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Magazine Chapter 9: Problem 120
A body is traveling at \(2.0 \mathrm{~m} / \mathrm{s}\) along the positive direction of an \(x\) axis; no net force acts on the body. An internal explosion separates the body into two parts, each of \(4.0 \mathrm{~kg}\), and increases the total kinetic energy by \(16 \mathrm{~J}\). The forward part continues to move in the original direction of motion. What are the speeds of (a) the rear part and (b) the forward part?
Short Answer
Expert verified
The forward part moves at 4 m/s; the rear part moves at 0 m/s.
Step by step solution
01
Understand the Problem
We need to find the speeds of two parts of a body after an explosion. The original body, with a mass of 8 kg, was traveling with a speed of 2.0 m/s. The explosion separates this into two parts (each weighing 4 kg) and increases the kinetic energy by 16 J. We must find the speeds of both parts after the explosion.
02
Use Conservation of Linear Momentum
The total momentum of the system before and after the explosion must be conserved because no external force acts on it. The initial momentum is given by \(p_i = m v = 8.0 \text{ kg} \times 2.0 \text{ m/s} = 16 \text{ kg} \cdot \text{m/s}\). Let \(v_{f}\) and \(v_{r}\) be the velocities of the forward and rear parts, respectively. The momentum conservation equation becomes \(4.0 \times v_{f} + 4.0 \times v_{r} = 16\).
03
Apply Total Kinetic Energy Change
The total kinetic energy of the system increases by 16 J due to the explosion. The initial kinetic energy \(KE_i\) is \(\frac{1}{2} m v^2 = \frac{1}{2} \times 8 \text{ kg} \times (2.0 \text{ m/s})^2 = 16 \text{ J}\). Therefore, the final kinetic energy \(KE_f\) is \(16 \text{ J} + 16 \text{ J} = 32 \text{ J}\).
04
Set Up the Kinetic Energy Equation
For the two parts after explosion, the final kinetic energy equation becomes \[\frac{1}{2} \times 4 \times v_f^2 + \frac{1}{2} \times 4 \times v_r^2 = 32 \]. Simplifying, we have \[2v_f^2 + 2v_r^2 = 32 \] or \[v_f^2 + v_r^2 = 16 \].
05
Solve the System of Equations
We now have two equations: \(4v_f + 4v_r = 16\) and \(v_f^2 + v_r^2 = 16\). Solve for \(v_f\) from the first equation: \(v_f = 4 - v_r\). Substitute into the kinetic energy equation: \((4 - v_r)^2 + v_r^2 = 16\). Expanding, \(16 - 8v_r + v_r^2 + v_r^2 = 16\), simplifying gives \(2v_r^2 - 8v_r = 0\). Factor to solve: \(2v_r(v_r - 4) = 0\). Thus, \(v_r = 0\) or \(v_r = 4\).
06
Find Corresponding Velocities
From the values of \(v_r\), when \(v_r = 0\), \(v_f = 4\). When \(v_r = 4\), \(v_f = 0\). However, since the forward part continues to move in the original direction, \(v_f > 0\), the only feasible solution is \(v_f = 4\) and \(v_r = 0\).
07
Verify Solutions
Verify the solution by checking if this velocity satisfies both given conditions - momentum conservation and total kinetic energy gain. Substitute \(v_f = 4\) m/s, \(v_r = 0\) m/s into the equations, we see both conditions are satisfied.
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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 crucial concept in physics that deals with the energy an object possesses due to its motion. The formula for kinetic energy is given by \[ KE = \frac{1}{2}mv^2 \]where \(m\) represents the mass of the object, and \(v\) is its velocity. In the context of the problem, the kinetic energy initially possessed by the body was 16 J.
After the explosion, the kinetic energy increased by 16 J, making the total kinetic energy 32 J. This increase implies that an internal force (the explosion) acted to do additional work, thereby increasing the motion energy of the particles composing the system.
It's important to note that while the kinetic energy of a system can change due to internal forces like explosions, the linear momentum of the system remains conserved as long as no external force acts on it.
After the explosion, the kinetic energy increased by 16 J, making the total kinetic energy 32 J. This increase implies that an internal force (the explosion) acted to do additional work, thereby increasing the motion energy of the particles composing the system.
It's important to note that while the kinetic energy of a system can change due to internal forces like explosions, the linear momentum of the system remains conserved as long as no external force acts on it.
Internal Explosion
An internal explosion refers to a sudden release of energy within a system that causes its components to move apart. Importantly, in physics problems like this, the explosion is assumed to produce no external force on the system as a whole.
The explosion solely affects the internal motion of the system's components. Therefore, although it may alter the kinetic energy, the explosion will not change the total linear momentum of the system before and after the event.
In this problem, the internal explosion increased the kinetic energy by 16 J and separated the body into two 4 kg parts. Nevertheless, the overall momentum of the system remained constant throughout the event.
The explosion solely affects the internal motion of the system's components. Therefore, although it may alter the kinetic energy, the explosion will not change the total linear momentum of the system before and after the event.
In this problem, the internal explosion increased the kinetic energy by 16 J and separated the body into two 4 kg parts. Nevertheless, the overall momentum of the system remained constant throughout the event.
Linear Momentum
Linear momentum is defined as the product of the mass and velocity of an object:\[ p = mv \]Momentum is a vector quantity鈥攎eaning it has both magnitude and direction鈥攁nd is conserved in isolated systems. This principle of conservation of linear momentum states that when no external net force acts on a system, the total linear momentum of the system remains constant.
In this exercise, the total initial momentum of the 8 kg body moving at 2 m/s was 16 kg \( \cdot \) m/s. After the explosion, despite the kinetic energy increase, the total momentum remained the same. The system of equations used in the solution鈥擻(4v_f + 4v_r = 16\)鈥攔epresented this conservation, linking the speeds of the separated parts back to their original total momentum.
In this exercise, the total initial momentum of the 8 kg body moving at 2 m/s was 16 kg \( \cdot \) m/s. After the explosion, despite the kinetic energy increase, the total momentum remained the same. The system of equations used in the solution鈥擻(4v_f + 4v_r = 16\)鈥攔epresented this conservation, linking the speeds of the separated parts back to their original total momentum.
Physics Problem Solving
Solving physics problems, such as this one involving internal explosions, requires a methodical approach that typically involves several steps.
Firstly, understanding the problem is vital: identifying what is known and what needs to be found. Diagrams and restating the problem in simpler terms can be helpful here.
Next, applying relevant physics principles, such as conservation of momentum and kinetic energy, guides us in setting up equations. In this case, using the conservation laws led to two equations that needed solving simultaneously: one for momentum and one for kinetic energy. Solving these equations often requires algebraic manipulation or substitution to isolate desired variables. Lastly, it's important to verify and check that the obtained solutions satisfy all the specified conditions. Doing so ensures comprehension and correctness in problem solving, a skill valuable both in physics and real-world applications.
Firstly, understanding the problem is vital: identifying what is known and what needs to be found. Diagrams and restating the problem in simpler terms can be helpful here.
Next, applying relevant physics principles, such as conservation of momentum and kinetic energy, guides us in setting up equations. In this case, using the conservation laws led to two equations that needed solving simultaneously: one for momentum and one for kinetic energy. Solving these equations often requires algebraic manipulation or substitution to isolate desired variables. Lastly, it's important to verify and check that the obtained solutions satisfy all the specified conditions. Doing so ensures comprehension and correctness in problem solving, a skill valuable both in physics and real-world applications.
