Problem 109 - University Physics with Modern Physics

Question

A fireworks rocket is fired vertically upward. At its maximum height of \(80.0 \mathrm{m},\) it explodes and breaks into two pieces: one with mass 1.40 \(\mathrm{kg}\) and the other with mass 0.28 \(\mathrm{kg} .\) In the explosion, 860 \(\mathrm{J}\) of chemical energy is converted to kinetic energy of the two fragments. (a) What is the speed of each fragment just after the explosion? (b) It is observed that the two fragments hit the ground at the same time. What is the distance between the points on the ground where they land? Assume that the ground is level and air resistance can be ignored.

Step-by-Step Solution

Answer

(a) Speeds are 30 m/s and 150 m/s; (b) Distance is 484.8 m.

Step 1: Understanding the Problem

We have an explosion at the maximum height of a firework rocket. The total energy from the explosion converts into kinetic energy of the two fragments. We need to calculate their speeds and then determine the distance between their landing positions.

Step 2: Conservation of Momentum

Before the explosion, the system's momentum is zero (the rocket was momentarily at a standstill at maximum height). Consequently, the momentum of the two fragments must be equal and opposite after the explosion. Let the speed of the 1.40 kg fragment be \( v_1 \), and the speed of the 0.28 kg fragment be \( v_2 \). By conservation of momentum: \[ 1.40 \cdot v_1 = 0.28 \cdot v_2 \].

Step 3: Kinetic Energy Distribution

The kinetic energy from the explosion is 860 J, which is shared between the two fragments: \[ \frac{1}{2} \times 1.40 \times v_1^2 + \frac{1}{2} \times 0.28 \times v_2^2 = 860. \] We have two equations and two unknowns from Steps 2 and 3.

Step 4: Solving the System of Equations

Substitute \( v_2 = \frac{1.40}{0.28}v_1 \) from Step 2 into Step 3's energy equation and solve for \( v_1 \). After solving the quadratic equation, we find \( v_1 = 30 \text{ m/s} \). Substitute back to find \( v_2 \): \( v_2 = 150 \text{ m/s} \).

Step 5: Determine the Horizontal Distance Traveled

Since the fragments hit the ground simultaneously, consider the projectile motion of both fragments under gravity for the vertical descent of 80.0 m. Time to fall can be computed using: \[ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 80}{9.81}}. \] Calculate it to find \( t \approx 4.04 \text{ s} \).

Step 6: Calculate Landing Positions

Assume horizontal components of velocities are equal to \( v_1 \) and \( v_2 \) calculated. Distances are \( d_1 = v_1 \cdot t \) and \( d_2 = v_2 \cdot t \). Compute these: \( d_1 = 30 \times 4.04 = 121.2 \text{ m} \) and \( d_2 = 150 \times 4.04 = 606.0 \text{ m} \).

Step 7: Finalize the Landing Distance

The difference in landing positions \( |d_2 - d_1| \) gives the distance between the fragments' landing points. This evaluates to \( 606.0 - 121.2 = 484.8 \text{ m} \).

Key Concepts

Kinetic Energy

Kinetic energy is the energy an object has because of its motion. It's one of the key concepts in understanding how energy from an explosion is transformed and utilized. When the fireworks rocket explodes, a portion of the chemical energy stored within is converted into kinetic energy.

Kinetic energy (\( KE \)) is expressed mathematically as: \[ KE = \frac{1}{2}mv^2 \]where:

• \( m \) is the mass of the object.
• \( v \) is the velocity of the object.

In the given problem, the total kinetic energy after the explosion is 860 J. This energy is distributed between the two fragments, giving them different speeds due to their differing masses. Understanding this distribution allows us to calculate each fragment's speed using the law of conservation of momentum and the kinetic energy equation. Conserving energy in this system means that energy does not disappear; it simply changes forms from chemical to kinetic.

Projectile Motion

Projectile motion describes the motion of an object thrown or projected into the air. It is subject to only the acceleration of gravity. In the exercise, the exploded fragments of the rocket behave like projectiles when they fall back to the ground.

Projectile motion can be analyzed in horizontal and vertical components:

• Vertically, the objects are acted upon by gravity, causing them to accelerate towards the earth. The time to fall can be calculated using the formula:\[ t = \sqrt{\frac{2h}{g}} \]where:
  • \( h \) is the height from which the object descends (80 m in this case).
  • \( g \) is the acceleration due to gravity (approximately 9.81 m/s²).
  This gives the time it takes for both fragments to fall to the ground.
• Horizontally, the velocity at the moment of the explosion determines how far the fragments travel, given by:\[ d = vt \]where:
  • \( v \) is the horizontal component of the velocity.
  • \( t \) is the time of flight calculated from the vertical motion.

By treating the motion of the fragments as independent in the vertical and horizontal directions, we can accurately predict where they will land on the ground.

Explosion Fragments

When the rocket explodes, it breaks into two pieces or fragments. Each fragment is subjected to the conservation of momentum, as the total momentum before and after the explosion must remain the same. This implies that the momentum gained by one fragment is equal and opposite to the other.

The law of conservation of momentum can be expressed as:\[ m_1v_1 = -m_2v_2 \]where:

• \( m_1 \) and \( v_1 \) are the mass and speed of the first fragment.
• \( m_2 \) and \( v_2 \) are the mass and speed of the second fragment.

This relationship helps us determine the speeds of the fragments post-explosion. This principle ensures that even though the rocket bursts apart, the system's total momentum is conserved.

Understanding the behavior of fragments also involves calculating how far each fragment travels once they hit the ground. This distance, as found from projectile motion analysis, gives us the separation between their landing points. Thus, fragmentation is key to exploring different aspects of motion and energy in this physics problem.