A particle of mass $m$ is projected with velocity $v$ making an angle of $45^{\circ}$ with the horizontal.…

A particle of mass $m$ is projected with velocity $v$ making an angle of $45^{\circ}$ with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be
  1. zero
  2. $2 m v$
  3. $m v / \sqrt{2}$
  4. $m v \sqrt{2}$

Solution

Momentum change $=2 \mathrm{mv} \sin \theta$

Asked in: NEET 2008 (Mains)

Practice more Center of Mass Momentum and Collision questions on Aicharya