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(D) The total mass of the bomb is $M = 1 \,kg$. The masses of the three fragments are in the ratio $1: 1: 3$. Thus,the masses are $m_1 = 0.2 \,kg$,$m_

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$5 \sqrt{2}$

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(D) The total mass of the bomb is $M = 1 \,kg$. The masses of the three fragments are in the ratio $1: 1: 3$. Thus,the masses are $m_1 = 0.2 \,kg$,$m_2 = 0.2 \,kg$,and $m_3 = 0.6 \,kg$.Initially,the bomb is at rest,so the initial momentum is $P_i = 0$. By the law of conservation of linear momentum,the final momentum must also be zero: $\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0$.The two fragments of mass $0.2 \,kg$ move perpendicular to each other with speed $v_1 = v_2 = 15 \,m / s$. Their combined momentum is $p_{12} = \sqrt{(m_1 v_1)^2 + (m_2 v_2)^2} = \sqrt{(0.2 	imes 15)^2 + (0.2 	imes 15)^2} = \sqrt{3^2 + 3^2} = 3 \sqrt{2} \,kg \cdot m / s$.For the total momentum to be zero,the momentum of the third fragment must be equal and opposite to $p_{12}$. Thus,$m_3 v_3 = 3 \sqrt{2}$.Substituting $m_3 = 0.6 \,kg$,we get $0.6 	imes v_3 = 3 \sqrt{2}$.$v_3 = \frac{3 \sqrt{2}}{0.6} = 5 \sqrt{2} \,m / s$.

$A$ bomb of mass $1 \,kg$ initially at rest,explodes and breaks into three fragments of masses in the ratio $1: 1: 3$. The two pieces of equal mass fly off perpendicular to each other with a speed $15 \,m / s$ each. The speed of the heavier fragment is ........... $m / s$.

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