Question

A body of mass 0.2 kg dropped from a height '6 m'. If $e=\frac{1}{\sqrt{6}}$ then K.E. lost during its first bounce from the ground is

• A. $1.96J$
• B. $9.8J$
• C. $19.6J$
• D. $zero$

Answer 1

Answer: (B)

$9.8J$

Thus P.E will convert to K.E:

$mgh=\frac{1}{2}m{v}^2$

$v=\sqrt{2\times 9.8\times 6}$

$v=2\times 7\sqrt{0.6}$

After collision velocity $=-ev$
$=2\times 7\times \sqrt{0.6}\times \frac{1}{\sqrt{6}}$

$=2\times 7\sqrt{\frac{1}{10}}$

K.E after first bounce $=\frac{1}{2}m{v}^2$

$=\frac{1}{2}\times 0.2\times \frac{196}{10}$

$=1.96J$

Energy lost $=11.76-1.96$$=9.8J$