Problem 110 - University Physics with Modern Physics

Question

A \(12.0-\mathrm{kg}\) shell is launched at an angle of \(55.0^{\circ}\) above the horizontal with an initial speed of 150 \(\mathrm{m} / \mathrm{s} .\) When it is at its highest point, the shell explodes into two fragments, one three times heavier than the other. The two fragments reach the ground at the same time. Assume that air resistance can be ignored. If the heavier fragment lands back at the same point from which the shell was launched, where will the lighter fragment land, and how much energy was released in the explosion?

Step-by-Step Solution

Answer

The lighter fragment lands 8565.6 m away from the launch point; explosion releases energy of 221.8 kJ.

Step 1: Analyze the Initial Situation

We are given a shell with mass \( m = 12.0 \) kg launched at an angle \( \theta = 55.0^{\circ} \) and speed \( v_i = 150.0 \text{ m/s} \). The shell reaches its highest point, which can be calculated, then explodes into two fragments.

Step 2: Break Down the Problem

The shell reaches its highest point, where only horizontal velocity \( v_{ix} \) remains, given by \( v_{ix} = v_i \cdot \cos(\theta) \). At this point, the vertical velocity component \( v_{iy} = v_i \cdot \sin(\theta) \) is zero.

Step 3: Conservation of Momentum Analysis

At the highest point, the total horizontal momentum before the explosion equals the total horizontal momentum immediately after. Let the masses of the heavier and lighter fragments be \( 3m \) and \( m \), respectively. If the heavier fragment lands back at the launch point, its horizontal velocity after explosion is zero. Thus, the lighter fragment moves at velocity \( v_l = 4 v_{ix} \) to maintain momentum conservation.

Step 4: Solve for the Lighter Fragment's Landing Point

The lighter fragment, with velocity \( v_l = 4v_{ix} \), travels horizontally while falling under gravity. Calculate \( v_{ix} = 150 \cdot \cos(55^{\circ}) \approx 86.0 \text{ m/s} \). The lighter fragment's horizontal velocity is \( v_l = 4 \times 86.0 \approx 344.0 \text{ m/s} \). Since the time of flight \( t \) remains unchanged by the explosion, \( t = \frac{2v_{iy}}{g} = \frac{2 \times 150 \cdot \sin(55^{\circ})}{9.81} \approx 24.9 \text{ s} \). The distance is \( d = v_l \cdot t \approx 344.0 \times 24.9 \approx 8565.6 \text{ m} \).

Step 5: Calculate Energy Released in the Explosion

The total mechanical energy at the highest point before the explosion is purely kinetic (only horizontal), \( K_i = \frac{1}{2} \, 12.0 \, v_{ix}^2 \). After the explosion, the energy consists of systems \( K_h = 0 \) and \( K_l = \frac{1}{2} \, 4 \, m \, v_{ix}^2 \). Energy released is the increase in kinetic energy, \( E_{released} = K_l - K_i \). Calculate \( K_i = \frac{1}{2} \, 12.0 \, (86.0)^2 \) and \( K_l = \frac{1}{2} \, 4 \, 86.0^2 \), giving the energy difference as released energy.

Key Concepts

Conservation of Momentum

In physics, conservation of momentum is a fundamental principle stating that the total momentum of a closed system remains constant if no external forces act on it. When analyzing projectile motion and explosions, like with our shell example, momentum conservation plays a crucial role.

Before the explosion of the shell, all momentum is found in its horizontal motion, given by the product of its mass and horizontal velocity component. After the explosion, even though the shell splits into two fragments, the total horizontal momentum must remain the same to satisfy conservation laws.

For the shell exploding into two fragments—one heavier and one lighter—momentum calculations show that the heavier fragment's post-explosion velocity is zero since it returns to the starting point. This dictates that the lighter fragment must move with increased velocity to keep the overall momentum balanced. By solving the equations, you'd find the lighter fragment must travel at four times the original horizontal velocity of the shell to maintain this balance.

Energy Conservation

Energy conservation is another foundational concept, particularly when dealing with explosions in physics. Before any explosion happens, the shell possesses kinetic energy solely due to its horizontal velocity at its peak height. When we refer to conservation of energy, initially we have only kinetic energy because all potential energies in projectile motion are handled separately.

As the shell explodes, some of the chemical potential energy stored within it is transformed, primarily into kinetic energy of the fragments moving in different directions. It is at this stage that examining their individual energies provides insight.

The amount of energy released during this explosion represents the increase in total kinetic energy of the fragments compared to the initial kinetic energy of the shell. By comparing both before and after scenarios, and solving the respective energy calculations, one can determine precisely how much energy was released in that explosion.

Explosions in Physics

Explosions offer fascinating scenarios for observing physical principles like momentum and energy conservation in action. In a controlled environment like our shell problem, an explosion essentially redistributes energy and momentum among newly created parts while conserving the total amount present in the system.

Explosions happen due to the rapid expansion of gases or release of stored energy. In our example, the explosion happens at the shell’s highest point, significantly affecting its trajectory and behavior.

After the explosion, the fragments might exhibit vastly different behaviors: one fragment remains nearby, demonstrating reduced horizontal velocity, while the other could travel a great distance. Understanding the dynamics and physics of explosions enables us to predict these behaviors and resolve real-world engineering problems.