AP EAPCET 23-Aug-2021 Shift 2 Paper

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 $F_{\text{net}}=Mg-F$ By work-energy theorem, work done by all forces is equal to change in $\text{KE}$, which gives $W = \Delta K$ $F_{\text{net}} \times x = K_f - K_i$ $(Mg-F) x = 0 - \frac{1}{2} M v^{2}$ $[\because \text{ Final KE, } K_f = 0]$ $(Mg-F) x = -\frac{1}{2} M v^{2}$ ...(ii) Substituting the value from Eq. (i) in Eq. (ii), we get $\Rightarrow Mg x - F x = -\frac{1}{2} M (2 g h)$ $\Rightarrow Mg x - F x = -Mg h$ $\Rightarrow F x = Mg x + Mg h$ $\Rightarrow F = Mg \left( \frac{x+h}{x} \right)$