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Work Energy and Power question

2004 · Q145
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Work Energy and Power question

2004 · Q145

NEETPhysicsWork Energy and PowerMCQ+4 / −1
A particle of mass m1 is moving with a velocity v1 and another particle of mass m2 is moving with a velocity v2. Both of them have the same momentum but their different kinetic energies are E1 and E2 respectively. If m1 > m2 then :
  1. A
    E1 < E2
  2. B
    E1E2=m1m2{{{E_1}} \over {{E_2}}} = {{{m_1}} \over {{m_2}}}E2​E1​​=m2​m1​​
  3. C
    E1 > E2
  4. D
    E1 = E2
View written solutionFree

Correct answer: A

Kinetic energy = p22m{{{p^2}} \over {2m}}2mp2​

E1E2=p12/2m1p22/2m2⇒E1E2=m2m1{{{E_1}} \over {{E_2}}} = {{p_1^2/2{m_1}} \over {p_2^2/2{m_2}}} \Rightarrow {{{E_1}} \over {{E_2}}} = {{{m_2}} \over {{m_1}}}E2​E1​​=p22​/2m2​p12​/2m1​​⇒E2​E1​​=m1​m2​​ as m1 > m2

∴\therefore∴ E1 < E2

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