91Ó°ÊÓ

A \(20.0 \mathrm{~kg}\) body is moving in the positive \(x\) direction with a speed of \(200 \mathrm{~m} / \mathrm{s}\) when, due to an internal explosion, it breaks into three parts. One part, with a mass of \(10.0 \mathrm{~kg}\), moves away from the point of explosion with a speed of \(100 \mathrm{~m} / \mathrm{s}\) in the positive \(y\) direction. A second fragment, with a mass of \(4.00 \mathrm{~kg}\), moves in the negative \(x\) direction with a speed of \(500 \mathrm{~m} / \mathrm{s}\). What is the velocity of the third \((6.00 \mathrm{~kg})\) fragment? 53\. Vessel at Rest Explodes A vessel at rest explodes, breaking into three pieces. Two pieces, having equal mass, fly off perpendicular to one another with the same speed of \(30 \mathrm{~m} / \mathrm{s}\). The third piece has three times the mass of each other piece. What are the magnitude and direction of its velocity immediately after the explosion?

Step by step solution

01

- Understand the problem

A 20.0 kg body moving with 200 m/s in the positive x direction explodes into three parts. We need to find the velocity of the third part.

02

- Identify given values

Mass of the original body, \(m = 20.0 \, \mathrm{kg}\). Initial velocity, \(v = 200 \, \mathrm{m/s}\). The masses and velocities of the fragments are: \[m_1 = 10.0 \, \mathrm{kg}, \; v_1 = 100 \, \mathrm{m/s}\, (in \, +y \, direction)\]\[m_2 = 4.00 \, \mathrm{kg}, \; v_2 = 500 \, \mathrm{m/s}\, (in \, -x \, direction)\]\[m_3 = 6.00 \, \mathrm{kg}, \; v_3 = ?\]

03

- Apply the principle of conservation of momentum

The total momentum before the explosion should equal the total momentum after the explosion. Before the explosion: Momentum, \(p_\mathrm{initial} = m \, v = 20.0 \, \mathrm{kg} \times 200 \, \mathrm{m/s} = 4000 \, \mathrm{kg \, m/s} \) in the x-direction.After the explosion: \[p_\mathrm{final,x} = m_1 v_{1,x} + m_2 v_{2,x} + m_3 v_{3,x} \]\[p_\mathrm{final,y} = m_1 v_{1,y} + m_2 v_{2,y} + m_3 v_{3,y} \]

04

- Break down momentum components

For the x-direction: \[m_1 v_{1,x} = 0 \, (100 \, \mathrm{m/s} \text{ is in the y-direction})\]\[m_2 v_{2,x} = 4.00 \, kg \times (-500 \, \mathrm{m/s}) = -2000 \, \mathrm{kg \, m/s}\]\[m_3 v_{3,x} = 6.00 \, kg \times v_{3,x} \]For the y-direction: \[m_1 v_{1,y} = 10.0 \, kg \times 100 \, \mathrm{m/s} = 1000 \, \mathrm{kg \, m/s}\]\[m_2 v_{2,y} = 0 \]\[m_3 v_{3,y} = 6.00 \, kg \times v_{3,y} \]

05

- Set up equations for conservation of momentum

In the x-direction: \[4000 = -2000 + 6.00 v_{3,x} \]\[6000 = 6.00 v_{3,x} \]\[v_{3,x} = 1000 / 3 \, \mathrm{m/s} \]Similarly, in the y-direction: \[0 = 1000 + 6.00 v_{3,y} \]\[6.00 v_{3_y} = -1000 \]\[v_{3,y} = -1000 / 6 = -166.67 \mathrm{m/s} \]

06

- Calculate the resultant velocity

Combine the x and y components to find the magnitude: \[v_3 = \sqrt{(v_{3,x})^2 + (v_{3,y})^2} \]\[v_3 = \sqrt{(1000/3)^2 + (-166.67)^2} \approx 344.65 \mathrm{m/s} \]Direction: \[\theta = \arctan\left(\frac{v_{3,y}}{v_{3,x}}\right) = \arctan\left(\frac{-166.67}{333.33}\right) = \arctan(-0.5) \approx -26.57^{\circ} \]

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Key Concepts

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

Collision Analysis

Collision analysis is a core aspect of understanding the exercise. In this problem, we are analyzing the explosion of a moving object that breaks into fragments. One way to approach collision analysis is by employing the principle of conservation of momentum, which states that the total momentum of a system remains constant if no external forces act on it. This principle helps us set up equations before and after the collision (or explosion). Specifically, we consider both the components of momentum in different directions (x and y). The momentum before the explosion must equal the combined momentum of all fragments after the explosion. By breaking down the momentum into both x and y components and ensuring they add up correctly, we systematically solve the problem.

Two-Dimensional Motion

Two-dimensional motion involves analyzing objects moving in a plane, rather than just along a straight line. In this exercise, after the explosion, we have fragments moving in different directions, requiring a breakdown of the motion into two dimensions: the x-direction and the y-direction. This adds complexity to the problem, as each fragment's velocity must be considered in both dimensions separately. By decomposing the movement into vector components, we can individually calculate the x and y components of velocity. Later, we combine these components to find the resultant speed and direction of the third fragment. Understanding how to break up vector quantities and analyze them in two dimensions is crucial for solving such problems efficiently.

Vector Components

Vector components play a significant role in understanding and solving the provided exercise. A vector has both magnitude and direction, and any vector can be broken down into perpendicular components (usually the x and y components in two-dimensional analysis). In the exercise, you start by determining the initial momentum and then the components of the velocities of the fragments post-explosion. These components help you set up the appropriate equations to solve for unknowns. For instance, the velocity of the third fragment is determined by solving for its x and y components separately. Once those components are found, the magnitude of the resultant velocity is calculated using the Pythagorean theorem, and the direction is found using trigonometric functions like arctangent. This step-by-step breakdown ensures a thorough understanding and accurate results.