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(A) Since the bomb is initially at rest,the initial momentum of the system is zero.According to the law of conservation of linear momentum,the total f

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$(-3\hat{i}+1\hat{j})$

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(A) Since the bomb is initially at rest,the initial momentum of the system is zero.According to the law of conservation of linear momentum,the total final momentum must also be zero.Let the mass of each fragment be $m$. The velocities of the three fragments are $\vec{v}_1 = (1\hat{i}+3\hat{j}) \ m/s$,$\vec{v}_2 = (2\hat{i}-4\hat{j}) \ m/s$,and $\vec{v}_3$ (unknown).The conservation of momentum equation is: $m\vec{v}_1 + m\vec{v}_2 + m\vec{v}_3 = 0$.Dividing by $m$,we get: $\vec{v}_1 + \vec{v}_2 + \vec{v}_3 = 0$.Substituting the given values: $(1\hat{i}+3\hat{j}) + (2\hat{i}-4\hat{j}) + \vec{v}_3 = 0$.$(3\hat{i}-1\hat{j}) + \vec{v}_3 = 0$.Therefore,$\vec{v}_3 = -(3\hat{i}-1\hat{j}) = (-3\hat{i}+1\hat{j}) \ m/s$.

$A$ bomb initially at rest explodes by itself into three equal mass fragments. The velocities of two fragments are $(1\hat{i}+3\hat{j}) \ m/s$ and $(2\hat{i}-4\hat{j}) \ m/s$. The velocity of the third fragment (in $m/s$) is

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