/*! 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 62 A \(5.00 \mathrm{~g}\) bullet tr... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(5.00 \mathrm{~g}\) bullet traveling horizontally at \(450 \mathrm{~m} / \mathrm{s}\) is shot through a \(1.00 \mathrm{~kg}\) wood block suspended on a string \(2.00 \mathrm{~m}\) long. If the center of mass of the block rises a distance of \(0.450 \mathrm{~cm},\) find the speed of the bullet as it emerges from the block.

Short Answer

Expert verified
The bullet's speed as it emerges from the block is approximately 391 m/s.

Step by step solution

01

Determine Potential Energy Increase

First, calculate the potential energy increase of the block as it rises. Use the formula for gravitational potential energy: \( \Delta U = mgh \), where \( m \) is the mass of the block, \( g \) is the acceleration due to gravity (approximately \( 9.81 \mathrm{~m/s^2} \)), and \( h \) is the height it rises. Convert 0.450 cm to meters: \( h = 0.00450 \mathrm{~m} \). Substitute in the values: \( \Delta U = 1.00 \mathrm{~kg} \times 9.81 \mathrm{~m/s^2} \times 0.00450 \mathrm{~m} = 0.044145 \mathrm{~J} \).
02

Use Conservation of Energy

By the conservation of energy, the initial kinetic energy transferred to the block by the bullet is equal to the potential energy at the highest point. The initial kinetic energy of the block \( K_i \) is \( \Delta U = 0.044145 \mathrm{~J} \). This is the kinetic energy imparted to the block by the bullet.
03

Calculate Initial Velocity of Block

Use the kinetic energy formula \( K = \frac{1}{2} mv^2 \) to find the initial velocity \( v \) of the block. Set \( K = 0.044145 \mathrm{~J} \) and solve for \( v \): \( 0.044145 = \frac{1}{2} \times 1.00 \times v^2 \). Therefore, \( v = \sqrt{2 \times 0.044145} \approx 0.297 \mathrm{~m/s} \).
04

Apply Conservation of Momentum

Apply the principle of conservation of momentum. Initial momentum of system = Final momentum of system. Initially, only the bullet contributes to momentum: \( p_i = 0.00500 \times 450 \). Final momentum = momentum of bullet after passing through + max momentum of block: \( p_f = 0.00500v_f + 1.00 \times 0.297 \). Equate and solve: \( 2.25 = 0.00500v_f + 0.297 \).
05

Solve for Final Bullet Speed

Rearrange and solve for \( v_f \): \( 2.25 - 0.297 = 0.00500v_f \), thus \( v_f = \frac{1.953}{0.00500} = 390.6 \mathrm{~m/s} \).

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. It is directly proportional to the mass of the object and the square of its velocity. This is why faster-moving objects and those with more mass have significantly higher kinetic energy levels. In physics, we calculate kinetic energy using the formula:
  • \( K = \frac{1}{2} mv^2 \)
where \( m \) is the mass and \( v \) is the velocity of the object.
In the exercise provided, when the bullet travels through the block, it initially transfers part of its kinetic energy to the block. This energy then gets transformed, leading to the block's movement. Therefore, analyzing changes in kinetic energy provides insights into how much energy is imparted or transformed in a dynamic system.
Gravitational Potential Energy
Gravitational potential energy refers to the energy an object has due to its position relative to the Earth or another gravitational field. It is determined by the height of the object and its mass, along with gravitational acceleration. The formula for gravitational potential energy is:
  • \( U = mgh \)
where \( m \) is the mass, \( g \) is the gravitational acceleration (approximately \( 9.81 \, \text{m/s}^2 \) on Earth), and \( h \) is the height.When the wooden block in the exercise is pushed upward due to the bullet, it gains gravitational potential energy. This energy is a result of climbing against gravity, translated from the initial kinetic energy given to it by the bullet. This principle helps us understand positional energy and its dependence on height and mass.
Conservation of Energy
The law of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In closed systems, the total energy remains constant. This principle is pivotal in solving many physics problems, including the one involving the bullet and block. When the bullet transfers energy to the block, it is transformed from kinetic to potential energy as the block rises. The conservation of energy principle tells us that the kinetic energy given to the block's mass equates to the potential energy at the block's highest point. This forms the basis of the calculations in the exercise, ensuring the total energy before and after the event remains constant, confirming our solution's correctness.
Physics Problem Solving
Solving physics problems efficiently requires a clear strategic approach, often utilizing fundamental principles. Start by analyzing the problem, identifying known variables, and noting the unknowns. Use diagrams if necessary to better understand the problem setup.
  • Apply relevant physics principles like conservation of momentum and energy.
  • Translate physical scenarios into mathematical equations.
  • Organize your solution into logical steps to trace how each part contributes to the overall solution.
In the exercise, these steps involved computing changes in potential energy and applying energy conservation principles. Then, using the conservation of momentum, we determined the bullet's final speed through mathematical equations. The careful balance of steps ensures that each principle is applied correctly, leading to accurate solutions.

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Most popular questions from this chapter

A 4.25 g bullet traveling horizontally with a velocity of magnitude \(375 \mathrm{~m} / \mathrm{s}\) is fired into a wooden block with mass \(1.12 \mathrm{~kg},\) initially at rest on a level frictionless surface. The bullet passes through the block and emerges with its speed reduced to \(122 \mathrm{~m} / \mathrm{s} .\) How fast is the block moving just after the bullet emerges from it?

A rocket is fired in deep space, where gravity is negligible. In the first second, it ejects \(1 / 160\) of its mass as exhaust gas and has an acceleration of \(15.0 \mathrm{~m} / \mathrm{s}^{2}\). What is the speed of the exhaust gas relative to the rocket?

On a frictionless air track, a \(0.150 \mathrm{~kg}\) glider moving at \(1.20 \mathrm{~m} / \mathrm{s}\) to the right collides with and sticks to a stationary \(0.250 \mathrm{~kg}\) glider. (a) What is the net momentum of this two-glider system before the collision? (b) What must be the net momentum of this system after the collision? Why? (c) Use your answers in parts (a) and (b) to find the speed of the gliders after the collision. (d) Is kinetic energy conserved during the collision?

2On a highly polished, essentially frictionless lunch counter, a \(0.500 \mathrm{~kg}\) submarine sandwich moving \(3.00 \mathrm{~m} / \mathrm{s}\) to the left collides with a \(0.250 \mathrm{~kg}\) grilled cheese sandwich moving \(1.20 \mathrm{~m} / \mathrm{s}\) to the right. (a) If the two sandwiches stick together, what is their final velocity? (b) How much mechanical energy dissipates in the collision? Where did this energy go?

A \(750 \mathrm{~kg}\) car is stalled on an icy road during a snowstorm. A \(1000 \mathrm{~kg}\) car traveling eastbound at \(10 \mathrm{~m} / \mathrm{s}\) collides with the rear of the stalled car. After being hit, the \(750 \mathrm{~kg}\) car slides on the ice at \(4 \mathrm{~m} / \mathrm{s}\) in a direction \(30^{\circ}\) north of east. (a) What are the magnitude and direction of the velocity of the \(1000 \mathrm{~kg}\) car after the collision? (b) Calculate the ratio of the kinetic energy of the two cars just after the collision to that just before the collision. (You may ignore the effects of friction during the collision.)

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