1. A body of mass 400g dropped from a height of 30m. If after rebounce the ball loses 2/3 of its energy, calculate: <a>.the velocity before bounce, .the height after rebounce, <c>.the velocity after rebounce
a. V^2 = Vo^2 + 2g*d.
V^2 = 0 + 19.6*30 = 588
V = 24.2 m/s.
To solve this problem, we can use the principles of conservation of energy and conservation of momentum. Let's break it down into parts:
a. The velocity before the bounce:
To calculate the velocity before bounce, we need to apply the principle of conservation of energy. The initial potential energy (mgh) is converted into kinetic energy (1/2 mv^2). Since energy is conserved, we can set these two equal:
mgh = (1/2)mv^2
Here, mass (m) is given as 400 grams, which is 0.4 kg. The height (h) is given as 30 meters. We can rearrange the equation to solve for velocity (v):
v^2 = 2gh
v = √(2gh)
Substituting the given values:
v = √(2 * 9.8 m/s^2 * 30 m)
v ≈ 17.15 m/s
Therefore, the velocity before the bounce is approximately 17.15 m/s.
b. The height after rebounce:
To calculate the height after rebounce, we can use the concept of conservation of momentum. The momentum before the bounce is equal to the momentum after the bounce. The momentum (p) is given by the product of mass (m) and velocity (v):
p = mv
Therefore, the momentum before the bounce (m1v1) is equal to the momentum after the bounce (m2v2). We can represent this as:
m1v1 = m2v2
Since the mass and velocity of the ball remain the same throughout (only the direction of velocity changes), we can write:
v1 = -v2
Substituting the values of v1 and v2, we have:
17.15 m/s = -v2
Since the ball bounces back with reverse velocity, v2 will be negative.
Now, to find the height after the rebound, we can use the formula for potential energy:
PE = mgh
Given that the ball loses 2/3 of its energy, we can write:
PE_after_rebounce = (1 - 2/3) * PE_before_rebounce
Let's substitute the values into the equation:
PE_after_rebounce = (1 - 2/3) * mgh
PE_after_rebounce = (1/3) * mgh
Substituting the known values:
PE_after_rebounce = (1/3) * 0.4 kg * 9.8 m/s^2 * 30 m
PE_after_rebounce ≈ 39.2 J
The height (h2) after the rebound can be calculated by rearranging the potential energy equation:
h2 = PE_after_rebounce / (mg)
Substituting the values:
h2 = 39.2 J / (0.4 kg * 9.8 m/s^2)
h2 ≈ 10 m
Therefore, the height after the rebound is approximately 10 meters.
c. The velocity after rebounce:
The velocity after the rebound (v2) can be calculated using the formula for potential energy:
PE = (1/2)mv^2
Given that the ball loses 2/3 of its energy, we can write:
PE_after_rebounce = (1 - 2/3) * PE_before_rebounce
Let's substitute the values into the equation:
PE_after_rebounce = (1 - 2/3) * (1/2)mv^2
PE_after_rebounce = (1/3) * (1/2)mv^2
Substituting the known values:
PE_after_rebounce = (1/3) * (1/2) * 0.4 kg * v^2
PE_after_rebounce = (1/12) mv^2
Since the mass and height remain the same, we can equate the before and after potential energy:
(mgh) = (1/12) mv^2
Canceling out the mass and solving for velocity (v):
gh = (1/12) v^2
v^2 = 12gh
v = √(12gh)
Substituting the known values:
v = √(12 * 9.8 m/s^2 * 10 m)
v ≈ 14.0 m/s
Therefore, the velocity after the rebound is approximately 14.0 m/s.