A body of mass 1,kg falls freely from a heigh | vedclass

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(D) $1$. Velocity of the $1\,kg$ body just before hitting the platform: $v = \sqrt{2gh} = \sqrt{2 	imes 10 	imes 100} = \sqrt{2000} = 20\sqrt{5}\,m/

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$2$

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(D) $1$. Velocity of the $1\,kg$ body just before hitting the platform: $v = \sqrt{2gh} = \sqrt{2 	imes 10 	imes 100} = \sqrt{2000} = 20\sqrt{5}\,m/s.$$2$. Using the Conservation of Linear Momentum $(COLM)$ during the perfectly inelastic collision: $m_1 v = (m_1 + m_2) v',$where $v'$ is the velocity of the combined mass $(1+3 = 4\,kg)$ immediately after the collision.$1 	imes 20\sqrt{5} = 4 	imes v' \implies v' = 5\sqrt{5}\,m/s.$$3$. Using the Conservation of Mechanical Energy for the system from the point of collision to the point of maximum compression $x$:The initial kinetic energy of the combined mass is converted into the potential energy of the spring and the gravitational potential energy change.Let the equilibrium position of the spring be the reference level. The platform is already compressed by $x_0 = \frac{m_2 g}{k} = \frac{3 	imes 10}{1.25 	imes 10^6} = 2.4 	imes 10^{-5}\,m,$ which is negligible.Applying energy conservation: $\frac{1}{2} M (v')^2 + M g x = \frac{1}{2} k x^2,$where $M = 4\,kg.$$\frac{1}{2} 	imes 4 	imes (5\sqrt{5})^2 + 4 	imes 10 	imes x = \frac{1}{2} 	imes 1.25 	imes 10^6 	imes x^2.$$2 	imes 125 + 40x = 6.25 	imes 10^5 x^2.$$6.25 	imes 10^5 x^2 - 40x - 250 = 0.$Since $x$ is very small,$40x$ is negligible compared to $250$ and $6.25 	imes 10^5 x^2.$$6.25 	imes 10^5 x^2 \approx 250 \implies x^2 \approx \frac{250}{6.25 	imes 10^5} = 40 	imes 10^{-5} = 4 	imes 10^{-4}.$$x = 0.02\,m = 2\,cm.$

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