A body of mass $M$ is dropped from a height $h$ on a sand floor. If the body penetrates $x \mathrm{~cm}$…
A body of mass $M$ is dropped from a height $h$ on a sand floor. If the body penetrates $x \mathrm{~cm}$ into the sand, then the average resistance offered by the sand to the body is
$M g\left(\frac{h}{x}\right)$
$M g\left(\frac{x+h}{x}\right)$
$M g(h+x)$
$M g\left(\frac{x-h}{x}\right)$
Solution
Given, mass of body $=M$
Height above the sand floor $=h$
Distance penetrated into sand $=x$
Let $v$ be the velocity with which body strikes the surface.
Then, by equation of motion, we get
$v^2=u^2+2 g h \Rightarrow v^2=2 g h$ ...(i)
$[\because$ Initial velocity at highest point, $u=0]$
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 body.
Then, net downward force will be
$F_{\text {net }}=M g-F$
By work-energy theorem, work done by all forces is equal to change in $\mathrm{KE}$, which gives
$\therefore \quad W=\Delta K$
$F_{\text {net }} \times x=K_f-K_i$
$(M g-F) x=0-\frac{1}{2} M v^2 \quad\left[\because\right.$ Final KE, $\left.K_f=0\right]$
$\Rightarrow \quad(M g-F) x=-\frac{1}{2} M v^2$ ...(ii)
Substituting the value from Eq. (i) in Eq. (ii), we get
$\Rightarrow \quad M g x-F x=-\frac{1}{2} M(2 g h)$
$M g x-F x=-M g h$
$\Rightarrow \quad F x=M g x+M g h \Rightarrow F=M g\left(\frac{x+h}{x}\right)$