Board and man sliding on ice problem

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A 250-kg board sliding at 21 m/s collides with a 63 kg man who grabs it, leading to a combined motion. The center of mass of the system moves at 16.77 m/s after the collision, calculated using conservation of momentum. The discussion highlights that kinetic energy is not conserved due to the inelastic nature of the collision, prompting the use of conservation of angular momentum for further analysis. Participants engage in clarifying concepts like center-of-mass coordinates and the distinction between translational and rotational motion. The conversation emphasizes the importance of understanding both momentum and angular momentum in collision scenarios.
Zach981
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Homework Statement


A 250-kg board, 2.4 m in length slides broadside along the surface of ice with a speed of 21m/s . A 63 kg man at rest grabs one end as it goes past and hangs on as both he and the beam go spinning down the ice. Assume frictionless motion, and assume that the man can be regarded as a point mass. [Editor's note: that does not seem like a realistic speed for anybody to catch and hold on to the board! Work the problem anyway.] (Figure 1)

Part A. How fast does the center of mass of the system move after the collision?


Part B. With what angular velocity does the system rotate about its new center of mass? (Note that this is not the original center of mass of the board.)


Homework Equations



L=Iω, TE = 1/2mv^2 + 1/2Iω^2

The Attempt at a Solution



The answer to part A is 16.77 m/s. I found this by using cons of momentum.
Is kinetic energy conserved in part B? Can someone set it up for me?
 
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Hi Zach981! :smile:
Zach981 said:
Is kinetic energy conserved in part B?

Certainly not … this is a completely inelastic collision.

Use conservation of angular momentum :wink:

(which, like ordinary momentum, is always conserved in collisions)​
 
Where else would the energy go? Note that there is translational and rotational motion.
Isn't angular momentum supposed to be conserved as well?
Have you been introduced to center-of-mass coordinates yet?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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