (0.004kg)(355)2 ]                   = --------------- Solve it I hope it helps you

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