A particle of mass $m$ is projected with a velocity $v$ at angle of $45^{o}$ with the horizontal. When the particle lands on the level ground the magnitude of 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 A $2 m v$
When it lands the velocity is equal in magnitude but opposite in direction to that at the time of projection. So change in momentum will be nothing but double of initial value so its magnitude will be $2 m v$

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A particle of mass m is projected with a velocity v at angle of 45o with the horizontal. When the particle lands on the level ground the magnitude of change in its momentum will be"

The correct option is A 2mvWhen it lands the velocity is equal in magnitude but opposite in direction to that at the time of projection. So change in momentum will be nothing but double of initial value so its magnitude will be 2mv

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

The correct option is A $2 m v$
When it lands the velocity is equal in magnitude but opposite in direction to that at the time of projection. So change in momentum will be nothing but double of initial value so its magnitude will be $2 m v$