A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments each of mass m are found to move with a speed 'v' each in mutually perpendicular directions. The minimum energy released in the process of explosion is
Text Solution
Verified by ExpertsThe correct answer is:
B
mv 
+ mv 
+ 2m 
3 = 0
= 
= – 
( 
+ 
) = – 
k f = 
mv 2 + 
mv 2 + 
2m 
k f = 
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