A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in contact with the pavement, the lower side of the ball is temporarily flattened. Suppose the maximum depth of the dent is on the order of 1 cm. Compute an order-of-magnitude estimate for the maximum acceleration of the ball while it is in contact with the pavement. Assume chest height is 1.50 m.
To estimate the maximum acceleration of the ball while it is in contact with the pavement, we can use the concept of conservation of energy.
First, let's calculate the gravitational potential energy (GPE) of the ball when it is at chest height. GPE is given by the formula:
GPE = mgh
where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height (chest height in this case, which is given as 1.50 m).
Next, let's consider the situation when the ball is in contact with the pavement. At the maximum depth of the dent (1 cm or 0.01 m), the ball comes to rest momentarily before bouncing back up. At this point, all the potential energy has been converted into kinetic energy (KE) and is then converted back into potential energy as the ball rebounds.
So, the equation for kinetic energy is:
KE = (1/2)mv^2
where v is the velocity of the ball.
At the maximum depth of the dent, the velocity of the ball is momentarily zero. Thus, the kinetic energy at this point is also zero.
Since the total mechanical energy (potential energy + kinetic energy) remains constant during the ball's motion, we can equate the GPE at chest height to the kinetic energy at the maximum depth of the dent:
mgh = (1/2)mv^2
Simplifying this equation, we find:
gh = (1/2)v^2
Now, we can calculate the order-of-magnitude estimate for the maximum acceleration (a) while the ball is in contact with the pavement. Acceleration is the change in velocity over time. Since the ball momentarily stops at the maximum depth of the dent, the change in velocity (Δv) is its final velocity (0 m/s).
We can use the formula:
a = Δv / Δt
where Δt is the time duration the ball takes to come to a stop and start moving back up.
Since the ball momentarily stops, Δt is extremely small, approaching zero. In this case, we can consider it to be on the order of microseconds (10^-6 seconds). Therefore, the denominator is small, making the acceleration very large.
Hence, the order-of-magnitude estimate for the maximum acceleration of the ball while it is in contact with the pavement is very high, likely on the order of meters per second squared (m/s^2).