A 4.00-g bullet is moving horizontally with a velocity of +365 m/s, where the si

Question

A 4.00-g bullet is moving horizontally with a velocity of +365 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 1243 g, and its velocity is +0.552 m/s after the bullet passes through it. The mass of the second block is 1547 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

When the bullet punches through the first block, it loses some speed. To find out how much, use

mbulletvibullet=mblockvfblock + mbulletvfbullet

and solve for the bullets final velocity:

(mbulletvibullet - mblockvfblock)/mbullet=(.00515kg*359m/s - 1.217kg*0.628m/s)/.00515kg=210.6m/s

with the second block, the bullet imbeds, so you know you have a completely inellastic collision, and can use:

mbulletvibullet + mblockvfblock=(mblock+mbullet)vf and solve for vf

mbulletvibullet/(mblock+mbullet)=vf -->block is stationary, so it has no initial momentum

a) vf=(.00515kg*210.6m/s)/(.00515kg+1.649kg)=.66m/s

b)before the collision, the bullet had 1/2mv^2=1/2(.00515kg)(359m/s)^2=331.9J

After both collisions, the kinetic energy was 1/2(.00515kg + 1.649kg)(.66m/s)^2=.36J

so the ration is 331.9/.36=921.9:1, or say 922:1

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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