A body of mass 5 kg initially at rest explod | vedclass

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(D) The total mass is $5 \ kg$. The masses of the fragments are $m_1 = 1 \ kg$,$m_2 = 1 \ kg$,and $m_3 = 3 \ kg$.Since the body is initially at rest,t

Vedclass · Accepted Answer

$9.89$

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(D) The total mass is $5 \ kg$. The masses of the fragments are $m_1 = 1 \ kg$,$m_2 = 1 \ kg$,and $m_3 = 3 \ kg$.Since the body is initially at rest,the initial momentum is zero. By the law of conservation of linear momentum,the final vector sum of the momenta must be zero: $\vec{P}_1 + \vec{P}_2 + \vec{P}_3 = 0$.Let the two $1 \ kg$ fragments move along the $X$ and $Y$ axes respectively: $\vec{P}_1 = (1 \ kg)(21 \ m/s) \hat{i} = 21 \hat{i} \ kg \cdot m/s$ and $\vec{P}_2 = (1 \ kg)(21 \ m/s) \hat{j} = 21 \hat{j} \ kg \cdot m/s$.The momentum of the third fragment must be $\vec{P}_3 = -(\vec{P}_1 + \vec{P}_2) = -(21 \hat{i} + 21 \hat{j}) \ kg \cdot m/s$.The magnitude of the momentum of the third fragment is $|\vec{P}_3| = \sqrt{21^2 + 21^2} = 21\sqrt{2} \ kg \cdot m/s$.Since $|\vec{P}_3| = m_3 \cdot v_3$,we have $3 \cdot v_3 = 21\sqrt{2}$.$v_3 = 7\sqrt{2} \approx 7 	imes 1.414 = 9.898 \ m/s$.

$A$ body of mass $5 \ kg$ initially at rest explodes into three fragments. The ratio of the masses of the three fragments is $1 : 1 : 3$. The two fragments of equal mass move in mutually perpendicular directions with a velocity of $21 \ m/s$. What is the velocity of the heaviest fragment in $m/s$?

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