Physics, asked by Bittu5962, 1 year ago

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 will be

Answers

Answered by ALYBSN
0
mg is the answer of this question
Answered by ltsmeAliya
0

Answer:

\sqrt{2} mv

Explanation:

Here the particle follows a projectile motion.

In projectile Motion the velocity along the x-axis does not change.

∴ Change in momentum along x-axis is zero.

we will only consider the velocity along the y-axis to get the change in momentum of the particle.

so,

velocity along the y-axis = vsin45°

|ΔP|= m(Δv)

|ΔP|= m[vsin45°-(-vsin45°)]

|ΔP|=2×mv×sin45°

|ΔP|=2×mv×1/\sqrt{2}

|ΔP|= \sqrt{2}mv

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