Q. A body of mass $M$ is dropped from a height $h$ on a sand floor. If the body penetrates $x$ cm into the sand, the average resistance offered by the sand to the body is
A) $Mg\left( \frac{h}{x} \right)$
B) $Mg\left( 1 + \frac{h}{x} \right)$
C) $Mgh + Mgx$
D) $Mg\left( 1 - \frac{h}{x} \right)$
Explanation
(b) If the body strikes the sand floor with a velocity $v$, then $Mgh = \frac{1}{2}mv^{2}$
With this velocity $v$, when body passes through the sand floor it comes to rest after travelling a distance $x.$ Let $F$ be the resisting force acting on the body. Net force in downwards direction
$= Mg - F$
Work done by all the forces is equal to change in KE
$(Mg - F)x = 0 - \frac{1}{2}Mv^{2}$
$(Mg - F)x = - Mgh$ or $Fx = Mgh + Mgx$
or $F = Mg\left( 1 + \frac{h}{x} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.