Why the conservation energy doesn't work out?

  • Thread starter cks
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  • #1
cks
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In the
first case, a ball with a mass m is thrown vertically upward,
second case, a ball with the same mass is thrown horizontally,
third case, a ball with the same mass is thrown vertically downward,

Compare the velocity v when the ball hit the ground for the first, second and third case.

If I use conservation of energy to work it, I find that the velocity at the bottom of the ground is the same for each case.

However, if I use equation of linear motion, I find that the second case, the ball hits the ground with a velocity smaller than the first and third case.

What is wrong??? I'm totally confused.
 

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Answers and Replies

  • #2
Doc Al
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cks said:
If I use conservation of energy to work it, I find that the velocity at the bottom of the ground is the same for each case.
Right! The speed of the ball as it hits the ground is the same in each case.

However, if I use equation of linear motion, I find that the second case, the ball hits the ground with a velocity smaller than the first and third case.
While the first and third cases only involve vertical motion, the second case also involves horizontal motion. But when you applied the kinematic equation to the second case, you ignored the horizontal speed. That adds an extra [itex]1/2 m u^2[/itex] to the energy.
 
  • #3
cks
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Thanks, I finally understand what went wrong. Since the final v is a vector, I should add up the velocity of x-component and y-component, it's found to be the same. Thank you again!
 

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