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A ball falls under gravity

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  • #1
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Homework Statement



a ball falls under gravity from a height of 10m with an initial downward velocity v.It collides with the ground and looses 50% of its energy in collision and then rises to same height .1) find the initial velocity v 2)the height to which the ball rises after the collision if the initial velocity v is directed upward instead of downward

Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
lewando
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What relevant equations do you think would apply?
 
  • #3
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... and then rises to same height
I think you have a typo there. Should it be "and then rises to some height"?
 
  • #4
lewando
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I think you have a typo there. Should it be "and then rises to some height"?
I think it is right but could be worded better. I read into the problem that the ball rises to the 10m height with zero velocity at that moment.
 
  • #5
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If an object is dropped from 10m it can NEVER bounce 10m's upwards again, it will always be slightly less so I think the wording is "some height"
 
  • #6
lewando
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If an object is dropped from 10m it can NEVER bounce 10m's upwards again, it will always be slightly less so I think the wording is "some height"
If you apply an initial non-zero downward velocity (not just dropping it), it is possible.
 
  • #7
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If you apply an initial non-zero downward velocity (not just dropping it), it is possible.
Thanks for the clarification, lewando. I see the error of my ways.

So, norrh, follow lewando's lead: what equations do you think would apply?
 
  • #8
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ooooooooo it has a velocity while falling and it cannot be equivalent to a freely falling bodyand i do not know how to start.And i do not know many formulas for ccollisions i forgot.i just need starting.
 
  • #9
lewando
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comments in brackets
ooooooooo it has a velocity while falling [it has an initial velocity v, and velocity increases as it free falls] and it cannot be equivalent to a freely falling body [it is a freely falling body] and i do not know how to start [the standard kinematic equations are a good place to start--also, since the word "energy" is used, consider energy equations: KE, PE].And i do not know many formulas for ccollisions i forgot.i just need starting.
 
  • #10
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Start with conservation of mechanical energy.
Make at height 10m as reference of potential energy equal zero.
 
  • #11
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Thanks for the clarification, lewando. I see the error of my ways.

So, norrh, follow lewando's lead: what equations do you think would apply?
Well yes if you shot a pingpong ball at the floor from a distance of say 1 meter hit would it the floor and bounce higher but after that it will always get lower. Otherwise we'd all have free energy and the planet would be saved lol
 
  • #12
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would it be possible to find out v0 using just the following equation? y=v0t-0.5a(t*t)
 
  • #13
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would it be possible to find out v0 using just the following equation? y=v0t-0.5a(t*t)
It is about energy.
50% of total energy before collision use to raise the object, conservative force work.
Another 50% goes to non-conservative force work.

From above you can find total energy of the object just before collision.
 
  • #14
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So, to do the conservative force work we have to equal the mass to zero, but still, for the non conservative work, how do we find the "v" in the final cinectic energy?
 
  • #15
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So, to do the conservative force work we have to equal the mass to zero, but still, for the non conservative work, how do we find the "v" in the final cinectic energy?
50% lost, another 50% converted to potential energy.
The total energy is total kinetic enery before collision.
Use SUVAT to find intial velocity.
 
Last edited:
  • #16
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Ok,thanks dude
 
  • #17
lewando
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...
Make at height 10m as reference of potential energy equal zero.
Why would you not use the ground as the zero PE reference?
 
  • #18
CAF123
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I attempted this question too and came up with the following;
Taking the reference level of potential, [itex] U_o = 0 [/itex] to be ground level,
[itex] \frac{1}{2}mv_o^2 + mgh = \frac{1}{2}mv_a^2, [/itex] where [itex] v_o [/itex] is the initial velocity and [itex] v_a [/itex] is the velocity prior to impact.

Since there is 50% dissipation, I said that half of the total energy would be converted to potential energy;
[itex] \frac{1}{2}(\frac{1}{2}mv_a^2) = mgh [/itex]

Solving for [itex] v_a [/itex] yields [itex] v_a = \sqrt{4gh} [/itex] but this gives an initial velocity of [itex] \sqrt{2gh} [/itex] which is what it would be were it dropped. So I feel there is an error somewhere. Any ideas?
 
  • #19
lewando
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To summarize you:

If V0 = 0, Va = [itex] \sqrt{2gh} [/itex]

If V0 = [itex] \sqrt{2gh} [/itex], Va = [itex] \sqrt{4gh} [/itex]

I don't see an error.
 
  • #20
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Why would you not use the ground as the zero PE reference?
Yes i made an error in earlier post.
 
  • #21
CAF123
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Many thanks lewando.
Can I ask if this is the general opinion of what the answer should be?
 
  • #22
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I attempted this question too and came up with the following;
Taking the reference level of potential, [itex] U_o = 0 [/itex] to be ground level,
[itex] \frac{1}{2}mv_o^2 + mgh = \frac{1}{2}mv_a^2, [/itex] where [itex] v_o [/itex] is the initial velocity and [itex] v_a [/itex] is the velocity prior to impact.

Since there is 50% dissipation, I said that half of the total energy would be converted to potential energy;
[itex] \frac{1}{2}(\frac{1}{2}mv_a^2) = mgh [/itex]

Solving for [itex] v_a [/itex] yields [itex] v_a = \sqrt{4gh} [/itex] but this gives an initial velocity of [itex] \sqrt{2gh} [/itex] which is what it would be were it dropped. So I feel there is an error somewhere. Any ideas?
your calculations are correct i have come up with the same thing
the answer for part 2 is pretty interesting too
 
Last edited:

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