You drop a ball from a height of 2.0 m and it bounces back to a height of 1.5 m (a) What fraction of its initial energy is lost during the bounce? (b) What is the ball’s speed just at it leaves the ground after the bounce?
(a) It equals the fraction of the potential energy that is lost, at max weight. That looks like 1/4 in this case.
(b) To achieve 3/4 of the previous max height, you need the bounceback speed to be sqrt(0.75) = 0.866 of the speed just before inpact
To find the fraction of the ball's initial energy lost during the bounce, we can use the law of conservation of energy. Since the ball is dropped from a height of 2.0 m and bounces back to a height of 1.5 m, we can assume that there is no loss of energy due to other factors such as air resistance.
(a) The potential energy of the ball when it is dropped from a height of 2.0 m is given by the equation: PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity (9.8 m/s²), and h is the height.
Initial potential energy (PE₁) = mgh = mg(2.0 m) = 2mg
After bouncing, the ball reaches a height of 1.5 m. Therefore, the potential energy (PE₂) of the ball at this height is PE₂ = mgh = mg(1.5 m) = 1.5mg
The fraction of the initial energy lost during the bounce is given by: (PE₁ - PE₂) / PE₁
Fraction of energy lost = (2mg - 1.5mg) / 2mg = 0.5mg / 2mg = 0.5 / 2 = 0.25 or 25%
Therefore, the fraction of its initial energy lost during the bounce is 25%.
(b) To find the ball's speed just as it leaves the ground after the bounce, we can use the principle of conservation of mechanical energy. Since there is no loss of energy due to other factors like air resistance, the sum of the kinetic energy (KE) and potential energy (PE) before the bounce will be equal to the sum of KE and PE after the bounce.
Initial mechanical energy (ME₁) = PE + KE = mgh + 0 (since it's initially at rest, KE = 0)
Final mechanical energy (ME₂) = PE + KE = mg(1.5 m) + 0.5mv² (where v is the velocity just at it leaves the ground after the bounce)
Since the mechanical energy is conserved, we can equate ME₁ and ME₂:
ME₁ = ME₂
mgh = mg(1.5 m) + 0.5mv²
mgh - mg(1.5 m) = 0.5mv²
gh - 1.5g = 0.5v²
v² = 2(gh - 1.5g)
v = √(2(gh - 1.5g))
Substituting the values, v = √(2(9.8 m/s²)(2.0 m - 1.5 m))
v = √(2(9.8 m/s²)(0.5 m))
v = √(2(4.9 m²/s²))
v = √(9.8 m²/s²)
v = 3.13 m/s (approximately)
Therefore, the ball's speed just as it leaves the ground after the bounce is approximately 3.13 m/s.
To answer these questions, we need to understand the concepts of conservation of energy and the principles of motion.
(a) To find the fraction of the ball's initial energy lost during the bounce, we can compare the potential energy at the initial height to the potential energy at the maximum height reached during the bounce.
The potential energy of an object at a certain height is given by the formula PE = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
In this case, the initial potential energy (PE_initial) is mgh_initial at the 2.0 m height, and the maximum potential energy (PE_max) is mgh_max at the 1.5 m height.
Since we drop the ball, we start with only potential energy, so the initial energy is equal to the initial potential energy.
The fraction of energy lost during the bounce can be calculated as: (PE_initial - PE_max) / PE_initial.
(b) To find the ball's speed just as it leaves the ground after the bounce, we can use the law of conservation of energy.
The total mechanical energy (E) of the ball is the sum of its kinetic energy (KE) and potential energy (PE).
Initially, the ball has only potential energy, which is given by PE = mgh.
After the bounce, the ball reaches its maximum height and loses all its initial kinetic energy. Therefore, its total mechanical energy is equal to the potential energy at maximum height, which is PE_max.
Using the conservation of energy, we can equate the initial potential energy to the final potential energy (PE_initial = PE_max) and solve for the ball's speed at the moment it leaves the ground after the bounce.
The formula to calculate the speed (v) is given by the equation: v = sqrt(2gh), where g is the acceleration due to gravity (approximately 9.8 m/s^2) and h is the distance fallen.
Now, let's calculate the answers:
(a) Fraction of energy lost:
- Find the initial potential energy: PE_initial = mgh_initial
- Find the maximum potential energy: PE_max = mgh_max
- Calculate the fraction of energy lost: (PE_initial - PE_max) / PE_initial.
(b) Ball's speed at the moment it leaves the ground after the bounce:
- Calculate the speed using the formula: v = sqrt(2gh), where g is the acceleration due to gravity and h is the distance fallen.