Question
A body explodes into three equal fragments, two of
which fly off perpendicular to each other with equal speeds. What is the direction of the third fragment?Solution
Given: β A body explodes into 3 equal fragments (mass = m each) β Two fragments move at equal speeds v , perpendicular to each other β Body was initially at rest β total momentum = 0 Assume: β First fragment moves along x-axis β momentum = mΒ·v β Second fragment moves along y-axis β momentum = mΒ·v Resultant momentum of first two fragments: = β[(mΒ·v)Β² + (mΒ·v)Β²] = mΒ·vΒ·β2 To conserve momentum, the third fragment must have equal and opposite momentum: β magnitude = mΒ·vΒ·β2 β direction = opposite to the resultant of the first two (i.e., 225Β° from x-axis)
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