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)

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

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

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