/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 A \(5.20 \mathrm{~g}\) bullet mo... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(5.20 \mathrm{~g}\) bullet moving at \(672 \mathrm{~m} / \mathrm{s}\) strikes a \(700 \mathrm{~g}\) wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to \(428 \mathrm{~m} / \mathrm{s}\). (a) What is the resulting speed of the block? (b) What is the speed of the bullet-block center of mass?

Short Answer

Expert verified
(a) The speed of the block is 1.8126 m/s. (b) The center of mass speed is 4.955 m/s.

Step by step solution

01

Understand Initial Momentum

To solve the problem, we need to consider the conservation of linear momentum. Before the collision, only the bullet is moving, so the total initial momentum of the system is due to the bullet alone. Use the formula for momentum: \( p = mv \), where \( m \) is mass and \( v \) is velocity. The initial momentum of the bullet is \( p_i = 5.20 \times 10^{-3} \times 672 \). Calculate this value.
02

Calculate Initial Momentum of the Bullet

Calculate the initial momentum of the bullet:\[p_i = 5.20 \text{ g} \times 672 \text{ m/s} = 5.20 \times 10^{-3} \text{ kg} \times 672 \text{ m/s} = 3.4944 \text{ kg m/s}\]
03

Determine Final Momentum of the Bullet After Collision

After the collision, the bullet's speed is reduced to \(428 \text{ m/s}\). Calculate its final momentum using the same formula for momentum:\[p_f = 5.20 \times 10^{-3} \times 428 = 2.2256 \text{ kg m/s}\]
04

Calculate Change in Momentum

Find the change in momentum of the bullet, which is transferred to the block:\[\Delta p = p_i - p_f = 3.4944 - 2.2256 = 1.2688 \text{ kg m/s}\]
05

Find the Resulting Speed of the Block

The change in momentum calculated is the momentum gained by the block. Since momentum of the block \( p_{block} = m v \), the velocity of the block is given by:\[v = \frac{\Delta p}{m_{block}} = \frac{1.2688}{0.700} = 1.8126 \text{ m/s}\]
06

Determine Speed of Bullet-Block Center of Mass

The speed of the center of mass is calculated by dividing the total momentum by the total mass of the system. The initial total momentum is \(3.4944 \text{ kg m/s}\). The total mass is \( 5.20 \times 10^{-3} + 0.700 = 0.7052 \text{ kg}\).\[V_{cm} = \frac{3.4944}{0.7052} = 4.955 \text{ m/s}\]
07

Verify and Discuss Results

Verify all steps for consistency with the principles of momentum conservation. The resulting block speed is \(1.8126 \text{ m/s}\), and the center of mass of the system moves at \(4.955 \text{ m/s}\). Ensure the problem setup is understood with no external forces considered (frictionless surface).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Momentum
Linear momentum is a fundamental concept in physics, representing the quantity of motion an object possesses. It's a vector quantity, meaning it has both magnitude and direction.
The equation for linear momentum is:
  • \( p = mv \), where \( p \) is momentum, \( m \) represents mass, and \( v \) is velocity.
In our problem, the bullet has a certain amount of linear momentum before and after the collision with the wooden block. Initially, as the bullet moves, it solely carries all the system’s momentum. The block is at rest, meaning its initial momentum is zero. After the collision, part of this momentum is transferred to the block.
Tracking these changes is key to solving for the block's speed and ensuring that the conservation of momentum holds true.
Understanding momentum is essential because it allows us to predict the outcome of dynamic interactions like collisions. In essence, momentum is all about understanding how motion gets shared and changed but never lost.
Collision Dynamics
Collision dynamics involve examining how moving objects interact, particularly how they exchange momentum and energy.
When a bullet hits a block, we witness a collision, a critical type of interaction in physics. Collisions can be classified into different types, depending on how momentum and kinetic energy are conserved.
  • **Elastic Collision:** Both momentum and kinetic energy are conserved.
  • **Inelastic Collision:** Only momentum is conserved, and kinetic energy is not.
In the context of our problem, the interaction is closer to inelastic because the bullet and block do not stick together, but the bullet's speed changes. In most physics problems, ideal conditions are considered, such as a frictionless surface, to simplify calculations and focus on the core dynamics of the collision.
By evaluating momentum before and after the collision, we solved for the resulting speed of the block. This highlights how collision dynamics and the conservation of momentum work hand-in-hand to provide predictable and consistent outcomes.
Center of Mass
The center of mass of a system is a crucial concept used to simplify the analysis of motion. It represents the average position of all the mass in the system.
The center of mass moves according to the total momentum of the system. To locate the center of mass in motion, consider the equation:
  • \( V_{cm} = \frac{p_{total}}{m_{total}} \), where \( V_{cm} \) is the velocity of the center of mass, \( p_{total} \) is the total momentum, and \( m_{total} \) is the total mass.
In our example, the bullet and block system had a total initial momentum from the bullet. After the collision, understanding the motion of the center of mass helps us comprehend the system's overall movement without tracking each mass individually. This is especially useful when studying larger systems with more components.
The calculation showed the center of mass progressing forward to a speed of approximately \( 4.955 \text{ m/s} \), carrying forward the essence of momentum conservation across the entire system.

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Most popular questions from this chapter

A cart with mass \(340 \mathrm{~g}\) moving on a frictionless linear air track at an initial speed of \(1.2 \mathrm{~m} / \mathrm{s}\) undergoes an elastic collision with an initially stationary cart of unknown mass. After the collision, the first cart continues in its original direction at \(0.66 \mathrm{~m} / \mathrm{s}\). (a) What is the mass of the second cart? (b) What is its speed after impact? (c) What is the speed of the twocart center of mass?

Two skaters, one with mass \(65 \mathrm{~kg}\) and the other with mass 40 \(\mathrm{kg}\), stand on an ice rink holding a pole of length \(10 \mathrm{~m}\) and negligible mass. Starting from the ends of the pole, the skaters pull themselves along the pole until they meet. How far does the \(40 \mathrm{~kg}\) skater move?

A rocket sled with a mass of \(2900 \mathrm{~kg}\) moves at \(250 \mathrm{~m} / \mathrm{s}\) on a set of rails. At a certain point, a scoop on the sled dips into a trough of water located between the tracks and scoops water into an empty tank on the sled. By applying the principle of conservation of linear momentum, determine the speed of the sled after 920 kg of water has been scooped up. Ignore any retarding force on the \(\mathrm{scoop}\)

Two \(2.0 \mathrm{~kg}\) bodies, \(A\) and \(B\), collide. The velocities before the collision are \(\vec{v}_{A}=(15 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}\) and \(\vec{v}_{B}=(-10 \hat{\mathrm{i}}+5.0 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}\). After the collision, \(\vec{v}_{A}^{\prime}=(-5.0 \hat{\mathrm{i}}+20 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}\). What are (a) the final velocity of \(B\) and (b) the change in the total kinetic energy (including sign )?

In the Olympiad of 708 B.C., some athletes competing in the standing long jump used handheld weights called halteres to lengthen their jumps (Fig. \(9-56\) ). The weights were swung up in front just before liftoff and then swung down and thrown backward during the flight. Suppose a modern \(78 \mathrm{~kg}\) long jumper similarly uses two \(5.50 \mathrm{~kg}\) halteres, throwing them horizontally to the rear at his maximum height such that their horizontal velocity is zero relative to the ground. Let his liftoff velocity be \(\vec{v}=(9.5 \hat{\mathrm{i}}+4.0 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}\) with or without the halteres, and assume that he lands at the liftoff level. What distance would the use of the halteres add to his range?

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