To calculate the impulse given to the ball by the floor, we need to use the principle of conservation of momentum. The impulse is defined as the change in momentum of an object.
The momentum of an object is given by the product of its mass and velocity. Initially, the ball is dropped from rest, so its initial velocity is zero. When it rebounds, we can consider its final velocity as the upward velocity after the rebound.
Given: Mass of the ball (m) = 0.15 kg Initial height (h1) = 1.25 m Final height (h2) = 0.96 m
Using the principle of conservation of energy, we can equate the potential energy at the initial height to the potential energy at the final height (assuming no energy losses):
m * g * h1 = m * g * h2
Where: m = mass of the ball g = acceleration due to gravity
Solving for g, we find: g = (m * g * h2) / (m * h1) g = h2 / h1
g = 0.96 m / 1.25 m g = 0.768
Now, to calculate the final velocity (v2) of the ball after the rebound, we can use the equation:
v2^2 = v1^2 + 2 * g * Δh
Where: v1 = initial velocity (which is 0 in this case) g = acceleration due to gravity Δh = change in height = h2 - h1
Plugging in the values:
v2^2 = 0 + 2 * 0.768 * (0.96 - 1.25) v2^2 = -0.9216
Since velocity cannot be negative, we take the positive square root:
v2 = 0.96 m/s
Now we can calculate the impulse (J) given to the ball by the floor using the formula:
J = m * Δv
Where: m = mass of the ball Δv = change in velocity = v2 - v1
Plugging in the values:
J = 0.15 kg * (0.96 m/s - 0) J = 0.144 kg·m/s
Therefore, the impulse given to the ball by the floor is 0.144 kg·m/s.