Question

A body of mass $M$ is dropped from a height $h$ on a sand floor. If the body penetrates $x\text{m}$ into the sand, the average resistance offered by the sand to the body is

• A. $Mg\left(\frac{h}{x}\right)$
• B. $Mg(1+\frac{h}{x})$
• C. $Mgh+Mgx$
• D. $Mg(1-\frac{h}{x})$

Answer 1

The correct option is B $Mg(1+\frac{h}{x})$

If the body strikes the sand floor with a velocity $v$, then
$Mgh=\frac{1}{2}M{v}^2$
Let $F$ be the resisting force acting on the body. Then, the resultant force = $F-mg$
Using work-energy theorem
$(F-Mg)x=\frac{1}{2}M{v}^2$
$(F-Mg)x=Mgh\Rightarrow F=Mg(1+\frac{h}{x})$
$\therefore \left(b\right)$