Question

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. 11.5 m/s
• B. 14.0 m/s
• C. 7.0 m/s
• D. 9.89 m/s

Answer 1

Answer: (D)

9.89 m/s

Answer 2

$P _ { x } = m \times v _ { x } = 1 \times 21 = 21 \mathrm {~kg} \mathrm {~m} / \mathrm { s }$

$P _ { y } = m \times v _ { y } = 1 \times 21 = 21 \mathrm {~kg} \mathrm {~m} / \mathrm { s }$

∴ 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