A particle of mass 4 m 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 energy released in the process of explosion is
Text Solution
Verified by ExpertsThe correct answer is:
A
Here momentum of third fragment is
p 3 = 
or p 3 = 
= 
Final K.E. of the system
= 
= 
mv 2 + 
mv 2 + 
mv
= 
mv 2
Since initial K.E. = 0 therefore energy released = 
mv 2 .
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