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

Question

A 4.00kg bullet is moving horizontally with a velocity of 355m/s, where the + sign indicates that it is 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 first block( an inelastic collision) and embeds itself in the second one. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1150g, and its velocity +0.550m/s after the bullet passes through it. The mass of the second block is 1530g. a) what is the velocity of the second block after the bullet imbeds itself? b) find the ratio of the totally kinetic energy after the collision to that before the collision.

Explanation / Answer

Initial momentum = 0.004*355 = 1.42 Final momentum = 1.15*0.550 + (1.534*v) equating v = 0.513 m/s Total kinetic energy before collision = 0.5*0.004*355^2 = 252.05 Final energy = 0.3758 Ratio = 1.49*10^-3

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