Two bodies A and B having masses in the ratio of 3 : 1 possess the same kinetic energy. The ratio of linear momentum of B to A is :

Let the masses of bodies A and B be represented by m_A and m_B respectively. Given that the masses are in the ratio of 3:1, we can write m_A/m_B = 3/1.

We know that kinetic energy (KE) is given by the formula KE = (1/2)mv^2, where m is the mass and v is the velocity.

Since the kinetic energy of both bodies is the same, we can write (1/2)m_Av_A^2 = (1/2)m_Bv_B^2.

Dividing both sides of the equation by (1/2)m_A, we get v_A^2 = (m_B/m_A)v_B^2.

Now, let's consider the ratio of linear momentum (p) of body B to body A. Linear momentum is given by the formula p = mv.

The ratio of linear momentum of B to A can be written as p_B/p_A = (m_Bv_B)/(m_Av_A).

Substituting the value of v_A^2 from the previous equation, we have p_B/p_A = (m_B/m_A)v_B^2/(m_Av_A).

Canceling out the common terms, we get p_B/p_A = (m_B/m_A)v_B/v_A.

Since we know that m_A/m_B = 3/1, we can substitute this value into the equation to get p_B/p_A = (1/3)v_B/v_A.

Therefore, the ratio of linear momentum of B to A is 1/3.