A bomb at rest explodes into three equal frag | vedclass

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(C) Since the bomb is initially at rest,the initial momentum is zero.According to the law of conservation of linear momentum,the final momentum must a

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$15$

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(C) Since the bomb is initially at rest,the initial momentum is zero.According to the law of conservation of linear momentum,the final momentum must also be zero.Let the mass of each fragment be $m$. The total mass of the bomb is $3m$.Let the velocities of the three fragments be $\vec{v}_1 = 9\hat{i} \ m s^{-1}$,$\vec{v}_2 = 12\hat{j} \ m s^{-1}$,and $\vec{v}_3$.Applying the conservation of momentum: $m\vec{v}_1 + m\vec{v}_2 + m\vec{v}_3 = 0$.Dividing by $m$,we get $\vec{v}_1 + \vec{v}_2 + \vec{v}_3 = 0$.$\vec{v}_3 = -(\vec{v}_1 + \vec{v}_2) = -(9\hat{i} + 12\hat{j})$.The magnitude of the velocity of the third fragment is $|\vec{v}_3| = \sqrt{(-9)^2 + (-12)^2}$.$|\vec{v}_3| = \sqrt{81 + 144} = \sqrt{225} = 15 \ m s^{-1}$.

$A$ bomb at rest explodes into three equal fragments. Two of the fragments move at right angles to each other with velocities of $9 \ m s^{-1}$ and $12 \ m s^{-1}$. The magnitude of the velocity of the third fragment is ....... $m s^{-1}$.

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