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(B) Given,mass of body $= M$.Height above the sand floor $= h$.Distance penetrated into sand $= x$.Let $v$ be the velocity with which the body strikes

Vedclass · Accepted Answer

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

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(B) Given,mass of body $= M$.Height above the sand floor $= h$.Distance penetrated into sand $= x$.Let $v$ be the velocity with which the body strikes the surface.By the equation of motion,$v^2 = u^2 + 2gh$. Since initial velocity $u = 0$,we get $v^2 = 2gh$ ... $(i)$.When the body passes through the sand,it comes to rest after traveling a distance $x$. Let $F$ be the average resisting force.The work done by the net force is equal to the change in kinetic energy (Work-Energy Theorem).The forces acting on the body inside the sand are gravity ($Mg$ downwards) and resistance ($F$ upwards).Net force $= Mg - F$.Work done $= (Mg - F)x$.Change in kinetic energy $= K_f - K_i = 0 - \frac{1}{2}Mv^2$.Equating them: $(Mg - F)x = -\frac{1}{2}Mv^2$.Substituting $v^2 = 2gh$: $(Mg - F)x = -\frac{1}{2}M(2gh) = -Mgh$.$Mgx - Fx = -Mgh$.$Fx = Mgx + Mgh$.$F = Mg\left(\frac{x+h}{x}\right)$.

$A$ body of mass $M$ is dropped from a height $h$ on a sand floor. If the body penetrates $x$ cm into the sand,then the average resistance offered by the sand to the body is

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