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Question

A stationary body explodes into three identical fragments such that two of them fly off mutually perpendicular to each other, each with same kinetic energy, Eo. The energy of the explosion is

A
3Eo
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B
3Eo
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C
4Eo
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D
(2+2)Eo
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Solution

The correct option is C 4Eo
Two fragments have same kinetic energy and mass, then their speed must be same.
Let speed of both fragments is v
Both are perpendicular, then consider velocity of first fragment is x axis direction while velocity of second is y axis direction.
Therefore, v1=v^i,v2=v^j
Third fragment will fly in a direction opposite to the direction of resultant of other two particles. So, the total momentum of the system after explosion also remains zero.
Applying momentum conservation for all the three fragments,
m1v1+m2v2+m3v3=0
v1+v2+v3=0
v^i+v^j+v3=0
v3=(v^i+v^j)|v3|=2v

Now, kinetic energy of the first and second fragment =E0=12mv2

Total K.E =12mv2+12mv2+12m(2v)2
=E0+E0+2E0
Total KE =4E0

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