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(C) The potential energy stored in the stretched rubber cord is converted into the kinetic energy of the missile.The elastic potential energy stored i

Vedclass · Accepted Answer

$250$

Vedclass · Answer

(C) The potential energy stored in the stretched rubber cord is converted into the kinetic energy of the missile.The elastic potential energy stored in a stretched wire is given by $U = \frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain} 	imes 	ext{volume} = \frac{1}{2} \frac{YA l^2}{L}$.Equating this to the kinetic energy of the missile: $\frac{1}{2} m v^2 = \frac{1}{2} \frac{YA l^2}{L}$.Solving for velocity $v$: $v = \sqrt{\frac{YA l^2}{mL}}$.Given values: $Y = 5 	imes 10^8\,N/m^2$,$A = 25 	imes 10^{-6}\,m^2$,$l = 5 	imes 10^{-2}\,m$,$L = 10 	imes 10^{-2}\,m$,$m = 5 	imes 10^{-3}\,kg$.Substituting the values: $v = \sqrt{\frac{5 	imes 10^8 	imes 25 	imes 10^{-6} 	imes (5 	imes 10^{-2})^2}{5 	imes 10^{-3} 	imes 10 	imes 10^{-2}}} = \sqrt{\frac{5 	imes 10^8 	imes 25 	imes 10^{-6} 	imes 25 	imes 10^{-4}}{5 	imes 10^{-4}}} = \sqrt{62500} = 250\,m/s$.

$A$ rubber cord catapult has a cross-sectional area of $25\,mm^2$ and an initial length of $10\,cm$. It is stretched by $5\,cm$ and then released to project a missile of mass $5\,g$. Taking $Y_{rubber} = 5 \times 10^8\,N/m^2$,the velocity of the projected missile is ......... $m/s$.

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