A 4.15-g bullet is moving horizontally with a velocity of +340 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 1196 g, and its velocity is +0.644 m/s after the bullet passes through it. The mass of the second block is 1518 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. +340m/s Block 1 (a) Before collision +0.644m/s mblock 1=1196% (b) After collision (a) Vblock2= Number V Block 2 mblock 2=1518 Mbullet =4.15g i block 2 1516 Units m/s

Principles of Physics: A Calculus-Based Text
5th Edition
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Chapter8: Momentum And Collisions
Section: Chapter Questions
Problem 6P: A girl of mass mg is standing on a plank of mass mp. Both are originally at rest on a frozen lake...
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Part 1
A 4.15-g bullet is moving horizontally with a velocity of +340 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 1196 g, and its velocity is +0.644 m/s after the
bullet passes through it. The mass of the second block is 1518 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.
+340m/s
Block 1
(a) Before collision
+0.644m/s
mblock 1=1196%
(b) After collision
(a) Vblock2= Number
(b) KE after/KEbefore = Number
Block 2
mblock 2=1518
Mbullet =4.15g
i
1516
block 2
1.597E-3
Units
Units
m/s
No units.
Transcribed Image Text:A 4.15-g bullet is moving horizontally with a velocity of +340 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 1196 g, and its velocity is +0.644 m/s after the bullet passes through it. The mass of the second block is 1518 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. +340m/s Block 1 (a) Before collision +0.644m/s mblock 1=1196% (b) After collision (a) Vblock2= Number (b) KE after/KEbefore = Number Block 2 mblock 2=1518 Mbullet =4.15g i 1516 block 2 1.597E-3 Units Units m/s No units.
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