Momentum and CollisionMedium
Question
A particle of mass 4m which is at rest explodes into four equal fragments. All 4 fragment s scattered in the same horizontal plane. Three fragments are found to move with velocity v each as in the fig. The total energy released in the process of explosion
Options
A.mv2(3 - √2)
B.mv2(3 - √2)/2
C.2mv2
D.mv2 (1 + √2)/2
Solution
COLM ⇒ mvî + mvĵ + m
+ 
= 0
= - n 
Total energy released
=
mv2 + 
mv2 + 
mv2 + m 
= mv2(3 - √2)
+ = 0 = - n Total energy released
=
mv2 + mv2 + mv2 + m = mv2(3 - √2)
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