A body of mass $m$ is projected at an angle of $45^{o}$ with horizontal. If air resistance is negligible total change in momentum when it strikes the ground is:

2 m v

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\sqrt{2} m v

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m v

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m v / \sqrt{2}

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Solution

The correct option is B $\sqrt{2} m v$
Momentum is defined as $P = m v$
$P_{i} = m v sin θ$
$P_{f} = - m v sin θ$
Change in momentum $= 2 m v s i n θ = 2 m v s i n \frac{π}{4} = \sqrt{2} m v$

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A body of mass m is projected at an angle of 45o with horizontal. If air resistance is negligible total change in momentum when it strikes the ground is:

The correct option is B √2mvMomentum is defined as P=mvPi=mvsinθPf=−mvsinθChange in momentum =2mvsinθ=2mvsinπ4=√2mv

A body of mass $m$ is projected at an angle of $45^{o}$ with horizontal. If air resistance is negligible total change in momentum when it strikes the ground is:

The correct option is B $\sqrt{2} m v$
Momentum is defined as $P = m v$
$P_{i} = m v sin θ$
$P_{f} = - m v sin θ$
Change in momentum $= 2 m v s i n θ = 2 m v s i n \frac{π}{4} = \sqrt{2} m v$