A body of mass M at rest explodes into three | vedclass

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(B) The masses of the three pieces are $m_1 = \frac{M}{4}$,$m_2 = \frac{M}{4}$,and $m_3 = \frac{2M}{4} = \frac{M}{2}$.Since the initial body is at res

Vedclass · Accepted Answer

$25$

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(B) The masses of the three pieces are $m_1 = \frac{M}{4}$,$m_2 = \frac{M}{4}$,and $m_3 = \frac{2M}{4} = \frac{M}{2}$.Since the initial body is at rest,the total initial momentum is zero. By the law of conservation of linear momentum,the final total momentum must also be zero: $\vec{P}_1 + \vec{P}_2 + \vec{P}_3 = 0$.This implies $\vec{P}_3 = -(\vec{P}_1 + \vec{P}_2)$.The magnitudes of the momenta of the two smaller pieces are $P_1 = m_1 v_1 = \frac{M}{4} 	imes 30 = 7.5M$ and $P_2 = m_2 v_2 = \frac{M}{4} 	imes 40 = 10M$.Since $\vec{P}_1$ and $\vec{P}_2$ are perpendicular,the magnitude of their resultant is $P_{12} = \sqrt{P_1^2 + P_2^2} = \sqrt{(7.5M)^2 + (10M)^2} = \sqrt{56.25M^2 + 100M^2} = \sqrt{156.25M^2} = 12.5M$.Since $\vec{P}_3 = -(\vec{P}_1 + \vec{P}_2)$,the magnitude $P_3 = P_{12} = 12.5M$.Given $P_3 = m_3 v_3 = \frac{M}{2} v_3$,we have $\frac{M}{2} v_3 = 12.5M$.Solving for $v_3$,we get $v_3 = 12.5 	imes 2 = 25 \, m/s$.

$A$ body of mass $M$ at rest explodes into three pieces,in the ratio of masses $1: 1: 2$. Two smaller pieces fly off perpendicular to each other with velocities of $30 \, m/s$ and $40 \, m/s$ respectively. The velocity of the third piece will be ............... $\, m/s$.

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