A ball is dropped from the initial height h1 above the concrete floor and rebounds to a height of equal to 67% of the initial height h1. What is the ratio of the speed of the ball right after the bounce to the speed of the ball just before the bounce? This ratio is called coefficient of restitution e=v2/v1.
To find the ratio of the speed of the ball right after the bounce to the speed of the ball just before the bounce, we can make use of the concept of conservation of energy.
Let's assume that the ball was dropped from an initial height h1. When the ball reaches the ground after the bounce, its potential energy is converted entirely into kinetic energy. The kinetic energy at this point is given by K1 = m * v^2 / 2, where m is the mass of the ball and v is its velocity.
After the bounce, the ball reaches a height of 67% of the initial height, which corresponds to a potential energy of E2 = m * g * h2, where g is the acceleration due to gravity and h2 is 67% of h1.
Since energy is conserved, the total energy before and after the bounce should be the same. Thus, we have:
K1 + E1 = E2
Substituting the equations for kinetic and potential energy, we have:
m * v^2 / 2 + m * g * h1 = m * g * h2
Canceling out the mass, we get:
v^2 / 2 + g * h1 = g * h2
Simplifying, we have:
v^2 / 2 = g * (h2 - h1)
Finally, solving for v^2, we have:
v^2 = 2 * g * (h2 - h1)
Now, let's assume that the speed of the ball just before the bounce is v1 and the speed of the ball right after the bounce is v2.
The ratio of v2 to v1, which is the coefficient of restitution (e), is given by:
e = v2 / v1
To find the ratio e, we need to find both v1 and v2. From the equation above, we can solve for v2 as follows:
v2^2 = 2 * g * (h2 - h1)
Taking the square root of both sides, we get:
v2 = sqrt(2 * g * (h2 - h1))
Therefore, the ratio e would be:
e = sqrt(2 * g * (h2 - h1)) / v1