Concept explainers
Review. Assume a certain liquid, with density 1 230 kg/m3, exerts no friction force on spherical objects. A ball of mass 2.10 kg and radius 9.00 cm is dropped from rest into a deep tank of this liquid from a height of 3.30 m above the surface. (a) Find the speed at which the hall enters the liquid. (b) Evaluate the magnitudes of the two forces that are exerted on the ball as it moves through the liquid. (c) Explain why the ball moves down only a limited distance into the liquid and calculate this distance. (d) With what speed will the ball pop up out of the liquid? (c) How does the time interval ∆tdown, during which the ball moves from the surface down to its lowest point, compare with the lime interval ∆tup for the return trip between the same two points? (f) What If? Now modify the model to suppose the liquid exerts a small friction force on the ball, opposite in direction to its motion. In this case, how do the time intervals ∆tdown and ∆tup compare? Explain your answer with a conceptual argument rather than a numerical calculation.
(a)
The speed of the ball which enters the liquid.
Answer to Problem 14.64AP
The speed of the ball which enters the liquid is
Explanation of Solution
The density of the liquid is
By the conservation of energy,
Here,
Substitute
Conclusion:
Therefore, the speed of the ball which enters the liquid is
(b)
The magnitudes of the two forces that are exerted on the ball as move through liquid.
Answer to Problem 14.64AP
The magnitude of the gravitational force that is exerted on the ball as move through liquid is
Explanation of Solution
Formula to calculate the gravitational force or weight of the ball is,
Here,
Substitute
Thus, the gravitational force exerted on the ball is
The buoyant force exerted on the ball is equal to the volume of water displaced by the ball.
Formula to calculate the buoyant force exerted on the ball is,
Here,
Formula to calculate the volume of the spherical ball is,
Here,
Substitute
Substitute
Conclusion:
Therefore, the magnitudes of the gravitational force that are exerted on the ball as move through liquid is
(c)
The distance covered by the ball in water.
Answer to Problem 14.64AP
The distance covered by the ball in water is
Explanation of Solution
The buoyant force exerted on the ball is greater than the weight of the ball, therefore the ball certain distance covered inside the water because it changes the direction of motion.
From third law of motion,
Here,
Formula to calculate the acceleration of the ball is,
Formula to calculate the net force acting on a ball is,
Substitute
Substitute
Substitute
The negative sign shows direction of the ball in downward direction.
Conclusion:
Therefore, the distance covered by the ball in water is
(d)
The speed of the ball pop up out of the liquid.
Answer to Problem 14.64AP
The speed of the ball which enters the liquid is
Explanation of Solution
The speed of the ball which enters the liquid is equal to the speed of the ball pop up out of the liquid because absence of friction, no energy losses occur in this system. Hence the speed of the ball pop up out of the liquid is
Conclusion:
Therefore, the speed will the ball pop up out of the liquid is
(e)
The result of comparison the time interval during which the ball moves from the surface to its lowest point with the time interval for return trip at the same point.
Answer to Problem 14.64AP
The time interval during which the ball moves from the surface to its lowest point is identical to the time interval for return trip at the same point.
Explanation of Solution
The time interval during which the ball moves from the surface to its lowest point is identical to the time interval for return trip at the same point because ball going down and up acceleration of the ball and distance covered by the ball is same.
Conclusion:
Therefore, the time interval during which the ball moves from the surface to its lowest point is identical to the time interval for return trip at the same point.
(f)
Compare the time interval during which the ball moves from the surface to its lowest point with the time interval for return trip at the same point.
Answer to Problem 14.64AP
The time interval during which the ball moves from the surface to its lowest point is not equal to the time interval for return trip at the same point.
Explanation of Solution
The time interval during which the ball moves from the surface to its lowest point is not equal to the time interval for return trip at the same point when friction is present because energy losses by the system.
Conclusion:
Therefore, the time interval during which the ball moves from the surface to its lowest point is not equal to the time interval for return trip at the same point.
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