Projectile question involving Law of Conservation of Energy

In summary, the Law of Conservation of Energy states that energy cannot be created or destroyed, but can only be transferred from one form to another. In projectile motion, the initial kinetic energy of the object is converted into potential energy as it gains height, and then back into kinetic energy as it falls back down. The amount of kinetic and potential energy in a projectile is affected by its mass, initial velocity, and launch height. This law cannot be violated in any scenario, including projectile motion, and can be used to solve for final velocity, height, or range of a projectile by equating initial kinetic energy to final potential energy, or vice versa.
  • #1
leighzer
10
0
A particle of mass 0.500 kg is shot from point P, 60 m above the ground. The particle has an initial velocity Vi, with a horizontal component of 30.0 m/s. The particle rises to a maximum height of 20.0 m above P. Using the Law of Conservation of Energy, determine the vertical component of Vi.
 
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  • #2
use lose in PE=gain in KE to start
 
  • #3


According to the Law of Conservation of Energy, the total energy of a system remains constant. In this case, the system consists of the particle and the Earth. The initial energy of the particle at point P is purely kinetic, given by 1/2 * m * (Vi)^2, where m is the mass of the particle and Vi is the initial velocity. As the particle rises to a maximum height of 20.0 m, it gains potential energy, given by m * g * h, where g is the acceleration due to gravity and h is the height. At the maximum height, the particle has zero kinetic energy, since it has come to a momentary stop. Therefore, the total energy at this point is purely potential energy.

Equating the initial kinetic energy to the final potential energy, we can solve for the vertical component of the initial velocity, Viy. Using the given information, we get:

1/2 * m * (Vi)^2 = m * g * h
1/2 * 0.500 kg * (30.0 m/s)^2 = 0.500 kg * 9.8 m/s^2 * 20.0 m
225 m^2/s^2 = 9.8 m/s^2 * 10.0 m
225 m^2/s^2 = 98 m^2/s^2

Solving for Viy, we get:

Viy = √(225 m^2/s^2 - 98 m^2/s^2) = √(127 m^2/s^2) = 11.3 m/s

Therefore, the vertical component of the initial velocity is 11.3 m/s. This means that the total initial velocity, Vi, can be found using the Pythagorean theorem:

Vi = √((Vix)^2 + (Viy)^2) = √((30.0 m/s)^2 + (11.3 m/s)^2) = √(900 m^2/s^2 + 127 m^2/s^2) = √(1027 m^2/s^2) = 32.0 m/s

In conclusion, using the Law of Conservation of Energy, we have determined that the vertical component of the initial velocity, Viy, is 11.3 m/s and the total initial velocity, Vi, is 32.0 m/s
 

Related to Projectile question involving Law of Conservation of Energy

1. What is the Law of Conservation of Energy?

The Law of Conservation of Energy states that energy cannot be created or destroyed, but can only be transferred from one form to another.

2. How does the Law of Conservation of Energy apply to projectile motion?

In projectile motion, the initial kinetic energy of the object is converted into potential energy as it gains height. As it falls back down, the potential energy is converted back into kinetic energy, with the total energy of the system remaining constant.

3. What factors affect the amount of kinetic and potential energy in a projectile?

The amount of kinetic and potential energy in a projectile is affected by its mass, initial velocity, and the height at which it is launched.

4. Can the Law of Conservation of Energy be violated in a projectile motion scenario?

No, the Law of Conservation of Energy is a fundamental law of physics and cannot be violated in any scenario, including projectile motion.

5. How is the Law of Conservation of Energy used to solve projectile motion problems?

The Law of Conservation of Energy can be used to determine the final velocity, height, or range of a projectile by setting the initial kinetic energy equal to the final potential energy, or vice versa.

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