A 8.63-g bullet is moving horizontally with a velocity of +355 m/s, where the si

Question

A 8.63-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 1226 g, and its velocity is +0.658 m/s after the bullet passes through it. The mass of the second block is 1580 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 is conserved

pi = pf………………………………….eq1

mb*vb = m1v1 + (m2+mb)v2

pi = mb*vb

pi = 0.00863kg*355m/s

pi = 3.06365 kg-m/s

pf = m1v1 + (m2+mb)v2
pf = 1.226kg*0.658m/s + (1.58+0.00863)v2
pf = 1.226kg*0.658m/s + (1.58+0.00863)v2
pf = 0.806708kg-m/s + 1.58863*v2

Calculating for v2 by using eq1

3.06365 kg-m/s = 0.806708kg-m/s + 1.58863*v2

v2 = 1.42 m/s

Part b:

Kafter = {½*1.226*0.6582}+{ ½ *(1.58863)*1.422)

Kafter = 0.265406932 + 1.601656766

Kafter = 1.867 J

Kbefore = ½*0.00863*3552

Kbefore = 543.797875J

So the ratio = 1.867/543.797

ratio = 0.003433

ratio = 3.43 x 10-3

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