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A 20.0 -kg projectile is fired at an angle of \(60.0^{\circ}\) above the horizontal with a speed of 80.0 \(\mathrm{m} / \mathrm{s}\) . At the highest point of its trajectory, the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. You can ignore air resistance. (a) How far from the point of firing does the other fragment strike if the terrain is level? (b) How much energy is released during the explosion?

Short Answer

Expert verified
(a) The other fragment lands 565.6 m from the firing point. (b) 8000 J of energy is released during the explosion.

Step by step solution

01

Calculate the Initial Velocity Components

First, let's find the initial horizontal (\(v_{0x}\)) and vertical (\(v_{0y}\)) components of the velocity. Given the projectile is fired at an angle of \(60^\circ\) with a speed of \(80.0 \text{ m/s}\):\[v_{0x} = 80.0 \cdot \cos(60^\circ) = 40.0 \text{ m/s}\\]\[v_{0y} = 80.0 \cdot \sin(60^\circ) = 69.28 \text{ m/s}\\]
02

Determine the Maximum Height and Time to Reach Maximum Height

Use the vertical component of velocity to determine the maximum height using \(v_{0y}\) and the time to reach this height. At the maximum height, the vertical component of the velocity is 0.First, find the time to reach the maximum height (\(t_\text{max}\)): \[t_\text{max}=\frac{v_{0y}}{g}=\frac{69.28}{9.8}\approx7.07 \text{ s}\\]Now calculate the maximum vertical height (\(h\)):\[h= v_{0y} \cdot t_\text{max} - \frac{1}{2}gt_\text{max}^2 = 69.28 \times 7.07 - 0.5 \times 9.8 \times (7.07)^2\approx244.16 \text{ m}\\]
03

Determine Total Time of Flight and Horizontal Distance Before Explosion

The total time of flight (\(T\)) is double the time to the maximum height, as the motion is symmetric. So: \[T = 2 \times t_\text{max} = 14.14 \text{ s}\\] Calculate the horizontal distance at the highest point:\[x_\text{max} = v_{0x} \cdot t_\text{max} = 40.0 \cdot 7.07 \approx 282.8 \text{ m}\\]
04

Analyze Post-Explosion Trajectory of Second Fragment

After the explosion, one fragment has zero initial velocity and falls vertically, while the other continues horizontally. Since the fragments have equal mass (10 kg each), the second fragment's horizontal velocity remains \(40.0 \text{ m/s}\).Therefore, the horizontal distance traveled by the fragment after the explosion is equal to the horizontal velocity \(v_{0x} = 40.0 \text{ m/s}\) multiplied by the remaining flight time \((t_\text{max} = 7.07 \text{ s})\):\[x_\text{frag} = 40.0 \times 7.07 \approx 282.8 \text{ m}\\]
05

Calculate Total Horizontal Distance from Firing Point to Impact

The total horizontal distance from the firing point where the second fragment lands (\(x_\text{total}\)) is the sum of the horizontal distance to the point of explosion (\(x_\text{max}\)) and the distance traveled by the fragment after explosion (\(x_\text{frag}\)):\[x_\text{total} = 282.8 + 282.8 \approx 565.6 \text{ m}\\]
06

Calculate Energy Released during Explosion

Use energy principles to calculate the energy released during the explosion:Initially, the energy in the two-mass system is just due to kinetic energy of horizontal movement since the other vertical velocity is 0 post-explosion. Energy of the system just before explosion:\[E_{\text{initial}} = \frac{1}{2} m \cdot v_{0x}^2 = \frac{1}{2} \times 20 \times 40^2 = 16000 \text{ J}\]Kinetic energy of the fragments after the explosion:1. Fragment falling vertically: \(0\) J.2. Horizontal fragment:\[E_{\text{final}} = \frac{1}{2} \times 10 \times 40^2 = 8000 \text{ J}\]Energy released:\[E_{\text{release}} = E_{\text{initial}} - E_{\text{final}} = 16000 - 8000 = 8000 \text{ J}\]

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Key Concepts

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

Energy Conservation
Energy conservation is an essential principle in physics, especially when analyzing projectile motion and explosions.
When a projectile is fired, it possesses kinetic energy due to its velocity and potential energy when at a height above the ground.
This total mechanical energy remains constant if we ignore air resistance and external forces.
  • Before the explosion, the projectile's total energy consists of both potential and kinetic energy.
  • At the highest point of the trajectory, the potential energy is maximized, while the vertical component of kinetic energy becomes zero.
During the explosion, energy is released, and it must be accounted for.
The law of energy conservation helps to calculate how much energy is transformed into different forms, such as the kinetic energy of the exploding fragments, post-explosion.
Understanding these energy transformations allows us to predict the projectile's behavior after the explosion.
Explosion Dynamics
Explosion dynamics explain how projectiles behave when an explosion occurs at certain points in their trajectory.
An explosion in a projectile can be seen as a sudden release of energy causing a disturbance.
  • In our exercise, the projectile breaks into two fragments of equal mass.
  • One fragment falls with zero initial velocity, while the other maintains horizontal motion.
These dynamics follow the principle of conservation of momentum, where the momentum before and after the explosion remains constant.
This forms the basis for calculating the post-explosion velocities and trajectories of the projectile's fragments.
The explosion's dynamics determine how the fragments travel after detachment and help us solve key questions, like where the second fragment lands.
Kinematics
Kinematics describes the motion of objects without considering the forces causing that motion.
In projectile motion, analyzing the object's path involves both horizontal and vertical motion components.
  • The horizontal motion has a constant velocity if air resistance is ignored.
  • The vertical motion is influenced by gravitational acceleration.
Essential kinematics concepts used in solving projectile problems involve calculating initial velocity components, time to reach maximum height, and total time of flight.
In our problem, understanding kinematics allows you to ascertain points such as the highest altitude reached and total distance covered by the projectile.
This knowledge goes hand-in-hand with energy conservation principles to analyze more complex situations like explosions during flight.
Velocity Components
Velocity components are crucial in understanding projectile motion, especially when an object is launched at an angle.
Velocity can be split into two orthogonal components: horizontal and vertical, which are easier to manage mathematically.
  • The horizontal component (\(v_{0x}\)) remains constant throughout the flight in absence of air resistance.
  • The vertical component (\(v_{0y}\)) changes over time due to gravity.
Calculating these components helps in determining essential attributes of the projectile's path, like its range and maximum height.
For the given problem, the horizontal velocity after the explosion helps determine how far the second fragment travels.
A solid grasp of velocity components is essential in resolving projectile problems, offering clear insights into the effect of initial conditions and external changes like explosions.

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