Answer:
Approximately .
Assumption: air resistance is negligible.
Explanation:
Make sure all the values are in standard units.
.
The bounce here is an inelastic collision between the ball and the surface. Some of the kinetic energy (KE) was lost. The exact value of energy loss would be equal to .
Before the bounce, all the kinetic energy of the ball would come from the drop from . That is:
.
After the bounce, the ball travels to a height of . All the potential energy gained in that process should come from the kinetic energy when the ball bounces back from the ground.
.
Hence, the size of energy loss due to the bounce would be equal to
.
Answer:
Explanation:
Given
mas of block
speed of block
spring constant
As the mass collides with the spring its kinetic energy is converted to the Elastic Potential energy of the spring
When you draw an illustration for this problem, you would come up with the same drawing as shown in the picture. As the hot-air balloon travels upwards, there is a slight time when the bag of sand rises up until it reaches the maximum height. Then, it goes back down to the ground. The total time would be t₁ + t₂. The solution is as follows:
H = v₀²/2g = (2.45)²/2(9.81) = 0.306 m
t₁ = H/v₀ = 0.306 m/2.45 m/s = 0.125 s
t₂ = √2(H + 98.8)/g = √2(0.306+ 98.8)/9.81
t₂ = 4.495 s
Total time = 0.125 s + 4.495 s = 4.62 seconds