A body of mass $5 \,kg$ explodes at rest into three fragments with masses in the ratio $1 : 1 : 3$. The fragments with equal masses fly in mutually perpendicular directions with speeds of $21 \,m/s$. The velocity of the heaviest fragment will be
A body of mass $5 \,kg$ explodes at rest into three fragments with masses in the ratio $1 : 1 : 3$. The fragments with equal masses fly in mutually perpendicular directions with speeds of $21 \,m/s$. The velocity of the heaviest fragment will be
$3\sqrt 2\;m/s$
$5\sqrt 2\;m/s$
$\sqrt 2\;m/s$
$7\sqrt 2\;m/s$
A body of mass $5 \,kg$ explodes at rest into three fragments with masses in the ratio $1 : 1 : 3$. The fragments with equal masses fly in mutually perpendicular directions with speeds of $21 \,m/s$. The velocity of the heaviest fragment will be
${P_x} = m \times {v_x} = 1 \times 21 = 21\;kg\;m/s$
${P_y} = m \times {v_y} = 1 \times 21 = 21\;kg\;m/s$
Resultant =$\sqrt {P_x^2 + P_y^2} = 21\sqrt 2 $kg m/s
The momentum of heavier fragment should be numerically equal to resultant of ${\vec P_x}$ and ${\vec P_y}$.
$3 \times v = \sqrt {P_x^2 + P_y^2} = 21\sqrt 2 $ $v = 7\sqrt 2 $
$= 9.89 \,m/s$
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