A 1 ; kg stationary bomb explodes into three | vedclass

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(D) Let the total mass be $M = 1 \; kg$. The masses of the three parts are $m_1 = 0.2 \; kg$,$m_2 = 0.2 \; kg$,and $m_3 = 0.6 \; kg$ (ratio $1:1:3$).S

Vedclass · Accepted Answer

$10\sqrt{2} \; m/s$

Vedclass · Answer

(D) Let the total mass be $M = 1 \; kg$. The masses of the three parts are $m_1 = 0.2 \; kg$,$m_2 = 0.2 \; kg$,and $m_3 = 0.6 \; kg$ (ratio $1:1:3$).Since the bomb is initially stationary,the initial momentum is $0$.By the law of conservation of linear momentum,the final momentum must also be $0$.Let the velocities of the two smaller parts be $\vec{v}_1 = 30\hat{i} \; m/s$ and $\vec{v}_2 = 30\hat{j} \; m/s$.The momentum of these two parts is $\vec{p}_{12} = m_1\vec{v}_1 + m_2\vec{v}_2 = 0.2(30\hat{i} + 30\hat{j}) = 6\hat{i} + 6\hat{j} \; kg \cdot m/s$.The magnitude of this momentum is $p_{12} = \sqrt{6^2 + 6^2} = 6\sqrt{2} \; kg \cdot m/s$.For the total momentum to be zero,the third part must have momentum $\vec{p}_3 = -\vec{p}_{12}$.Thus,$m_3 v_3 = 6\sqrt{2}$.$0.6 \cdot v_3 = 6\sqrt{2}$.$v_3 = \frac{6\sqrt{2}}{0.6} = 10\sqrt{2} \; m/s$.

$A$ $1 \; kg$ stationary bomb explodes into three parts having mass ratio $1:1:3$. The parts with equal mass move in perpendicular directions with a velocity of $30 \; m/s$. What is the velocity of the third (heavier) part?

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