A body of mass 14 kg initially at rest explo | vedclass

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(B) The total mass of the body is $M = 14 \ kg$. The masses of the three fragments are in the ratio $2:2:3$. Let the masses be $m_1 = 4 \ kg$, $m_2 =

Vedclass · Accepted Answer

$12\sqrt{2}$

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(B) The total mass of the body is $M = 14 \ kg$. The masses of the three fragments are in the ratio $2:2:3$. Let the masses be $m_1 = 4 \ kg$, $m_2 = 4 \ kg$, and $m_3 = 6 \ kg$ (since $2x + 2x + 3x = 7x = 14 \ kg$, so $x = 2 \ kg$).Initially, the body is at rest, so the initial momentum is zero. By the law of conservation of linear momentum, the final total momentum must also be zero: $\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0$.Let the two equal fragments move along the negative $x$-axis and negative $y$-axis respectively: $\vec{v}_1 = -18 \hat{i} \ m/s$ and $\vec{v}_2 = -18 \hat{j} \ m/s$.The momentum of the first two fragments is $\vec{p}_1 + \vec{p}_2 = m_1 \vec{v}_1 + m_2 \vec{v}_2 = 4(-18 \hat{i}) + 4(-18 \hat{j}) = -72 \hat{i} - 72 \hat{j} \ kg \cdot m/s$.Since $\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0$, we have $\vec{p}_3 = -(\vec{p}_1 + \vec{p}_2) = 72 \hat{i} + 72 \hat{j} \ kg \cdot m/s$.The velocity of the third (heavier) fragment is $\vec{v}_3 = \frac{\vec{p}_3}{m_3} = \frac{72 \hat{i} + 72 \hat{j}}{6} = 12 \hat{i} + 12 \hat{j} \ m/s$.The magnitude of the velocity is $|\vec{v}_3| = \sqrt{12^2 + 12^2} = \sqrt{144 + 144} = \sqrt{288} = 12\sqrt{2} \ m/s$.

$A$ body of mass $14 \ kg$ initially at rest explodes and breaks into three fragments of masses in the ratio $2:2:3$. The two pieces of equal masses fly off perpendicular to each other with a speed of $18 \ m/s$ each. The velocity of the heavier fragment is . . . . . . $m/s$.

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