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

11.5 m/s

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14.0 m/s

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7.0 m/s

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9.9 m/s

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Solution

The correct option is D
9.9 m/s

$_{1} +_{2} +_{3} = 0_{1} = m \times v_{x} = 1 \times 21 = 21 k g m / s_{2} = m \times v_{y} = 1 \times 21 = 21 k g m / s ∴ R e s u l t a n t = \sqrt{P_{x}^{2} + P_{y}^{2}} = 21 \sqrt{2} k g m / s$

The momentum of heavier fragment should be equal to resultant of $_{1} a n d_{2}$

$3 \times v = \sqrt{P_{x}^{2} + P_{y}^{2}} = 21 \sqrt{2} ∴ v = 7 \sqrt{2} = 9.89 m / s$

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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

The correct option is D 9.9 m/s →P1+→P2+→P3=0→P1=m×vx=1×21=21kg m/s→P2=m×vy=1×21=21kg m/s∴ Resultant=√P2x+P2y=21√2kg m/s The momentum of heavier fragment should be equal to resultant of →P1 and →P2 3× v=√P2x+P2y=21√2∴ v=7√2=9.89 m/s