A ball of mass 0.150 kg is dropped from rest to a height of 0.125 m. It rebounds from the floor to reach a height of 0.960m. What impulse was given by the floor?
please help/explain. Thanks!
I know that momemtum (p) is equal to mass times velocity, but I am not sure how to get the velocity.
use the same formula you used to get the velocity falling from 1.25 meters to get the velocity it would get falling from 0.96m. The last is identical to the velocity needed to make it go to that height, by laws of conservation of energy.
In other words, if you dropped an object from a height H and it hit with a speed of V, that is the same set of numbers as if you propelled it upwards with a velocity of V to a height of H.
v = �ã(2gh) = �ã(2(9.8)(1.25)) = 4.95 m/s when it hits the ground
from 0.96 that is �ã(2(9.8)(0.96)) = 4.34 m/s
since the mass doesn't change
I = P1 - P2 = m(V1 - V2) = 0.0915
To find the impulse given by the floor, we can use the principle of conservation of momentum.
Impulse is defined as the change in momentum, which is equal to the product of the force applied and the time interval over which it acts. In this case, the force is exerted by the floor during the collision with the ball, and the time interval is the duration of the collision.
The principle of conservation of momentum states that the total momentum of a system before a collision is equal to the total momentum after the collision, provided there are no external forces acting on the system.
Let's first find the initial momentum of the ball just before it hits the floor. The ball is dropped from rest, so its initial velocity is zero. Therefore, the initial momentum is also zero.
Next, let's find the final momentum of the ball just before it rebounds off the floor. The momentum is given by the product of mass and velocity. Since the ball rebounds, its final velocity will be in the opposite direction compared to its initial velocity.
Using the equation for conservation of mechanical energy, we can relate the heights of the ball on each bounce to their corresponding velocities.
Initial potential energy (before the first bounce):
mgh = (0.150 kg)(9.8 m/s^2)(0.125 m) = 0.18375 J
Final potential energy (after the rebound):
mgh' = (0.150 kg)(9.8 m/s^2)(0.960 m) = 1.4112 J
The difference in potential energy is equal to the change in kinetic energy, which is equal to the impulse given by the floor.
Change in kinetic energy = Final kinetic energy - Initial kinetic energy
= 1.4112 J - 0.18375 J
= 1.22745 J
Therefore, the impulse given by the floor is 1.22745 J
To find the impulse given by the floor, we need to use the principle of conservation of momentum. The impulse is defined as the change in momentum, which can be calculated using the following equation:
Impulse = Change in momentum
The change in momentum is given by:
Change in momentum = Final momentum - Initial momentum
In this case, the ball is dropped from rest, so the initial momentum is zero. The final momentum can be calculated using the formula for momentum:
Momentum = mass * velocity
First, let's find the velocity of the ball when it hits the floor. We can use the principle of conservation of energy to relate potential energy and kinetic energy. The initial potential energy of the ball when it is dropped is:
Initial potential energy = mass * gravitational acceleration * initial height
= 0.150 kg * 9.8 m/s^2 * 0.125 m
Next, we can calculate the final potential energy of the ball when it rebounds:
Final potential energy = mass * gravitational acceleration * final height
= 0.150 kg * 9.8 m/s^2 * 0.960 m
Since the ball rebounds, its velocity reverses direction. Therefore, the kinetic energy is given by the final potential energy. The formula for kinetic energy is:
Kinetic energy = (1/2) * mass * velocity^2
Setting the final potential energy equal to the kinetic energy equation, we can solve for the velocity of the ball when it hits the floor:
(1/2) * mass * velocity^2 = mass * gravitational acceleration * final height
(1/2) * 0.150 kg * velocity^2 = 0.150 kg * 9.8 m/s^2 * 0.960 m
Now we can solve for the velocity by canceling out the mass and rearranging the equation:
velocity^2 = (2 * 9.8 m/s^2 * 0.960 m) / 0.125 kg
velocity^2 = 15.552 m^2/s^2
velocity ≈ √15.552 ≈ 3.95 m/s
Now that we have the velocity, we can calculate the change in momentum:
Change in momentum = final momentum - initial momentum
Since the initial momentum is zero, the change in momentum is equal to the final momentum, which is given by:
Final momentum = mass * velocity
= 0.150 kg * 3.95 m/s
Finally, we have:
Change in momentum ≈ 0.5925 kg·m/s
Therefore, the impulse given by the floor is approximately 0.5925 kg·m/s.