Part 1 **Problem Description and Solution Steps**A 4.15-g bullet is moving horizontally with a velocity of +340 m/s. The positive sign indicates it is moving to the right (refer to diagram part a). The bullet approaches 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 shown in part b. After the collision, both blocks are moving. The first block has a mass of 1196 g and a velocity of +0.644 m/s. The second block has a mass of 1518 g. **Diagram and Explanation**- **(a) Before collision:** - A bullet is approaching Block 1 with a velocity of +340 m/s. - Block 2 is stationary. - **(b) After collision:** - Block 1 is moving with a velocity of +0.644 m/s. - Block 2 is moving with a velocity labeled as \( v_{\text{block2}} \).**Masses:**- \( m_{\text{block1}} = 1196 \, \text{g} \)- \( m_{\text{block2}} = 1518 \, \text{g} \)- \( m_{\text{bullet}} = 4.15 \, \text{g} \)**Questions:**(a) What is the velocity of the second block after the bullet embeds itself?(b) Find the ratio of the total kinetic energy after the collision to that before the collision.**Solutions:**(a) \( v_{\text{block2}} = \) 1516 m/s (b) Ratio \( \frac{KE_{\text{after}}}{KE_{\text{before}}} = 1.597 \times 10^{-3} \) No units for the ratio as it is dimensionless.

Answered: Image /qna-images/answer/21c5756a-33a6-4d48-b43b-6851aef41eaf.jpg

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

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

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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