Question

A particle of mass m is projected with velocity V making an angle of 45 degree with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be?
ans :- mV 1.414

Answer 1

Answer on Question #60256-Physics-Mechanics-Relativity

A particle of mass $m$ is projected with velocity $V$ making an angle of 45 degree with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be?

Solution

The horizontal component of momentum of projectile doesn't changes. So,

$$\Delta P = \Delta P_{vertical} = \left\lfloor mv \sin(-45) - mv \sin 45 \right\rfloor = 2 \left\lfloor mv \sin 45 \right\rfloor = 2mv \frac{1}{\sqrt{2}} = \sqrt{2}mv.$$

Answer: $\sqrt{2}mv$.

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