Board and man sliding on ice problem

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SUMMARY

The discussion revolves around a physics problem involving a 250-kg board sliding on ice at 21 m/s and a 63 kg man who grabs the board. The center of mass of the system moves at 16.77 m/s after the collision, calculated using the conservation of momentum. The problem also addresses the angular velocity of the system post-collision, emphasizing that kinetic energy is not conserved due to the inelastic nature of the collision. Participants highlight the importance of using conservation of angular momentum to analyze the system's rotational dynamics.

PREREQUISITES
  • Understanding of conservation of momentum
  • Familiarity with angular momentum and its conservation
  • Knowledge of inelastic collisions
  • Basic principles of rotational motion
NEXT STEPS
  • Study the conservation of angular momentum in inelastic collisions
  • Learn about center-of-mass calculations in multi-body systems
  • Explore the differences between elastic and inelastic collisions
  • Investigate rotational dynamics and angular velocity calculations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of momentum and energy conservation principles.

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?
 

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