# Two bodies of masses m1 and m2 have kinetic energies in the ratio of 4 ∶ 1. The ratio of their linear momenta will be-

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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 : $\sqrt[2]{{{m_1}}} \sim \sqrt {{m_2}}$

Concept:

Momentum

• 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)

Calculation:

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)

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

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

$\implies \ \frac{p_1}{p_2} = \frac{m_1}{2 m_2}$