A 5.58-g bullet is moving horizontally with a velocity of +353 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 1210 g, and its velocity is +0.558 m/s after the bullet passes through it. The mass of the second block is 1557 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. ### Conservation of Momentum: An ExampleThis educational material illustrates the principle of conservation of momentum using a collision scenario involving two blocks and a bullet.#### Description of the Scenario:**(a) Before Collision:**- **Block 1** and **Block 2** are at rest on a surface.- A bullet is moving toward **Block 1** with a velocity of \(+353 \text{ m/s}\).**(b) After Collision:**- After the bullet collides with and passes through **Block 1**, **Block 1** moves to the right with a velocity of \(+0.558 \text{ m/s}\).- The bullet continues its path and collides with **Block 2**.- The velocity of **Block 2** after the collision (denoted as \(v_{\text{block 2}}\)) is to be determined.#### Masses of the Objects:- Mass of **Block 1** (\(m_{\text{block 1}}\)) = 1210 g- Mass of **Block 2** (\(m_{\text{block 2}}\)) = 1557 g- Mass of the bullet (\(m_{\text{bullet}}\)) = 5.58 g#### Explanation:This example demonstrates the application of the conservation of momentum principle in collisions. In such systems, the total momentum before the collision is equal to the total momentum after the collision, provided no external forces act on the system. The momentum of each object is the product of its mass and velocity. By accounting for the masses and velocities before and after the collision, students can learn how to apply the conservation of momentum to predict the resulting velocities of the objects involved.

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Description: A 5.58-g bullet is moving horizontally with a velocity of +353 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 resistan

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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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A 5.58-g bullet is moving horizontally with a velocity of +353 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 1210 g, and its velocity is +0.558 m/s after the bullet passes through it. The mass of the second block is 1557 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.

### Conservation of Momentum: An Example

This educational material illustrates the principle of conservation of momentum using a collision scenario involving two blocks and a bullet.

#### Description of the Scenario:

**(a) Before Collision:**
- **Block 1** and **Block 2** are at rest on a surface.
- A bullet is moving toward **Block 1** with a velocity of \(+353 \text{ m/s}\).

**(b) After Collision:**
- After the bullet collides with and passes through **Block 1**, **Block 1** moves to the right with a velocity of \(+0.558 \text{ m/s}\).
- The bullet continues its path and collides with **Block 2**.
- The velocity of **Block 2** after the collision (denoted as \(v_{\text{block 2}}\)) is to be determined.

#### Masses of the Objects:

- Mass of **Block 1** (\(m_{\text{block 1}}\)) = 1210 g
- Mass of **Block 2** (\(m_{\text{block 2}}\)) = 1557 g
- Mass of the bullet (\(m_{\text{bullet}}\)) = 5.58 g

#### Explanation:

This example demonstrates the application of the conservation of momentum principle in collisions. In such systems, the total momentum before the collision is equal to the total momentum after the collision, provided no external forces act on the system.

The momentum of each object is the product of its mass and velocity. By accounting for the masses and velocities before and after the collision, students can learn how to apply the conservation of momentum to predict the resulting velocities of the objects involved.

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