A 4.00-g bullet is moving horizontally with a velocity vector v = 343 m/s, where

Question

A 4.00-g bullet is moving horizontally with a velocity vector v = 343 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 +0.743 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

Momentum conservation

M1u1 + m2u2 = m1v1 + m2v2

For the first collision,

M1 (bullet)= 0.004 , u1 = 343 , v1 = ?

M2 (first block)= 1.150 , u2 = 0 , v2 = 0.743

For the second collision,

M1 (bullet)= 0.004 , u1 = 129.3875 , v1 = v

M2 (second block)= 1.530 , u2 = 0 , v2 = v

Total kinetic energy before the collision = kinetic energy of bullet

= 0.5 mv2

= 0.5 * 0.004 * 3432

= 235.298 J

Total kinetic energy after the collision

= kinetic energy of (bullet + first block + second block)

= 0.5 *0.004* 0.33732 + 0.5*1.150*0.7432 + 0.5*1.53*0.33732

= 0.40469 J

Energy before / energy after

= 0.44069 / 235.298

= 1.72 * 10-3

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