Question

A body of mass $m$ has a kinetic energy equal to one-fourth the kinetic energy of another body of mass $\frac{m}{4}$. If the speed of the heavier body is increased by $4m/s$, its new kinetic energy equals the original kinetic energy of the lighter body. The original speed of the heavier body in $m/s$ is

• A. $8$
• B. $6$
• C. $4$
• D. $2$

Answer 1

The correct option is C $4$

$\frac{1}{2}m{u}^2=\frac{1}{4}\left[\frac{1}{2}\frac{m}{4}{V}^2\right]\Rightarrow 4u=V$

$\frac{1}{2}m(u+4{)}^2=\frac{1}{2}\frac{m}{4}{V}^2$

$(u+4{)}^2=\frac{V^2}{4}$

$u+4=\frac{V}{2}$

$2u+8=4u$

$u=4$