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NEET Physics QUESTION #7334
Question 1

A particle of mass $m$ is projected with speed $v$ at $45°$ to the horizontal. When it returns to ground level, the magnitude of the change in momentum is:

  • $mv\sqrt{2}$✔️
  • Zero
  • $2mv$
  • $\dfrac{mv}{\sqrt{2}}$
Correct Answer Explanation

Initial vertical component: $v_y = v\sin45° = \dfrac{v}{\sqrt{2}}$ (upward)

On landing (by symmetry): $v_y' = \dfrac{v}{\sqrt{2}}$ (downward)

Horizontal component unchanged. Change in momentum:

$|\Delta\vec{p}| = m\left|\Delta v_y\right| = m\cdot\dfrac{2v}{\sqrt{2}} = \mathbf{mv\sqrt{2}}$