A body of mass M is dropped from a height h on a sand floor. If body penet

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5.Work, Energy, Power and Collision

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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 of the body is

A

$Mg\left( {\frac{h}{x}} \right)$

B

$Mg\left( {1\,+\,\frac{h}{x}} \right)$

C

$Mgh\,+\,Mgx$

D

$Mg\left( {1\,-\,\frac{h}{x}} \right)$

Solution

If the body strikes the sand floor with a velocity $v,$ then $m g h=\frac{1}{2}$ $m v^{2} .$

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. Then the resultant force $=$ $\mathrm{F}-\mathrm{Mg}$ using work $energy-theorem$ $:$

$(F-M g) x=\frac{1}{2} M v^{2}-0$

or $\quad(F-M g) x=M g h$

or $\mathrm{F} x=\mathrm{Mgh}+\mathrm{Mgx}$

$F=M g\left(1+\frac{h}{x}\right)$

Standard 11

Physics

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