A ball (5 cm diameter, 10 kg mass) is dropped from a height of 2 m above a concrete surface (5 cm thickness). The top of the ball rebounds to a height of 60 cm above the level of the concrete surface. Using the testing procedures recommended in your laboratory manual, the coefficient of restitution would be CLOSEST to:Question 20Select one:a.0.27b.0.30c.0.52d.1.83e.0.80
Question
A ball (5 cm diameter, 10 kg mass) is dropped from a height of 2 m above a concrete surface (5 cm thickness). The top of the ball rebounds to a height of 60 cm above the level of the concrete surface. Using the testing procedures recommended in your laboratory manual, the coefficient of restitution would be CLOSEST to:Question 20Select one:a.0.27b.0.30c.0.52d.1.83e.0.80
Solution
The coefficient of restitution (e) is a measure of how much kinetic energy is conserved in a collision. It is defined as the ratio of the final relative velocity to the initial relative velocity. In this case, it is the ratio of the rebound speed to the initial speed.
The initial speed of the ball can be found using the equation for the motion under gravity:
v = sqrt(2gh)
where:
- g is the acceleration due to gravity (approximately 9.81 m/s^2)
- h is the initial height (2 m)
So, the initial speed is sqrt(29.812) = 6.26 m/s.
The rebound speed is the speed the ball has when it reaches the maximum height of its bounce. This can also be found using the equation for motion under gravity, rearranged to solve for v:
v = sqrt(2gh)
where h is now the rebound height (0.6 m). So, the rebound speed is sqrt(29.810.6) = 3.43 m/s.
Finally, the coefficient of restitution is the ratio of these two speeds:
e = rebound speed / initial speed e = 3.43 / 6.26 = 0.55
So, the closest answer is c. 0.52.
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