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/sYou need to login to perform this action.
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