Question

A particle of mass m , strikes on ground with angle of incidence 45° . If coefficient of restitution e=1/√2 , the velocity of reflection is

Answer 1

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Answer 2

Dear Student,

◆ Answer -

v = √3 u/2

θ = 54.74°

● Explanation -

Let u be initial velocity.

ux = u.cos45 = u/√2

uy = u.sin45 = u/√2

After collision,

vx = ux = u/√2

But coefficient of restitution is -

e = vy/uy

vy = e.uy

vy = 1/√2 × u/√2

vy = u/√2

Velocity after impact is -

v = √(vx^2 + vy^2)

v = √(u^2/2 + u^2/4)

v = √3 u/2

Angle of reflection is -

θ = arctan(vx/vy)

θ = arctan[(u/√2)/(u/2)

θ = arctan(√2)

θ = 54.74°

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