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A fireworks rocket is launched vertically into the night sky with an initial speed of \(44.2 \mathrm{m} / \mathrm{s}\). The rocket coasts after being launched, then explodes and breaks into two pieces of equal mass 2.50 s later. (a) If each piece follows a trajectory that is initially at \(45.0^{\circ}\) to the vertical, what was their speed immediately after the explosion? (b) What is the velocity of the rocket's center of mass before and after the explosion? (c) What is the acceleration of the rocket's center of mass before and after the explosion?

Short Answer

Expert verified
(a) 27.9 m/s; (b) 19.7 m/s (unchanged); (c) -9.8 m/s^2 (unchanged).

Step by step solution

01

Understand the Problem

We need to determine (a) the speed of each piece immediately after the explosion, (b) the center of mass velocity before and after the explosion, and (c) the center of mass acceleration before and after explosion.
02

Analyze the Rocket's Initial Movement

The rocket is launched with an initial speed of \(44.2 \mathrm{m/s}\), moving vertically. Using the motion equation \(v = u + at\), and considering only gravitational force \(-9.8 \mathrm{m/s^2}\), find the velocity just before the explosion after 2.5 seconds.
03

Compute Velocity Before Explosion

Calculate: \[v = 44.2 \mathrm{m/s} - (9.8 \mathrm{m/s^2})(2.5\, \mathrm{s}) = 44.2 \mathrm{m/s} - 24.5 \mathrm{m/s} = 19.7 \mathrm{m/s}.\]This is the velocity of the rocket immediately before the explosion.
04

Determine Speed after Explosion

Since the explosion sends each piece out at a \(45^\circ\) angle to the vertical, consider the vertical and horizontal components are equal. Hence, each piece's speed \(v_f\) is \(v/ \cos(\theta)\) where \(\theta=45^\circ\).\[v_f = \frac{19.7 \mathrm{m/s}}{\cos(45^\circ)} = 19.7 \mathrm{m/s} \times \sqrt{2} = 27.9 \mathrm{m/s}.\]
05

Velocity of the Center of Mass

Before the explosion, the center of mass moves with the entire rocket's velocity, \(19.7 \mathrm{m/s}\). After the explosion, the center of mass continues at the same velocity because of conservation of momentum in vertical direction.
06

Acceleration of the Center of Mass

Before the explosion, the acceleration of the rocket's center of mass is due to gravity: \(-9.8 \mathrm{m/s^2}\). After the explosion, no horizontal forces have been introduced, so the vertical acceleration remains \(-9.8 \mathrm{m/s^2}\).

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Key Concepts

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

Center of Mass
The concept of the center of mass is crucial to understanding motion in physics, especially for objects like rockets that move and explode.
In the scenario of a rocket, the center of mass refers to the point where the entire mass of the rocket can be considered to concentrate for analysis.
When the rocket is in motion, this center moves along a predictable path, influenced only by external forces.
- Before the explosion, the rocket's center of mass moves uniformly because the rocket is a single object with a defined velocity. - After the explosion, even though the rocket splits into parts that move in different directions, the center of mass continues in the path that was determined by the initial momentum. This consistency is possible due to the law of conservation of momentum, which states that if no external force is acting, the center of mass's motion remains unchanged.
This explains why, even after the explosive split, the center of mass travels at the same velocity as it did right before the explosion.
Understanding this helps explain why no significant force disrupts the center of mass's path and acceleration post-explosion.
Explosion Physics
Explosion physics is fascinating when observing how a single object, like a fireworks rocket, fragments into multiple pieces.
An explosion is a rapid increase in volume and energy, moving objects violently in various directions.
In our case, once the rocket explodes, it divides into two equal mass fragments.
- Initially, the rocket moves upward at a consistent velocity. - During the explosion, a sudden force redistributes this velocity across the fragments. The pieces move outward at equal angles of 45 degrees to the vertical, illustrating a symmetry often found in explosive events.
This angle indicates that kinetic energy from the explosion launches both fragments equally in all directions, maintaining the rocket's previous momentum.
To calculate each piece's speed after the explosion, one needs to consider the conservation of momentum and energy.
This determines how the speed of pieces (\(v_f = \frac{19.7}{\cos(45^\circ)} = 27.9 \, \mathrm{m/s}\)) links directly to the forces exerted during the explosion.
Projectile Motion
Projectile motion describes the path of objects propelled through the air, like the rocket fragments after an explosion.
Understanding projectile motion involves analyzing both vertical and horizontal components for any moving body's trajectory.
- Initially, the rocket moves upward until the explosion. - Once divided, each piece follows a parabolic path, reflective of typical projectile motion. Since the pieces shoot out at a 45-degree angle, their initial velocities are equal in vertical and horizontal directions due to the symmetry in the angle.
This results in predictable paths informed by projectile motion principles: gravity only affects vertical motion while horizontal motion persists until halted by another force, like friction.
Overall, each fragment forms a path determined by initial velocity and gravitational pull, a key feature of projectile motion.
By understanding these basic principles, one can predict and understand the movement of any object launched through the air.

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