MHT CET · Physics · Center of Mass Momentum and Collision
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
- A \(2 m v\)
- B \(m v / \sqrt{2}\)
- C \(m v \sqrt{2}\)
- D zero
Answer & Solution
Correct Answer
(C) \(m v \sqrt{2}\)
Step-by-step Solution
Detailed explanation
The horizontal momentum does not change. The change in vertical momentum is
\(\begin{aligned}m v \sin \theta-(-m v \sin \theta) &=2 m v \frac{1}{\sqrt{2}} \\&=\sqrt{2} \mathrm{mv}\end{aligned}\)
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