A block of wood with mass M = 0.440kg hangs at rest (Vi=0) from the ceiling by a string and a m = 0.0720kg elastic ball is thrown straight upward striking the bottom of the block with a speed of vi=5.74m/s. The ball rebounds from the block in a perfectly elastic collision. How high, H, does this block rise above the original position?
To find the height (H) to which the block rises above the original position, we need to consider the conservation of momentum and conservation of energy.
First, let's analyze the collision between the ball and the block. Since the collision is perfectly elastic, the total kinetic energy before and after the collision remains the same.
The initial kinetic energy of the ball is given by: KE_initial = (1/2) * m * vi^2
where m is the mass of the ball and vi is its initial speed.
The final kinetic energy of the ball after the collision is given by: KE_final = (1/2) * m * vf^2
where vf is the final speed of the ball.
Since the collision is elastic, the final speed of the ball can be determined using the principle of conservation of momentum. Since the ball rebounds in the opposite direction, the change in momentum is doubled. So,
Change in momentum of the ball = 2 * (m * vf - m * vi)
Now, let's consider the block's motion. As the block rises, its potential energy increases. The potential energy gained by the block is equal to the loss in kinetic energy of the ball.
Potential energy gained by the block = KE_initial - KE_final = (1/2) * m * vi^2 - (1/2) * m * vf^2
Since the block is at rest initially, its initial potential energy is zero. Therefore,
Potential energy gained by the block = m * g * H
where m is the mass of the block, g is the acceleration due to gravity, and H is the height it rises.
Now, let's set up the equations to solve for the height (H).
m * g * H = (1/2) * m * vi^2 - (1/2) * m * vf^2
Since the collision is perfectly elastic, the final speed of the ball can be calculated by equating the change in momentum with the loss in kinetic energy:
2 * (m * vf - m * vi) = (1/2) * m * vi^2 - (1/2) * m * vf^2
Simplifying the equation:
2 * vf - 2 * vi = vi^2 - vf^2
Now, substitute the values given in the problem:
m = 0.440 kg
vi = 5.74 m/s
vf = ? (unknown)
g = 9.8 m/s^2
By solving the equation, you can find the value of vf (final speed of the ball) and then calculate H (the height to which the block rises above the original position).