Question

The kinetic energy of a particle of mass $mkg$ is half of that of another particle of mass $m/2kg$ . If the speed of heavier particle is increased by $3m{s}^{-1}$ , 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. $3m{s}^{-1},6m{s}^{-1}$
• B. $2m{s}^{-1},4m{s}^{-1}$
• C. $2m{s}^{-1},6m{s}^{-1}$
• D. $4m{s}^{-1},8m{s}^{-1}$

Answer 1

Answer: (A)

According to the question,
$K_m=\frac{1}{2}{K}_{m/2}$
$\Rightarrow \frac{1}{2}m{v}_1^2=\frac{1}{2}\times \frac{1}{2}(\frac{m}{2})v_2^2$
$\Rightarrow \frac{v_1^2}{v_2^2}=\frac{1}{4}$ initially
Speed of heavier particle is increased by $3m/s$
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
$v_2=3m/s$
$v_1=6m/s$