A 5.81-g bullet is moving horizontally with a velocity of +359 m/s, where the si

Question

A 5.81-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 1187 g, and its velocity is +0.664 m/s after the bullet passes through it. The mass of the second block is 1522 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

given, mass of bullet, m = 5.81*10^-3 kg

speed of bullet, v = 359 m/s

mass of first block, m1 = 1.187 kg

mass of second block, m2 = 1.522 kg

velocitu of first block after bullet passes througfh it, u = 0.664 m/s

a. let speed of second block after collision with the bullet be v'

then from conservation of momentum

mv = m1u + (m + m2)v'

5.81*10^-3*359 = 1.187*0.664 + (1.522+5.81*10^-3)*v'

v' = 0.84933 m/s

b. initial KE = 0.5mv^2 = 374.399305 J

final KE = 0.5m1u^2 + 0.5(m + m2)v'^2 = 0.8127 J

ratio of total KE after collision to total KE before collision = 0.00217

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.