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A spaceship of mass \(2.0 \times 10^{6} \mathrm{kg}\) is cruising at a speed of \(5.0 \times 10^{6} \mathrm{m} / \mathrm{s}\) when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of \(5.0 \times\) \(10^{5} \mathrm{kg},\) is blown straight backward with a speed of \(2.0 \times\) \(10^{6} \mathrm{m} / \mathrm{s} .\) A second piece, with mass \(8.0 \times 10^{5} \mathrm{kg},\) continues forward at \(1.0 \times 10^{6} \mathrm{m} / \mathrm{s} .\) What are the direction and speed of the third piece?

Short Answer

Expert verified
The speed of the third piece is \(6.0 \times 10^{6} m/s\) in the original direction of the spaceship.

Step by step solution

01

Calculate initial momentum

Calculate the initial momentum of the spaceship before the explosion using the formula \(p = mv\), where \(m\) is the mass of the spaceship and \(v\) is the velocity of the spaceship. That's \(p = (2.0 \times 10^{6} \mathrm{kg}) \times (5.0 \times 10^{6} \mathrm{m} / \mathrm{s}) = 1.0 \times 10^{13} \mathrm{kg.m/s}.\)
02

Calculate final momentum of each piece

Calculate the final momentum of the first and second pieces after the explosion: \n1. \(p_1 = (5.0 \times 10^{5} \mathrm{kg}) \times (-2.0 \times 10^{6} \mathrm{m} / \mathrm{s}) = -1.0 \times 10^{12} \mathrm{kg.m/s}\) (negative because it’s going in the opposite direction).\n2. \(p_2 = (8.0 \times 10^{5} \mathrm{kg}) \times (1.0 \times 10^{6} \mathrm{m} / \mathrm{s}) = 8.0 \times 10^{11} \mathrm{kg.m/s}\).
03

Calculate velocity of the third piece

To find the velocity of the third piece, we subtract momentum of the first and second piece from the total momentum. Let’s denote the momentum of the third piece as \(p_3\) and its mass as \(m_3\). We have:\(p_3 = p_{\text{total}} - p_1 - p_2.\)Thus, the velocity \(v_3\) of the third piece is given by:\(v_3 = p_3 / m_3 = (p_{\text{total}} - p_1 - p_2) / m_3.\)Substitute the known values into the equation:\(v_3 = (1.0 \times 10^{13} - (-1.0 \times 10^{12}) - (8.0 \times 10^{11})) / (2.0 \times 10^{6} - 5.0 \times 10^{5} - 8.0 \times 10^{5}) = 6.0 \times 10^{6} m/s.\)

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Key Concepts

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

Spaceship Motion
Spaceship motion involves understanding how an object's movement changes due to external forces. In outer space, without the resistance of air or friction, a spaceship will continue moving at a constant speed and direction unless acted upon by another force. This behavior is an example of Newton's First Law of Motion — or the principle of inertia. As seen in our exercise, the spaceship first travels seamlessly through space at a speed of \(5.0 \times 10^{6} \mathrm{m/s}\).
An essential aspect of spaceship motion is calculating momentum, which helps predict how the motion changes when influenced by forces such as explosions. In our example, the spaceship's momentum is the product of its mass \(2.0 \times 10^{6} \mathrm{kg}\) and its initial velocity \(5.0 \times 10^{6} \mathrm{m/s}\). Understanding these dynamics is crucial as it forms the basis for analyzing any changes to the motion, like when the ship experiences an explosion.
Explosion Dynamics
Explosion dynamics concentrates on the effects of a rapid expansion of energy, causing an object to break into multiple parts. In the case of our spaceship, the explosion results in the ship splitting into three pieces. This type of explosion illustrates the conservation of momentum, where the total momentum before the explosion equals the total momentum after the explosion.
When the spaceship's antimatter reactor fails, energy is released dramatically. This energy influences the speed and direction of the space pieces. It's crucial to know that despite the massive forces involved, the explosion does not create or destroy momentum; it merely redistributes it among the pieces.
  • The first piece is affected by the explosion by being pushed backward with a speed of \(2.0 \times 10^{6} \mathrm{m/s}\).
  • The second piece continues its journey forward at \(1.0 \times 10^{6} \mathrm{m/s}\).
Understanding explosion dynamics is key to calculating and predicting the behavior of each piece after the event.
Momentum Calculation
Momentum calculation is a fundamental concept for solving the problem of how the pieces of the exploded spaceship move. Momentum, a vector quantity, is calculated by multiplying an object's mass by its velocity.
In the context of the spaceship, the total momentum before the explosion (\(1.0 \times 10^{13} \mathrm{kg.m/s}\)) must equal the total momentum after. Calculating momentum for each piece involves accounting for both magnitude and direction.
  • The first piece has a negative momentum because it moves in the opposite direction \((-1.0 \times 10^{12} \mathrm{kg.m/s})\).
  • The momentum of the second piece remains positive \(8.0 \times 10^{11} \mathrm{kg.m/s}\).
Finally, finding the velocity of the third piece involves using the momentum equation, where the sum of the momentum of the first two pieces is subtracted from the total initial momentum. Incorporating these calculations ensures we respect the conservation of momentum principle, ultimately allowing us to solve for unknown velocities and directions effectively.

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Most popular questions from this chapter

A 20 g ball of clay is shot to the right (in the positive \(x\) direction) at \(12 \mathrm{m} / \mathrm{s}\) toward a \(40 \mathrm{g}\) ball of clay at rest. The two balls of clay collide and stick together. Call this reference frame \(S\). a. What is the total momentum in frame S? b. What is the velocity \(\vec{V}\) of a reference frame \(S^{\prime}\) in which the total momentum is zero? c. After the collision, what is the velocity in frame S' of the resulting \(60 \mathrm{g}\) ball of clay? Answering this question requires only thought, no calculations. d. Use your answer to part \(c\) and the Galilean transformation of velocity to find the post-collision velocity of the \(60 \mathrm{g}\) ball of clay in reference frame S.

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