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

$\frac{mv}{\sqrt{2}}$

3. C

$\mathrm{mv}\sqrt{2}$

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)

$=mv\mathrm{cos}\theta -mv\mathrm{cos}\theta =0$

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

$mv\mathrm{sin}\theta -\left(-mv\mathrm{sin}\theta \right)$

$=2mv\mathrm{sin}\theta =2mv×\mathrm{sin}45°$

$=2\mathrm{mv}×\frac{1}{\sqrt{2}}=\sqrt{2}\mathrm{mv}$

Hence, resultant change in momentum $=\sqrt{2}\mathrm{mv}$

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