A particle of mass m is projected with velocity v making an angle of 45° with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be:

A particle of mass m is projected with velocity v making an angle of 45° with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be:

  1. A

    2mv

  2. B

    mv2

  3. C

    mv2

  4. D

    zero

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    Solution:

    The magnitude of the resultant velocity at the point of projection and the landing point is same.

    Clearly, change in momentum along horizontal (i.e along x-axis)

    =mvcosθ-mvcosθ=0

    Change in momentum along vertical (i.e. along y–axis) =

    mvsinθ-(-mvsinθ)

    =2mvsinθ=2mv×sin45°

    =2mv×12=2mv

    Hence, resultant change in momentum =2mv

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