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

2 m v

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m v / \sqrt{2}

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m v \sqrt{2}

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Solution

The correct option is C $m v \sqrt{2}$

The horizontal momentum does not change.
The change in vertical momentum is
$m v sin θ - (- m v sin θ) = 2 m v \frac{1}{\sqrt{2}} = \sqrt{2} m v$

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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 :

The correct option is C mv√2The horizontal momentum does not change.The change in vertical momentum ismvsinθ−(−mvsinθ)=2mv1√2=√2mv

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

The correct option is C $m v \sqrt{2}$

The horizontal momentum does not change.
The change in vertical momentum is
$m v sin θ - (- m v sin θ) = 2 m v \frac{1}{\sqrt{2}} = \sqrt{2} m v$