How Do Kinetic Energies Compare When Two Bodies Have Equal Momentum?

  • Thread starter alijan kk
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In summary, in order to find the ratio of kinetic energies between two masses, you must first find the momentum of each mass. Next, you must find the ratio of the momentum of each mass. Finally, you must use the given equation to find the ratio of kinetic energies.
  • #1
alijan kk
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Homework Statement



4. Two bodies of mass m1 and m2 have equal momentum their kinetic energies E1 and E2 are in the ratio.

ans. m2 : m1[/B]

Homework Equations


E1=E2
P= mv

The Attempt at a Solution



E1=E2
1/2M1v^2=1/2M2v^2
(Mv^2)/2=(M2v^2)/2
multiply both side by 2
mv^2=2(m2v^2)/2
mv^2= m2v^2

what should i do now? or the attemp is full wrong.
 
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  • #2
Your attempt is wrong. How did you get E1 = E2?

By the given question we have equal momentum [itex] m_1v_1 = m_2v_2 [/itex] We have to find out the ratio [itex] \frac{\frac{1}{2}m_1v^2_1}{\frac{1}{2}m_2v^2_2} [/itex] . Try to get the second ratio using the first equation.
 
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  • #3
Mastermind01 said:
Your attempt is wrong. How did you get E1 = E2?

By the given question we have equal momentum [itex] m_1v_1 = m_2v_2 [/itex] We have to find out the ratio [itex] \frac{\frac{1}{2}m_1v^2_1}{\frac{1}{2}m_2v^2_2} [/itex] . Try to get the second ratio using the first equation.
i don't know how to get the second ratio
 
  • #4
alijan kk said:
i don't know how to get the second ratio

Well notice how you have to get [itex]v^2_1[/itex] and [itex]v^2_2[/itex] . So first off you square both sides and then try to remove the extra [itex] m_1 [/itex] and [itex]m_2[/itex]
 
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  • #5
Mastermind01 said:
Well notice how you have to get [itex]v^2_1[/itex] and [itex]v^2_2[/itex] . So first off you square both sides and then try to remove the extra [itex] m_1 [/itex] and [itex]m_2[/itex]
i got

m1v^2=m2v^2 right?

should i square both side now
 
  • #6
alijan kk said:
i got

m1v^2=m2v^2 right?

By squaring the equation you should get [itex] m^2_1v^2_1 = m^2_2v^2_2 [/itex]
 
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  • #7
Mastermind01 said:
By squaring the equation you should get [itex] m^2_1v^2_1 = m^2_2v^2_2 [/itex]
how can i get :-'
 
  • #8
give me the equation that i should try to solve
 
  • #9
Well using the equation [itex] m^2_1v^2_1 = m^2_2v^2_2 [/itex] you should get [itex] \frac{m_1v^2_1}{m_2v^2_2} = \frac{m_2}{m_1} [/itex]
 
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  • #10
alijan kk said:
give me the equation that i should try to solve
In your original attempt your very first equation was wrong. You wrote E1=E2. As Mastermind explained, that is not what you are given. You are given p1=p2, where pi=mivi. I see no evidence that you have understood that.
Start again with the given information and show all your steps. Do not write "I got ...".
 

1. What is the equation for work-energy ratio?

The equation for work-energy ratio is W = Fd, where W represents work, F represents force, and d represents distance.

2. How is work-energy ratio related to mechanical advantage?

Work-energy ratio is directly related to mechanical advantage, as it is a measure of how efficiently a machine can do work. A higher work-energy ratio indicates a higher mechanical advantage, meaning the machine can do more work with less effort.

3. Can the work-energy ratio be greater than 1?

Yes, the work-energy ratio can be greater than 1. This means that the machine is able to do more work than the amount of energy put into it.

4. How does mass affect the work-energy ratio?

The work-energy ratio is affected by the mass of the object being moved. Heavier objects require more work to be done, resulting in a lower work-energy ratio.

5. What units are used to measure work-energy ratio?

The work-energy ratio is measured in the unit of Joules (J). This unit is a combination of units for force (Newton) and distance (meter), as seen in the equation W = Fd.

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