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(A) Given: Mass of the body,$m = 0.15 \ kg$.Velocity of the body,$v = 15 \ ms^{-1}$.Spring constant,$k = 1500 \ Nm^{-1}$.According to the law of conse

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$0.15$

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(A) Given: Mass of the body,$m = 0.15 \ kg$.Velocity of the body,$v = 15 \ ms^{-1}$.Spring constant,$k = 1500 \ Nm^{-1}$.According to the law of conservation of energy,the initial kinetic energy of the body is converted into the elastic potential energy of the spring at maximum compression $x$.$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$Substituting the values:$\frac{1}{2} 	imes 0.15 	imes (15)^2 = \frac{1}{2} 	imes 1500 	imes x^2$$0.15 	imes 225 = 1500 	imes x^2$$33.75 = 1500 	imes x^2$$x^2 = \frac{33.75}{1500} = 0.0225$$x = \sqrt{0.0225} = 0.15 \ m$.

$A$ body of mass $0.15 \ kg$ moving with a velocity of $15 \ ms^{-1}$ comes to rest when it hits a spring that is fixed at the other end. If the force constant of the spring is $1500 \ Nm^{-1}$,then the compression in the spring is: (in $m$)

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