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

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 (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 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 embeds itself? (b) Find the ratio of the total kinetic energy after the collisions to that before the collisions.

Explanation / Answer

apply law of conservation of momentum

momentum before collision = momentum after collision wi the first block

4*355 = 1150*v1

v1 = 4.355/1150 = 1.24 m/s

again after embedding with second block

1150*1.24 = (1150+1530)*V

V = 1150*1.24/(1150+1530) = 0.53 m/s..............is teh answer for A)

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B) here in the question it was not clearly mention that to whom we need to calculte the required ratio

here i am finding the ratio of KE of the bullet

KE after the collision KEf = 0.5*4*10^-3*0.53^2 = 0.0005618 J

KE before the collisions is KEi = 0.5*4*10^-3*355^2 = 252.05 J

So KEf/KEi = 0.0005618/252.05 = 2.23

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