Question

Two bodies of masses m and 4m have kinetic energies in the ratio 1: 2. Their momenta p₁ and p₂ are in the ratio

• A. 1:2√2
• B. 1:2√3
• C. 2√2:1
• D. 3√2:1
• E. 1:3√2

Answer 1

The Correct Option is A

We know that the kinetic energy \( K \) of a body is given by the formula: \[ K = \frac{1}{2} mv^2 \] Where: - \( m \) is the mass, - \( v \) is the velocity of the body. For the two bodies, the kinetic energies are in the ratio 1 : 2, so: \[ \frac{K_1}{K_2} = \frac{1}{2} \quad \text{(given)} \] Substitute the kinetic energy formula: \[ \frac{\frac{1}{2} m v_1^2}{\frac{1}{2} (4m) v_2^2} = \frac{1}{2} \] Simplifying: \[ \frac{v_1^2}{4v_2^2} = \frac{1}{2} \quad \Rightarrow \quad v_1^2 = 2 v_2^2 \quad \Rightarrow \quad v_1 = \sqrt{2} v_2 \] The momentum \( p \) of a body is given by: \[ p = mv \] Thus, the ratio of the momenta \( p_1 \) and \( p_2 \) is: \[ \frac{p_1}{p_2} = \frac{m v_1}{4m v_2} = \frac{v_1}{4 v_2} = \frac{\sqrt{2} v_2}{4 v_2} = \frac{\sqrt{2}}{4} = 1 : 2\sqrt{2} \]

Thus, the momenta are in the ratio \( 1 : 2\sqrt{2} \).

Answer 2

Let the kinetic energies of the two bodies be: 
For mass m, kinetic energy = \( KE_1 = \frac{1}{2}mv_1^2 \) 
For mass 4m, kinetic energy = \( KE_2 = \frac{1}{2}(4m)v_2^2 \) 

Given: \[ \frac{KE_1}{KE_2} = \frac{1}{2} \Rightarrow \frac{\frac{1}{2}mv_1^2}{\frac{1}{2}(4m)v_2^2} = \frac{1}{2} \Rightarrow \frac{v_1^2}{4v_2^2} = \frac{1}{2} \Rightarrow \frac{v_1^2}{v_2^2} = 2 \times 4 = 8 \Rightarrow \frac{v_1}{v_2} = \sqrt{8} = 2\sqrt{2} \] 
Now, momenta: 
\[ p_1 = mv_1,\quad p_2 = 4mv_2 \] 
So, \[ \frac{p_1}{p_2} = \frac{mv_1}{4mv_2} = \frac{v_1}{4v_2} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2} \Rightarrow \frac{p_1}{p_2} = \frac{1}{2\sqrt{2}} \] 
Therefore, the ratio is: \[ p_1 : p_2 = 1 : 2\sqrt{2} \] 
Answer: 1:2√2