A body of mass 1 ,kg, initially at rest, expl | vedclass

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(A) The total mass is $M = 1 \,kg$. The ratio of masses is $1:1:3$, so the masses are $m_1 = \frac{1}{5} \,kg$, $m_2 = \frac{1}{5} \,kg$, and $m_3 = \

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$10 \sqrt{2}$

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(A) The total mass is $M = 1 \,kg$. The ratio of masses is $1:1:3$, so the masses are $m_1 = \frac{1}{5} \,kg$, $m_2 = \frac{1}{5} \,kg$, and $m_3 = \frac{3}{5} \,kg$.Since the body is initially at rest, the initial momentum is $0$.By the law of conservation of momentum, the final momentum must also be $0$.Let the two equal masses move along the $x$ and $y$ axes. Their momentum vectors are $\vec{p}_1 = m_1 v_1 \hat{i} = (\frac{1}{5} 	imes 30) \hat{i} = 6 \hat{i} \,kg \cdot m/s$ and $\vec{p}_2 = m_2 v_2 \hat{j} = (\frac{1}{5} 	imes 30) \hat{j} = 6 \hat{j} \,kg \cdot m/s$.The resultant momentum of these two parts is $\vec{p}_{12} = \vec{p}_1 + \vec{p}_2 = 6 \hat{i} + 6 \hat{j}$.The magnitude is $|\vec{p}_{12}| = \sqrt{6^2 + 6^2} = 6 \sqrt{2} \,kg \cdot m/s$.For the total momentum to be zero, the momentum of the third part $\vec{p}_3$ must satisfy $\vec{p}_{12} + \vec{p}_3 = 0$, so $\vec{p}_3 = -\vec{p}_{12}$.The magnitude is $|\vec{p}_3| = m_3 v_3 = \frac{3}{5} v_3 = 6 \sqrt{2}$.Solving for $v_3$: $v_3 = \frac{6 \sqrt{2} 	imes 5}{3} = 10 \sqrt{2} \,m/s$.

$A$ body of mass $1 \,kg$, initially at rest, explodes and breaks into three parts. The masses of the parts are in the ratio $1: 1: 3$. The two pieces of equal mass fly off perpendicular to each other with a speed of $30 \,m/s$ each. The velocity of the heavier part in $m/s$ is

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