Question

Two solids A and B of mass 1 kg and 2 kg respectively are moving with equal linear momentum. The ratio of their kinetic energies (K.E.)A : (K.E.)B will be A/1, so the value of A will be ______

Hint:

For equal momentum, the lighter body always has higher kinetic energy. For equal kinetic energy, the heavier body has higher momentum.

Answer 1

Correct Answer: 2

Step 1: The relationship between Kinetic Energy ($K$) and Momentum ($p$) is given by $K = \frac{p^2}{2m}$.
Step 2: Since momentum $p$ is equal for both, $K \propto \frac{1}{m}$.
Step 3: Therefore, $\frac{K_A}{K_B} = \frac{m_B}{m_A} = \frac{2}{1}$. Comparing this to $A/1$, we find $A = 2$.