A ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.960 m. What impulse was given to the ball by the floor?
To find the impulse given to the ball by the floor, you can use the principle of conservation of mechanical energy and the impulse-momentum theorem.
First, let's find the initial velocity of the ball just before it hits the floor. We can use the equation of motion:
v^2 = u^2 + 2as,
where v is the final velocity (0 m/s because the ball comes to rest), u is the initial velocity (unknown), a is the acceleration due to gravity (-9.8 m/s^2), and s is the distance traveled (1.25 m). Plugging in these values, the equation becomes:
0^2 = u^2 + 2*(-9.8)*1.25.
Simplifying it:
0 = u^2 - 24.5.
Rearrange the equation to solve for u^2:
u^2 = 24.5.
Taking the square root:
u = √(24.5).
Using a calculator, u ≈ 4.95 m/s.
Now that you know the initial velocity of the ball, you can find the change in momentum using the impulse-momentum theorem:
Δp = m * (v - u),
where Δp is the change in momentum, m is the mass of the ball (0.150 kg), and v is the final velocity of the ball after rebounding (also unknown).
Let's calculate the final velocity of the ball using the same equation of motion:
v^2 = u^2 + 2as,
where v is the final velocity (unknown), u is the initial velocity (4.95 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and s is the distance traveled (0.960 m). Plugging in these values, the equation becomes:
v^2 = (4.95)^2 + 2*(-9.8)*0.960,
v^2 = 24.5 - 18.816,
v^2 = 5.684.
Taking the square root:
v = √(5.684).
Using a calculator, v ≈ 2.38 m/s.
Now, we can find the change in momentum:
Δp = 0.150 * (2.38 - 4.95),
Δp = 0.150 * (-2.57),
Δp ≈ -0.386 kg m/s.
The negative sign indicates that the direction of the impulse is opposite to the initial motion, as the ball rebounds upward.
Therefore, the impulse given to the ball by the floor is approximately -0.386 kg m/s.