A body of mass 5 ,kg explodes at rest into th | vedclass

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(D) Given total mass $M = 5 \,kg$. The masses of the fragments are in the ratio $1:1:3$,so the masses are $m_1 = 1 \,kg$,$m_2 = 1 \,kg$,and $m_3 = 3 \

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$7\sqrt{2} \,m/s$

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(D) Given total mass $M = 5 \,kg$. The masses of the fragments are in the ratio $1:1:3$,so the masses are $m_1 = 1 \,kg$,$m_2 = 1 \,kg$,and $m_3 = 3 \,kg$.The two fragments of mass $1 \,kg$ move in mutually perpendicular directions with speed $v = 21 \,m/s$.The momentum of the first fragment is $P_x = m_1 v = 1 	imes 21 = 21 \,kg \cdot m/s$ along the $x$-axis.The momentum of the second fragment is $P_y = m_2 v = 1 	imes 21 = 21 \,kg \cdot m/s$ along the $y$-axis.The resultant momentum of these two fragments is $P_{res} = \sqrt{P_x^2 + P_y^2} = \sqrt{21^2 + 21^2} = 21\sqrt{2} \,kg \cdot m/s$.According to the law of conservation of linear momentum,the initial momentum of the body at rest is zero. Therefore,the momentum of the third (heaviest) fragment must be equal and opposite to the resultant momentum of the first two fragments.$P_3 = P_{res} = 21\sqrt{2} \,kg \cdot m/s$.Since $P_3 = m_3 v_3$,we have $3 	imes v_3 = 21\sqrt{2}$.$v_3 = \frac{21\sqrt{2}}{3} = 7\sqrt{2} \,m/s$.

$A$ body of mass $5 \,kg$ explodes at rest into three fragments with masses in the ratio $1 : 1 : 3$. The fragments with equal masses fly in mutually perpendicular directions with speeds of $21 \,m/s$. The velocity of the heaviest fragment will be

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