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

Question

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) mv √ 2  (B) zero (C) 2 mv (D) mv / √ 2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/e54813c619d70f22174c7b40e294a5f1-.jpeg" /> The horizontal momentum does not change. The change in vertical momentum is m v sin θ-(-m v sin θ)=2 m v (1/√2)=√2 m v

Q. A particle of mass m is projected with velocity v making an angle of $4 5^{\circ}$ with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be

$m v 2$

zero

$2 m v$

$m v / 2$

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

Q. 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

mv2​