Question

The kinetic energy of a particle of mass $ m \,kg $ is half of that of another particle of mass $ m/2\, kg $ . If the speed of heavier particle is increased by $ 3 \,ms^{-1} $ , its kinetic energy becomes equal to the original kinetic energy of the lighter particle. The original speeds of the heavier and lighter particles are

• A. $ 3 \,ms^{-1}, 6\, ms^{-1} $
• B. $ 2 \,ms^{-1}, 4\, ms^{-1} $
• C. $ 2 \,ms^{-1}, 6\, ms^{-1} $
• D. $ 4 \,ms^{-1}, 8\, ms^{-1} $

Answer 1

Answer: (A)

According to the question,
$ K_m = \frac{1}{2} K_{m/2}$
$\Rightarrow \frac{1}{2} mv_1^2 = \frac{1}{2} \times \frac{1}{2} (\frac{m}{2}) v_2^2$
$\Rightarrow \frac{v_1^2}{v_2^2} = \frac{1}{4}$ initially
Speed of heavier particle is increased by $3\,m/s$
Then, $ \frac{1}{2} m(v_1 + 3)^2 = \frac{1}{2}(\frac{m}{2}) v_2^2$
$\Rightarrow \frac{1}{2} m(v_1 + 3)^2 = \frac{1}{4} m \cdot 4v_1^2$
$(v_1 + 3)^2 = 2v_1^2$
Solving these equations, we get
$v_2 = 3\,m/s$
$v_1 = 6\,m/s$