A mass of 0.5, kg moving with a speed of 1.5, | vedclass

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(D) According to the law of conservation of energy,the initial kinetic energy of the mass is completely converted into the elastic potential energy of

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$0.15$

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(D) According to the law of conservation of energy,the initial kinetic energy of the mass is completely converted into the elastic potential energy of the spring at the point of maximum compression.The kinetic energy of the mass is $K = \frac{1}{2} m v^2$.The elastic potential energy stored in the spring is $U = \frac{1}{2} k x^2$,where $x$ is the maximum compression.Equating the two energies:$\frac{1}{2} m v^2 = \frac{1}{2} k x^2$Solving for $x$:$x = \sqrt{\frac{m v^2}{k}}$Substituting the given values ($m = 0.5\, kg$,$v = 1.5\, m/s$,$k = 50\, N/m$):$x = \sqrt{\frac{0.5 	imes (1.5)^2}{50}}$$x = \sqrt{\frac{0.5 	imes 2.25}{50}}$$x = \sqrt{\frac{1.125}{50}}$$x = \sqrt{0.0225}$$x = 0.15\, m$

$A$ mass of $0.5\, kg$ moving with a speed of $1.5\, m/s$ on a horizontal smooth surface,collides with a nearly weightless spring of force constant $k=50\, N/m$. The maximum compression of the spring would be ................. $m$.

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