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A \(20.0 \mathrm{~kg}\) body is moving through space in the positive direction of an \(x\) axis 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 part, with a mass of \(4.00 \mathrm{~kg}\), moves in the negative \(x\) direction with a speed of \(500 \mathrm{~m} / \mathrm{s}\). (a) In unit-vector notation, what is the velocity of the third part? (b) How much energy is released in the explosion? Ignore effects due to the gravitational force.

Short Answer

Expert verified
Velocity: \(1000 \hat{i} - 166.67 \hat{j} \, \text{m/s}\), Energy released: \(658,333.3 \, \text{J}\).

Step by step solution

01

Determine Initial Momentum

First, calculate the initial momentum of the system. The body is moving in the positive x-direction, so the initial momentum \( \vec{p}_i \) is given by:\[\vec{p}_i = m v_x = 20.0 \, \text{kg} \times 200 \, \text{m/s} = 4000 \, \text{kg} \cdot \text{m/s} \hat{i}.\]The initial momentum along the \(y\)-axis is zero because there is no initial motion in that direction.
02

Calculate Momentum After Explosion

Determine the momentum of two known fragments after the explosion:- For the first fragment moving in the positive \(y\)-direction: \[ \vec{p}_{1} = 10.0 \, \text{kg} \times 100 \, \text{m/s} \hat{j} = 1000 \, \text{kg} \cdot \text{m/s} \hat{j}. \]- For the second fragment moving in the negative \(x\)-direction: \[ \vec{p}_{2} = 4.0 \, \text{kg} \times (-500 \, \text{m/s}) \hat{i} = -2000 \, \text{kg} \cdot \text{m/s} \hat{i}. \]
03

Apply Conservation of Momentum

To find the third fragment's velocity, apply the conservation of momentum principle. The sum of momentum vectors of all fragments must equal the initial momentum.\[\vec{p}_i = \vec{p}_{1} + \vec{p}_{2} + \vec{p}_{3}.\]Rearranging for \(\vec{p}_{3}\):\[\vec{p}_{3} = \vec{p}_i - \vec{p}_{1} - \vec{p}_{2}.\]Substitute the known vectors:\[\vec{p}_{3} = (4000 \hat{i} - 0 \hat{j}) - (-2000 \hat{i} + 0 \hat{j}) - (0 \hat{i} + 1000 \hat{j}) = 6000 \hat{i} - 1000 \hat{j} \, \text{kg} \cdot \text{m/s}.\]
04

Solve for Third Part's Velocity

The third fragment's mass is \(6.00 \, \text{kg}\) (since the total mass must remain 20.0 kg, and the known fragments' masses are 10.0 kg + 4.0 kg).The velocity \(\vec{v}_{3}\) is the momentum divided by the mass:\[\vec{v}_{3} = \frac{\vec{p}_{3}}{m_3} = \frac{6000 \hat{i} - 1000 \hat{j}}{6.00} = 1000 \hat{i} - 166.67 \hat{j} \, \text{m/s}.\]
05

Calculate Energy Released

Evaluate the energy before and after the explosion to find the energy released. Initially, the kinetic energy \(K_i\) is:\[K_i = \frac{1}{2} m v_x^2 = \frac{1}{2} \times 20.0 \, \text{kg} \times (200 \, \text{m/s})^2 = 400,000 \, \text{J}.\]Calculate the final kinetic energy \(K_f\) of all fragments:\[K_{1f} = \frac{1}{2} \times 10.0 \, \text{kg} \times (100 \, \text{m/s})^2 = 50,000 \, \text{J},\]\[K_{2f} = \frac{1}{2} \times 4.0 \, \text{kg} \times (500 \, \text{m/s})^2 = 500,000 \, \text{J},\]\[K_{3f} = \frac{1}{2} \times 6.0 \, \text{kg} \times (\sqrt{1000^2 + 166.67^2} \, \text{m/s})^2 \, \text{J} \approx 508,333.3 \, \text{J}.\]Total final energy:\[K_f = 50,000 + 500,000 + 508,333.3 = 1,058,333.3 \, \text{J}.\]The energy released \(\Delta E\) is the difference:\[\Delta E = K_f - K_i = 1,058,333.3 - 400,000 = 658,333.3 \, \text{J}.\]
06

Conclusion

(a) The velocity of the third fragment is \(\vec{v}_3 = 1000 \hat{i} - 166.67 \hat{j} \, \text{m/s}\). (b) The amount of energy released in the explosion is approximately \(658,333.3 \, \text{J}\).

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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 a vital concept in physics that helps us understand motion and energy transformations. In simple terms, it's the energy that a body possesses due to its motion. This energy can be calculated using the expression \( K = \frac{1}{2} m v^2 \), where \( m \) represents the mass and \( v \) is the velocity of the object.
In the context of the problem, the initial kinetic energy of the whole body is substantial. It's determined by the mass of the body and its velocity before the explosion. When the explosion occurs, the kinetic energy is redistributed among the fragments that result from the explosion.
For each fragment, we calculate their own kinetic energy using the same formula but adjusting the mass and velocity according to each fragment's motion.
The calculation of kinetic energy before and after the explosion helps in determining the amount of energy released during the explosion. The energy released is simply the increase in total kinetic energy post-explosion compared to the initial state.
Explosion Physics
Explosions are fascinating phenomena in physics, characterized by a sudden and rapid release of energy. They often lead to dramatic changes in mass distribution and velocity of the involved components.
In the given problem, an internal explosion breaks a body moving through space into three parts. Each of these parts moves in different directions and velocities, showcasing classic examples of explosion physics.
The total energy released during an explosion can be calculated by observing the change in kinetic energy of the system. In this context, we first calculate the initial kinetic energy of the body before breaking up. After the explosion, the kinetic energies of all newly formed particles are computed. The energy released is the difference between the sum of these energies and the body's initial kinetic energy.
This release of energy can be attributed to potential energy stored within the system being transformed into kinetic energy as the body fragments, often resulting in elevated energies compared to the initial state.
Vector Analysis
Vector analysis is a powerful tool used in physics to solve problems involving direction and magnitude, especially when dealing with forces and motion.
In this problem, understanding how to break down and handle vector quantities is crucial. The initial momentum of the body moving along the x-axis must be distributed among the fragments resulting from the explosion. Each fragment has its own momentum vector, characterized by both a magnitude and a direction.
Applying the conservation of momentum, which states that momentum is conserved in a closed system unless acted upon by external forces, we break this problem into components. We analyze the x and y direction separately to ensure accuracy.
By using these principles of vector analysis, one can find unknown variables like the velocity of one of the fragments. This approach emphasizes why understanding and correctly handling vectors can simplify complex physical scenarios effectively.

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