Doing some physics homework and am kind of stumped here! **Dynamic Systems: Bullet and Block Interaction**In this physics problem, an 8.0-g bullet is fired into a 4.0-kg block that is initially at rest on a frictionless horizontal surface. As illustrated in the accompanying diagram, the bullet remains embedded in the block. The combined bullet-block system subsequently compresses an ideal, massless spring by a distance of 8.7 cm. Relevant parameters for the scenario include:- Spring constant (\(k\)) = 2400 N/m.The question asks for the initial velocity of the bullet. The possible options are:- a) 900 m/s- b) 1200 m/s- c) 1100 m/s- d) 1000 m/s- e) 1300 m/s**Diagram Explanation:**The diagram consists of three main elements:1. **Bullet and Block Setup**: On the left, an 8.0-g bullet (depicted as an arrow) approaches a 4.0-kg block positioned on a horizontal track.2. **Spring System**: A spring is shown to the right of the block, representing the system into which the bullet-block combination will move, causing compression.3. **Compression and Rest Position**: A wedge labeled with the spring constant \(k\) shows the final state where the spring is compressed by the block.This scenario serves as a practical demonstration of the principles of conservation of momentum and energy in dynamic systems.

Given Mass of the bullet is m=8.0 g=8.0×10-3 kg Mass of the block is M=4.0 kg Compression of the…

An 8.0-g bullet is shot into a 4.0-kg block, at rest on a frictionless horizontal surface (see the figure). The bullet remains lodged in the block. The block moves into an ideal massless spring and compresses it by 8.7 cm. The spring constant of the spring is 2400 N/m. The initial velocity of the bullet is closest to мей O O O O O 4.0 kg O 900 m/s. O1200 m/s. O 1100 m/s. O1000 m/s. 1300 m/s.

An 8.0-g bullet is shot into a 4.0-kg block, at rest on a frictionless horizontal surface (see the figure). The bullet remains lodged in the block. The block moves into an ideal massless spring and compresses it by 8.7 cm. The spring constant of the spring is 2400 N/m. The initial velocity of the bullet is closest to мей O O O O O 4.0 kg O 900 m/s. O1200 m/s. O 1100 m/s. O1000 m/s. 1300 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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Concept explainers

Spring Potential Energy

The potential energy is the energy possessed by a body by virtue of its position or the energy held by the object due to its position relative to other objects, its force like stresses, charge, and other factors affecting the particles of the body. The potential energy is the measure of the net work done by or on the body. The spring potential energy is the potential energy stored due to the deformation of an elastic spring and this elasticity is due to the stretched spring. The elasticity is the ability of a solid-state matter to come back to its equilibrium position or original position after any kind of external force is acted on it. It is also known as Elastic potential energy. The potential energy here is caused due to the displacement of the spring from its equilibrium position to another position and this potential energy is equal to the net work done due to the stretching of the spring. It also resembles the same mechanics of the oscillation of a simple pendulum, where due to the external force, the object moves from its equilibrium position to its maximum ends and the periodic motion continues till it comes back to the equilibrium position to rest. 

Question

Doing some physics homework and am kind of stumped here!

**Dynamic Systems: Bullet and Block Interaction**

In this physics problem, an 8.0-g bullet is fired into a 4.0-kg block that is initially at rest on a frictionless horizontal surface. As illustrated in the accompanying diagram, the bullet remains embedded in the block. The combined bullet-block system subsequently compresses an ideal, massless spring by a distance of 8.7 cm.

Relevant parameters for the scenario include:
- Spring constant (\(k\)) = 2400 N/m.

The question asks for the initial velocity of the bullet. The possible options are:

- a) 900 m/s
- b) 1200 m/s
- c) 1100 m/s
- d) 1000 m/s
- e) 1300 m/s

**Diagram Explanation:**

The diagram consists of three main elements:
1. **Bullet and Block Setup**: On the left, an 8.0-g bullet (depicted as an arrow) approaches a 4.0-kg block positioned on a horizontal track.
2. **Spring System**: A spring is shown to the right of the block, representing the system into which the bullet-block combination will move, causing compression.
3. **Compression and Rest Position**: A wedge labeled with the spring constant \(k\) shows the final state where the spring is compressed by the block.

This scenario serves as a practical demonstration of the principles of conservation of momentum and energy in dynamic systems.

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