A ball of mass 0.280 kilograms is thrown at 24.5 meters per second to a skater standing on ice with arms outstretched. The skater has mass 60.5 kilograms and moment of inertia 1.37 kilogram·meters squared. After catching the ball, at a distance of 0.607 meters from the skater's center, the skater both recoils and rotates as shown in part b of the figure below. a. What is the skater's final linear speed? Include units in your answer. b. What is the skater's final angular speed? Include units in your answer. More information. Hint: Solve this using the rotational equivalent of the concept you used for solving part a. Granted, it seems like initially the ball is moving only translationally. But at the instant the skater catches the ball, the ball has a velocity perpendicular to a radius of rotation. Thus, we can think of its motion as rotational also. c. How much kinetic energy is lost in the collision? Include units in your answer.

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Chapter1: Units, Trigonometry. And Vectors
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A ball of mass 0.280 kilograms is thrown at 24.5 meters per second to a skater standing on ice with arms outstretched. The skater has mass 60.5 kilograms and moment of inertia 1.37 kilogram·meters squared. After catching the ball, at a distance of 0.607 meters from the skater's center, the skater both recoils and rotates as shown in part b of the figure below.

 

a. What is the skater's final linear speed? Include units in your answer.


 
b. What is the skater's final angular speed? Include units in your answer. More information. Hint: Solve this using the rotational equivalent of the concept you used for solving part a. Granted, it seems like initially the ball is moving only translationally. But at the instant the skater catches the ball, the ball has a velocity perpendicular to a radius of rotation. Thus, we can think of its motion as rotational also.



 
c. How much kinetic energy is lost in the collision? Include units in your answer. 


PLEASE ANSWER IN HANDWRITING

(a)
(b)
Transcribed Image Text:(a) (b)
Expert Solution
Step 1

a)

Apply the momentum conservation equation and substitute the corresponding values to calculate the final linear speed of the skater.

pi=pfmbub+msus=(mb+ms)v(0.280 kg)(24.5 m/s)+(60.5 kg)(0)=(0.280 kg+60.5 kg)vv=0.113 m/s

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