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Center of Mass and Collision question

2008 · Q149
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Center of Mass and Collision question

2008 · Q149

NEETPhysicsCenter of Mass and CollisionMCQ+4 / −1
A particle of mass m is projected with velocity v making an angle of 45o with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be :
  1. A
    mv2mv\sqrt 2mv2​
  2. B
    zero
  3. C
    2mv
  4. D
    mv / 2\sqrt 22​
View written solutionFree

Correct answer: A

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

AIPMT 2008 Physics - Center of Mass and Collision Question 38 English Explanation

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

= mvcosθ\theta θ – mvcosθ\theta θ = 0

Change in momentum along vertical (i.e. along y–axis) = mv sinθ\theta θ – (–mv sinθ\theta θ)

= 2 mvsinθ\theta θ = 2mv × sin 45°

= 2mv×12=2mv2mv \times {1 \over {\sqrt 2 }} = \sqrt 2 mv2mv×2​1​=2​mv

Hence, resultant change in momentum = 2mv\sqrt 2 mv2​mv

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