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    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?
    A At 45° to the two Correct Answer Incorrect Answer
    B Opposite to the resultant of the two Correct Answer Incorrect Answer
    C Along the bisector of the two Correct Answer Incorrect Answer
    D Perpendicular to both Correct Answer Incorrect Answer

    Solution

    Given: – A body explodes into 3 equal fragments (mass = m each) – Two fragments move at equal speeds v , perpendicular to each other – Body was initially at rest → total momentum = 0 Assume: – First fragment moves along x-axis → momentum = m·v – Second fragment moves along y-axis → momentum = m·v Resultant momentum of first two fragments: = √[(m·v)² + (m·v)²] = m·v·√2 To conserve momentum, the third fragment must have equal and opposite momentum: → magnitude = m·v·√2 → direction = opposite to the resultant of the first two (i.e., 225° from x-axis)

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