SSC JE Mechanical Engineering 2014 Set 1 Paper

Question : 118 of 200

Marks: +1, -0

A ball is dropped vertically downwards, it hits the floor with a velocity of 9 m/s and bounces to a distance of 1.2 m. Coefficient of restitution between the floor and the ball is

0.54
zero
1
0.27

Validate

Solution:

Coefficient of restitution (e) = Velocity after rebound (v) / Velocity before rebound (u) ; and

v = (2.g.H) ^{1\/2}

; So,

e = [ (2gH) ^{1\/2} / u ].

Thus,

e = (2 × 9.81 × 1.2) ^{1\/2} / 9

= 4.85/9 = 0.54

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