A particle of mass m is projected with velocity v making an angle of 45° with the…
Physics · JEE Main · NTA Exams — Kinematics
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 \((\bar{p}=m \vec{v})\) will be:
- 2 mv
- \(\frac{m v}{\sqrt{2}}\)
- \(\mathrm{mv} \sqrt{2}\)
- zero
Answer
(C) mv 2
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