question_answer

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            [CBSE PMT 1991]

A)          11.5 m/s

B)                       14.0 m/s

C)          7.0 m/s

D)                       9.89 m/s

Correct Answer: D

Solution :

                           \[{{P}_{x}}=m\times {{v}_{x}}=1\times 21=21\ kg\ m/s\]               \[{{P}_{y}}=m\times {{v}_{y}}=1\times 21=21\ kg\ m/s\]               \ 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