a ball mass 800g is dropped from height of 5m and rebounds to a hight of 3.8m
the air resistance is negligible. calculate
a. the kinetic energy of the ball just before impact
b. the initial rebound speed of the ball
c. the energy transferred to the ground during impact
0.5 x 0.8 x 5
To solve these calculations, we can use the principle of conservation of mechanical energy.
a. The kinetic energy of the ball just before impact can be calculated using the equation:
Kinetic Energy = Potential Energy at the initial height
Potential Energy = mass * gravitational acceleration * height
Given:
mass = 800g = 0.8 kg
gravitational acceleration = 9.8 m/s^2
height = 5m
Potential Energy = 0.8 kg * 9.8 m/s^2 * 5m
Potential Energy = 39.2 Joules
Therefore, the kinetic energy of the ball just before impact is 39.2 Joules.
b. To find the initial rebound speed of the ball, we need to consider the conservation of mechanical energy. At the highest point of the rebound, all the potential energy is converted to kinetic energy. The kinetic energy just before impact is equal to the kinetic energy just after rebounding. Using the equation:
Kinetic Energy = 1/2 * mass * (speed)^2
Rearranging the equation for speed:
speed = sqrt((2 * Kinetic Energy) / mass)
Given:
Kinetic Energy = 39.2 Joules
mass = 0.8 kg
speed = sqrt((2 * 39.2 Joules) / 0.8 kg)
speed ≈ 8.78 m/s
Therefore, the initial rebound speed of the ball is approximately 8.78 m/s.
c. The energy transferred to the ground during impact can be calculated as the difference between the initial kinetic energy and the final potential energy.
Energy transferred to the ground = Initial Kinetic Energy - Final Potential Energy
Given:
Initial Kinetic Energy = 39.2 Joules
Final Potential Energy = mass * gravitational acceleration * height
Final Potential Energy = 0.8 kg * 9.8 m/s^2 * 3.8m
Final Potential Energy ≈ 29.7 Joules
Energy transferred to the ground = 39.2 Joules - 29.7 Joules
Energy transferred to the ground ≈ 9.5 Joules
Therefore, the energy transferred to the ground during impact is approximately 9.5 Joules.
To solve this problem, we will use the concepts of mechanical energy, potential energy, and kinetic energy.
a. The kinetic energy of the ball just before impact can be calculated using the principle of conservation of mechanical energy. The total mechanical energy of the ball at any given point is the sum of its potential energy and kinetic energy. As the ball is dropped from a height of 5m, it has potential energy initially. Since the air resistance is negligible, we can neglect any energy losses due to air resistance.
The potential energy (PE) of the ball at a height of 5m is given by the formula:
PE = m * g * h
Where:
m = mass of the ball = 800g = 0.8kg (converting grams to kilograms)
g = acceleration due to gravity = 9.8 m/s² (approximately)
h = height = 5m
PE = 0.8kg * 9.8 m/s² * 5m
PE = 39.2 J
At the height just before impact, the potential energy is completely converted into kinetic energy.
Therefore, the kinetic energy (KE) just before impact is also 39.2 J.
b. To find the initial rebound speed of the ball, we can use the principle of conservation of mechanical energy again. At the maximum height of rebound (3.8m), the potential energy is equal to the initial kinetic energy just before impact.
PE = KE
m * g * h = 0.8kg * 9.8 m/s² * 3.8m
Solving for KE:
KE = 0.8kg * 9.8 m/s² * 3.8m
KE = 29.568 J
Since kinetic energy is given by the formula:
KE = (1/2) * m * v²
Solving for v:
v = √(2 * KE / m)
v = √(2 * 29.568 J / 0.8kg)
v ≈ √93.21
v ≈ 9.657 m/s (rounded to three decimal places)
c. The energy transferred to the ground during impact is equal to the difference in potential energy before and after the impact. The potential energy just before impact was 39.2 J, and the potential energy just after the impact at a height of 3.8m is given by:
PE = m * g * h
PE = 0.8kg * 9.8 m/s² * 3.8m
PE = 29.568 J
Therefore, the energy transferred to the ground during impact is 39.2 J - 29.568 J = 9.632 J.