A 0.580-kg ball is dropped from rest at a point 1.60 m above the floor. The ball rebounds straight upward to a height of 0.650 m. What are the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor?
To find the magnitude and direction of the impulse, we need to consider the change in momentum of the ball during the collision. Impulse is defined as the change in momentum, and momentum is the product of an object's mass and velocity.
To calculate the momentum before and after the collision, we need to determine the velocity of the ball at each point. We can use the principle of conservation of energy to find the initial velocity.
First, let's find the initial velocity using the equation for potential energy:
Potential Energy (PE) = mgh
Where:
m = mass of the ball = 0.580 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height = 1.60 m
PE = mgh
PE = (0.580 kg)(9.8 m/s^2)(1.60 m)
PE = 9.0552 J
Since the potential energy is converted to kinetic energy when the ball reaches the bottom, the kinetic energy at the bottom is equal to the potential energy at the top.
Kinetic Energy (KE) = 1/2 mv^2
Where:
v = velocity of the ball
KE = 9.0552 J
KE = 1/2 (0.580 kg)v^2
v^2 = (2)(9.0552 J) / 0.580 kg
v^2 = 2(15.6379310345 J/kg)
v^2 = 31.275862069 J/kg
v = √(31.275862069 J/kg)
v ≈ 5.592 m/s
The initial velocity of the ball is approximately 5.592 m/s.
Now, let's find the final velocity using the principle of conservation of energy. At the highest point, the final velocity will be zero since the ball momentarily comes to a stop.
Now that we have the initial and final velocities, we can calculate the change in momentum and the impulse.
Change in momentum (Δp) = m * Δv
Where:
Δp = change in momentum
m = mass of the ball = 0.580 kg
Δv = change in velocity = final velocity - initial velocity
Δp = (0.580 kg) * (0 m/s - 5.592 m/s)
Δp = -(3.23856 kg.m/s)
The change in momentum is approximately -3.23856 kg·m/s. The negative sign indicates that the direction of the impulse is opposite to the initial velocity.
Therefore, the magnitude of the impulse of the net force applied to the ball during the collision with the floor is approximately 3.23856 kg·m/s, and the direction is upward (opposite to the initial velocity).