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Two bodies of masses m1 and m2 have kinetic energies in the ratio of 4 ∶ 1. The ratio of their linear momenta will be-
1. \(\sqrt {{m_1}} \sim \sqrt {{m_2}} \)
2. \(\sqrt[2]{{{m_1}}} \sim \sqrt {{m_2}} \)
3. \(\sqrt {{m_1}} \sim \sqrt[2]{{{m_2}}}\)
4. 4 m1 ∶ m2

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Correct Answer - Option 2 : \(\sqrt[2]{{{m_1}}} \sim \sqrt {{m_2}} \)




  • Momentum is defined as the product of the mass and velocity of the body. 
  • It is a vector quantity directed toward velocity. 
  • It is given as 

p = mv      ----(1)

m is mass, v is the velocity, p is momentum

Kinetic Energy

  • The energy of a body in motion due to its state of motion is called kinetic energy. 
  • It is given as 

\(K = \frac{1}{2}mv^2\)        ----(2)

K is kinetic energy, m is mass, v is speed.

Relationship between Momentum and Kinetic Energy

If we combine equation (1) and (2) we will get the relationship between momentum and kinetic energy as

\(K = \frac{p^2}{2m}\)     ----(3)


So, the Kinetic energy of bodies are in ratio 1 : 4

\(\frac{K_1}{K_2} = \frac{P_1}{P_2}\)

Kinetic Energy of first body is 

\(K _1= \frac{p_1^2}{2m_1}\) --- (5)

Kinetic Energy of the second body is

\(K _2= \frac{p_2^2}{2m_2}\) --- (6)

Dividing (5) by (6)

\(\frac{K_1}{K_2} = \frac{\frac{p_{1}^{2}}{m_1}}{\frac{p_{2}^{2}}{m_2}}\)

\(\implies \frac{K_1}{K_2} =( \frac{p_1}{p_2} )^2\ \frac{m_2}{m_1}\) -- (7)

Putting (4) in (7)

\(\implies \frac{1}{4} = (\frac{p_1}{p_2})^2 \ \frac{m_2}{m_1}\)

\(\implies \ \frac{m_1}{m_2} \frac{1}{4} = (\frac{p_1}{p_2})^2\)

\(\implies \ \frac{p_1}{p_2} = \frac{m_1}{2 m_2}\)

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