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

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(C) Initial momentum of the body is $0$ because it is at rest.Let the masses of the three pieces be $m_1 = M/4$,$m_2 = M/4$,and $m_3 = M - (M/4 + M/4)

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(C) Initial momentum of the body is $0$ because it is at rest.Let the masses of the three pieces be $m_1 = M/4$,$m_2 = M/4$,and $m_3 = M - (M/4 + M/4) = M/2$.The momenta of the first two pieces are $p_1 = (M/4) 	imes 3$ and $p_2 = (M/4) 	imes 4$ in perpendicular directions.The resultant momentum of these two pieces is $p_{12} = \sqrt{p_1^2 + p_2^2} = \sqrt{(\frac{3M}{4})^2 + (\frac{4M}{4})^2} = \sqrt{\frac{9M^2}{16} + \frac{16M^2}{16}} = \sqrt{\frac{25M^2}{16}} = \frac{5M}{4}$.According to the law of conservation of linear momentum,the total momentum must remain zero. Therefore,the third piece must have a momentum $p_3$ equal and opposite to $p_{12}$.$p_3 = \frac{5M}{4} = m_3 	imes v = \frac{M}{2} 	imes v$.Solving for $v$: $v = \frac{5M/4}{M/2} = \frac{5}{4} 	imes 2 = 2.5 \, m/s$.

$A$ body of mass $M$ at rest explodes into three pieces,two of which of mass $M/4$ each are thrown off in perpendicular directions with velocities of $3 \, m/s$ and $4 \, m/s$ respectively. The third piece will be thrown off with a velocity of .......... $m/s$

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