/*! 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 60 An unstable nucleus of mass \(1.... [FREE SOLUTION] | 91Ó°ÊÓ

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An unstable nucleus of mass \(1.7 \times 10^{-26} \mathrm{~kg}\), initially at rest at the origin of a coordinate system, disintegrates into three particles. One particle, having a mass of \(m_{1}=5.0 \times 10^{-27} \mathrm{~kg}\), moves in the positive \(y\) -direction with speed \(v_{1}=6.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Another particle, of mass \(m_{2}=8.4 \times 10^{-27} \mathrm{~kg}\), moves in the positive \(x\) -direction with speed \(v_{2}=4.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Find the magnitude and direction of the velocity of the third particle.

Short Answer

Expert verified
The magnitude of the velocity of the third particle is given by the ratio of the magnitude of the total momentum of the first two particles to the mass of the third particle, and its direction is opposite to the direction of the total momentum of the first two particles

Step by step solution

01

Compute the momentum of the first and second particles

This is done simply by multiplying their respective masses and velocities. Let's denote these momenta as \(p_1 = m_1 * v_1\) and \(p_2 = m_2 * v_2\)
02

Compute the total momentum of the first and second particles

The total momentum of the first and second particles can be represented as a 2D vector \(P_{total} = (-p_2, p_1)\) where 'negative' in front of \(p_2\) denotes that \(p_2\) is in the negative x-direction
03

Compute the magnitude and direction of the total momentum

The magnitude of the total momentum \(|P_{total}|\) can be computed using the Pythagorean theorem as \(|P_{total}| = \sqrt{{(-p_2)}^2 + p_1^2}\). The direction can be computed using the inverse tangent function, let's denote it with \(\theta\), \(\theta = atan2(p_1, -p_2)\)
04

Compute the mass and momentum of the third particle

The mass \(m_3\) of the third particle can be computed by subtracting the masses of the first and second particles from the mass of the original nucleus as \(m_3 = M - m_1 - m_2\). The momentum of the third particle \(p_3\) is equal to the magnitude of the total momentum of the first two particles, but is directed in the opposite direction, so \(p_3 = -|P_{total}|\)
05

Compute the velocity and direction of the third particle

The magnitude of the velocity of the third particle \(v_3\) can be computed by dividing its momentum by its mass as \(v_3 = |p_3| / m_3\). The direction of the velocity of the third particle will be the opposite of \(\theta\), namely \(180 - \theta\) degrees

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Key Concepts

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

Nuclear Decay: Understanding the Process
Nuclear decay is a fascinating process that involves the transformation of an unstable atomic nucleus into one or more different nuclei. When an unstable nucleus decays, it can emit various particles such as alpha particles, beta particles, or gamma rays. In our exercise, the nucleus decays into three smaller particles. This type of decay often happens because the original nucleus has too much energy or is imbalanced in terms of its constituent forces.
  • The goal of the decay is to reach a more stable state, a lower energy configuration.
  • The process also releases energy through the movement of particles.
This change in energy and state is a key point in understanding nuclear reactions and their impacts. In practical scenarios, such as in nuclear reactors or during certain types of medical treatments, controlling nuclear decay is crucial.
Particle Velocity: Calculating Speed and Direction
When studying nuclear decay, one interesting aspect is the velocities of the emitted particles. Velocity is a vector quantity, meaning it has both magnitude and direction.
  • The velocity of a particle dictates how fast it is moving and the path it follows.
  • In the given exercise, the velocity of each particle was calculated to understand the details of the decay event.
To calculate velocity:- Use the formula: \[ v = \frac{p}{m} \]where \( p \) is the momentum and \( m \) is the mass of the particle.
Understanding the velocity not only helps in knowing the speed of the particles but also provides insights into the distribution of energy during the decay. This is vital for applications like radiation shielding and dosimetry, where knowing the direction and speed of particles ensures safety and effectiveness.
Coordinate System: Mapping Particle Movement
In physics, a coordinate system helps us map the movement of particles. It provides a standardized way to describe the position and direction of particles post-decay. A coordinate system consists typically of multiple axes:
  • - The x-axis (horizontal) and y-axis (vertical) are the standard reference axes.
  • In more complex cases, a z-axis may be used for three-dimensional movement.
For the exercise, the particles' movements are mapped on a 2D plane. The importance of a coordinate system increases as it allows for clear and precise description of events and calculations: - Each particle’s direction is easily tracked by labeling its movement along these axes. - The origin (0,0) helps to pinpoint the starting position of a particle, like the original nucleus before decay.
Using a coordinate system enhances understanding and communication in studies involving motion and forces, making it an essential tool for physicists and engineers alike.

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Most popular questions from this chapter

A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to \(5.20 \mathrm{~m} / \mathrm{s}\) in \(0.832 \mathrm{~s}\). What are the magnitudes of the linear impulse and the average total force experienced by a \(70.0\) -kg passenger in the car during the time the car accelerates?

A cue ball traveling at \(4.00 \mathrm{~m} / \mathrm{s}\) makes a glancing, elastic collision with a target ball of equal mass that is initially at rest. The cue ball is deflected so that it makes an angle of \(30.0^{\circ}\) with its original direction of travel. Find (a) the angle between the velocity vectors of the two balls after the collision and (b) the speed of each ball after the collision.

ecp Two blocks collide on a frictionless surface. After the collision, the blocks stick together. Block A has a mass \(M\) and is initially moving to the right at speed \(v\). Block B has a mass \(2 M\) and is initially at rest. System \(\mathrm{C}\) is composed of both blocks. (a) Draw a free-body diagram for each block at an instant during the collision. (b) Rank the magnitudes of the horizontal forces in your diagram. Explain your reasoning. (c) Calculate the change in momentum of block A, block B, and system C. (d) Is kinetic energy conserved in this collision? Explain your answer. (This problem is courtesy of Edward \(F\). Redish. For more such problems, visit http://www.physics.umd.edu/perg.)

Two automobiles of equal mass approach an intersection. One vehicle is traveling with velocity \(13.0 \mathrm{~m} / \mathrm{s}\) toward the east, and the other is traveling north with speed \(v_{2 i}\). Neither driver sees the other. The vehicles collide in the intersection and stick together, leaving parallel skid marks at an angle of \(55.0^{\circ}\) north of east. The speed limit for both roads is \(35 \mathrm{mi} / \mathrm{h}\), and the driver of the northward-moving vehicle claims he was within the limit when the collision occurred. Is he telling the truth?

Calculate the magnitude of the linear momentum for the following cases: (a) a proton with mass \(1.67 \times 10^{-27} \mathrm{~kg}\), moving with a speed of \(5.00 \times 10^{6} \mathrm{~m} / \mathrm{s} ;\) (b) a \(15.0-\mathrm{g}\) bullet moving with a speed of \(300 \mathrm{~m} / \mathrm{s} ;\) (c) a \(75.0-\mathrm{kg}\) sprinter (d) the Earth (mass running with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\); \(=5.98 \times 10^{24} \mathrm{~kg}\) ) moving with an orbital speed equal to \(2.98 \times 10^{4} \mathrm{~m} / \mathrm{s}\)

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