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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 :
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- 2mv
- mv / √2
- mv √2
- zero
Correct Option: C
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) = mv cos θ – mv cos θ = 0
Change in momentum along vertical (i.e. along y–axis) = mv sinθ – (–mv sinθ) = 2 mvsinθ = 2mv × sin 45°
| = 2mv × | = √2 mv | |
| √2 |
Hence, resultant change in momentum = √2 mv
