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Question

A bomb of mass 4 m at rest, explodes into three pieces of masses in the ratio 1:1:2. The two identical pieces fly in mutually perpendicular directions, each with a velocity v. The third piece is making an angle α with the line of motion of each piece. Find α.

A
450
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B
900
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C
1350
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D
1800
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Solution

The correct option is B 1350
Let the two identical pieces fly along ^i and ^j directions,
Using momentum conservation
Initial momentum =0, therefore,
mv^i+mv^j+2mx=0

x=v2(^i+^j)

Let α be the angle between x and ^i.
By taking dot product of x and ^i, we get,

x.^i=|x||^i|cosα

v2=(v2)2+(v2)2cosα

v2=v2cosα

cosα=12

α=1350

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