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Question

(A) Let the total mass be $M = 7k$. The masses of the three parts are $m_1 = k$,$m_2 = 3k$,and $m_3 = 3k$.Since the body is initially stationary,the t

Vedclass · Accepted Answer

$45\sqrt{2}\,m/s$

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(A) Let the total mass be $M = 7k$. The masses of the three parts are $m_1 = k$,$m_2 = 3k$,and $m_3 = 3k$.Since the body is initially stationary,the total initial momentum is $0$. By the law of conservation of linear momentum,the vector sum of the momenta of the three parts must be zero: $\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0$.Given that the two parts of mass $3k$ move at right angles with velocity $v_2 = v_3 = 15\,m/s$,their combined momentum is $\vec{p}_2 + \vec{p}_3$. The magnitude of this resultant momentum is $P_{23} = \sqrt{(3k 	imes 15)^2 + (3k 	imes 15)^2} = 3k 	imes 15 	imes \sqrt{2} = 45k\sqrt{2}$.For the system to be in equilibrium,the momentum of the first part $(m_1 = k)$ must be equal and opposite to this resultant: $p_1 = P_{23}$.$k 	imes v_1 = 45k\sqrt{2}$.Therefore,$v_1 = 45\sqrt{2}\,m/s$.

$A$ stationary body of mass $m$ explodes into $3$ parts having masses in the ratio $1 : 3 : 3$. The two parts having equal mass move at right angles to each other with a velocity of $15\,m/s$. What is the velocity of the third part?

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