A body of mass 1 kg is suspended from a spri | vedclass

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(A) $1$. Initial state: The mass $M = 1 \ kg$ is in equilibrium. The extension in the spring is $x_0 = \frac{Mg}{k} = \frac{1 	imes 10}{600} = \frac{

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$5$

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(A) $1$. Initial state: The mass $M = 1 \ kg$ is in equilibrium. The extension in the spring is $x_0 = \frac{Mg}{k} = \frac{1 	imes 10}{600} = \frac{1}{60} \ m$.$2$. Collision: The body of mass $m = 0.5 \ kg$ hits the mass $M$ with velocity $v = 3 \ m \ s^{-1}$. By conservation of linear momentum,$(M+m)V = mv$,where $V$ is the common velocity after collision.$3$. $V = \frac{mv}{M+m} = \frac{0.5 	imes 3}{1 + 0.5} = \frac{1.5}{1.5} = 1 \ m \ s^{-1}$.$4$. New equilibrium position: After the collision,the total mass is $M' = 1.5 \ kg$. The new equilibrium extension is $x_0' = \frac{M'g}{k} = \frac{1.5 	imes 10}{600} = \frac{15}{600} = 0.025 \ m = 2.5 \ cm$.$5$. Displacement from new equilibrium: The collision occurs at the old equilibrium position $x_0 = \frac{1}{60} \ m \approx 1.67 \ cm$. The displacement from the new equilibrium is $x = x_0' - x_0 = 2.5 \ cm - 1.67 \ cm = 0.833 \ cm$.$6$. Amplitude calculation: Using energy conservation,$\frac{1}{2}k A^2 = \frac{1}{2}(M+m)V^2 + \frac{1}{2}k x^2$.$7$. $600 A^2 = 1.5(1)^2 + 600(0.00833)^2 \implies 600 A^2 = 1.5 + 0.0416 = 1.5416$.$8$. $A^2 = \frac{1.5416}{600} \approx 0.002569 \implies A \approx 0.0506 \ m \approx 5 \ cm$.

$A$ body of mass $1 \ kg$ is suspended from a spring of force constant $600 \ N \ m^{-1}$. Another body of mass $0.5 \ kg$ moving vertically upwards hits the suspended body with a velocity of $3 \ m \ s^{-1}$ and gets embedded in it. The amplitude of motion is (in $cm$)

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