A 6.60-g bullet is moving horizontally with a velocity of +359 m/s, where the si

Question

A 6.60-g bullet is moving horizontally with a velocity of +359 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 1173 g, and its velocity is +0.650 m/s after the bullet passes through it. The mass of the second block is 1601 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

here ,

let the speed of bullet after passing through the first block is v1

Using conservation of momentum after first collision

a)

1173 * 0.65 = 6.6 * ( 359 - v1 )

v1 = 243.5 m/s

Now, let the final speed is v

using conservation of momentum

(1601 + 6.6) * v2 = 243.5 * 6.6

v2 = 1 m/s

the velocity of second block after the bullet is 1 m/s

b)

ratio of kinetic energy = 0.5 * ( 1.601 * 1^2 + 1.173 * 0.65^2 + 0.0066 * 1^2)/(0.5 * 0.0066 * 359^2)

ratio of kinetic energy = 0.00247

the ratio of kinetic energy after and before collision is 0.00247

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