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Answer: 3/4vo 9/32(v0*v0/g)

the answer is given but i need to know how this problem is actually solved. Any help would be greatly appreciated.

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- Thread starter jjf5122
- Start date

- #1

- 13

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Answer: 3/4vo 9/32(v0*v0/g)

the answer is given but i need to know how this problem is actually solved. Any help would be greatly appreciated.

- #2

- 2,063

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Have you tried to solve using conservation laws?

- #3

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- #4

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- #5

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v'= m1v1+m2v2/ total mass.....but i have no numbers, am i missing something?

- #6

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- #7

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- #8

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Have you come across the terms 'kinetic energy' and 'potential energy' before?

- #9

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also... K= mgh

U = -mgh

- #10

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Exactly.Am i suppossed to use, delta K + delta Ug(ptoential gravitational force) = 0?

Kinetic energy is (1/2)mvAnd if so...how would i actually set this up?

also... K= mgh

U = -mgh

Delta {something} refers to change in the something, usually at two different times. You know what the velocity of the combined mass is after colission. Consider the point to be at the 'zero' height. Now can you solve for the height to which the mass will go to?

- #11

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1/2mv2'-1/2mv2 + mgh'-mgh = 0

0 - 1/2m(3/4vo)2 + mgh' - mg(0)= 0

-1/2m(3/4vo)2 + mgh'

-1/2m(3/4vo)2/mg = -h

this doesnt seem right...do u have any suggestions as to how to set the equation up?

- #12

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That is correct.wouldnt the velocity i solved for actually be the initial velocity, right after the collosion occured, and the final velocity be 0(because it will come to a stop briefly when it reaches its max height)?

So im still confused as to how this is actually set up.

1/2mv2'-1/2mv2 + mgh'-mgh = 0

0 - 1/2m(3/4vo)2 + mgh' - mg(0)= 0

-1/2m(3/4vo)2 + mgh'

-1/2m(3/4vo)2/mg = -h

this doesnt seem right...do u have any suggestions as to how to set the equation up?

Why doesn't it seem right? It's perfect*. You just have to simplify it to get it to look like it is given in your book.

*-1/2m(3/4vo)

- #13

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2m(9/16)/mg( this is where im confused...g would be 9.8m/s^2?) if i multiply 4m(9.8) i get 39.2 m^2s^2

none of this seems to reduce down to the answer in my book of 9/32 v0^2/g

none of this seems to reduce down to the answer in my book of 9/32 v0^2/g

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- #14

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[tex]\frac{\frac{1}{2}m\left(\frac{9v_{o}^{2}}{16}\right)}{mg}[/tex]

Does that make more sense?

Does that make more sense?

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- #15

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hmm...so i dont actually have to plug a mass(inthe denominator) or 9.8 m/s^2 into the formula?

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