Ok. I have a sort of... eccentric lecturer for my physics class. He's always diving about the room and setting us all kinds of "mind-tasking" questions to ponder. But i REALLY cant get my head around this one. I'd appreciate if somebody could explain to me how on earth this works. The scenario is as follows:
A man is standing in the middle of a perfectly smooth ice-rink. Smooth in the sense that there is absolutely no friction whatsoever. He weighs 70kg, and is holding a bowling ball which weighs 10kg. I know that makes no sense; but i didn't come up with this question.
He throws his bowling ball to the right at 2.0 m/s. From this, i am asked to calculate the velocity of the man. WHAT THE HECK!?
I'm not a physics major or anything... but I figure the ball would go at 1 m/s and the guy would go at 1 m/s... since the force he used to throw it would be held against the ground, and since there's no friction the force would be divided between the man and the force used to push the ball.
I'm not a physics major or anything... but I figure the ball would go at 1 m/s and the guy would go at 1 m/s... since the force he used to throw it would be held against the ground, and since there's no friction the force would be divided between the man and the force used to push the ball.Taxpaying_Acorn
[QUOTE="Taxpaying_Acorn"]I'm not a physics major or anything... but I figure the ball would go at 1 m/s and the guy would go at 1 m/s... since the force he used to throw it would be held against the ground, and since there's no friction the force would be divided between the man and the force used to push the ball.Ninja-Vox
But he is seven times heavier than the ball. So take 6 parts of 2 m/s? I don't know. Or care anymore :P
I just asked my dad, who is a scientist and he told me how to do it. This is basically the law of conservation of energy. You have to set 10kg * 2.0 m/s =Â 70 kg * the unknown or X and solve that.
ahh.. frictionless.. thats the way to go :) anyway the force being applied to push the 10kg ball in one direction at 2m/s is also being applied to the man (by the ball) in the opposite direction. so figure out what force is required to give a stationary ball of mass 10kg a velocity of 2ms. then apply that force to the man to determine how fast he is moving.
Thanks for the help guys. You're right that it is indeed just a simple application of a law, but i think the obsurd circumstances of the example make your brain go screwey. Thanks again. :)
Thanks for the help guys. You're right that it is indeed just a simple application of a law, but i think the obsurd circumstances of the example make your brain go screwey. Thanks again. :)Ninja-Vox
If you think about what they're asking, it's not that absurd. They really mean to say "at the moment of release." In that case, the 70kg man and the 10kg ball are two separate entities (not one mass added to another) and can be considered into the equation as such.
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