The mass of the Earth is 5.98 × 1024 kg. A 6.83 kg bowling ball initially at rest is dropped from a height of 2.12 m. The acceleration of gravity is 9.8 m/s^2. What is the speed of the Earth coming up to meet the ball just before the ball hits the ground?
How do you solve this?
To solve this problem, we can use the principle of conservation of energy. The initial potential energy of the bowling ball, when it is dropped from a height, will be converted into kinetic energy just before it hits the ground.
1. Calculate the potential energy of the bowling ball at the initial height:
The potential energy (PE) is given by the formula PE = mgh, where m is the mass (6.83 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (2.12 m).
PE = (6.83 kg)(9.8 m/s^2)(2.12 m) = 142.0856 J
2. Calculate the kinetic energy of the bowling ball just before it hits the ground:
The kinetic energy is given by the formula KE = 0.5mv^2, where m is the mass (6.83 kg) and v is the velocity (speed) just before the ball hits the ground.
Since the bowling ball started from rest, its initial velocity is zero. Therefore:
KE = 0.5(6.83 kg)(v^2)
KE = 3.415 kg(v^2)
3. Apply the conservation of energy principle:
According to the principle of conservation of energy, the potential energy at the initial height is equal to the kinetic energy just before hitting the ground:
PE = KE
142.0856 J = 3.415 kg(v^2)
v^2 = 142.0856 J / 3.415 kg
v^2 = 41.6 m^2/s^2
4. Find the velocity (speed) of the Earth coming up to meet the ball:
To find the velocity, take the square root of both sides:
v = sqrt(41.6 m^2/s^2)
v ≈ 6.45 m/s
Therefore, the speed of the Earth coming up to meet the ball just before the ball hits the ground is approximately 6.45 m/s.
To solve this problem, we need to apply the principles of conservation of energy and motion under gravity.
Step 1: Determine the potential energy of the bowling ball at the initial height using the formula:
Potential Energy (PE) = mass (m) × acceleration due to gravity (g) × height (h)
Given:
Mass of the bowling ball (m) = 6.83 kg
Acceleration due to gravity (g) = 9.8 m/s^2
Height (h) = 2.12 m
So, PE = 6.83 kg × 9.8 m/s^2 × 2.12 m
Step 2: Calculate the potential energy converted to kinetic energy just before the ball hits the ground. At that point, all of the ball's potential energy is converted to kinetic energy, as the ball is at its lowest height.
Kinetic Energy (KE) = Potential Energy (PE)
Step 3: Use the kinetic energy formula to find the velocity of the ball just before it hits the ground:
Kinetic Energy (KE) = (1/2) × mass (m) × velocity^2 (v^2)
Substituting the value of Kinetic Energy from Step 2, we get:
Potential Energy (PE) = (1/2) × mass (m) × velocity^2 (v^2)
Step 4: Rearrange the equation to solve for the velocity:
velocity (v) = √(2 × Potential Energy (PE) / mass (m))
Step 5: Substitute the values into the equation and calculate the velocity:
velocity (v) = √(2 × PE / m)
velocity (v) = √(2 × (6.83 kg × 9.8 m/s^2 × 2.12 m) / 6.83 kg)
By simplifying and solving the equation, you will get the speed of the Earth coming up to meet the ball just before it hits the ground.
the force of the Earth on the ball is the same as the ball on the Earth
f = m a
use the drop time to find the velocity
v = a t