Question

A ball is dropped from height of 20 m on a floor. It is found that the ball rises up
to 5 m after second bounce, then coefficient of restitution for the collision
between the ball and the floor is
1/4
1/2
1/√2
1/2√2
​

Answer 1

Answer:

initial velocity of ball, u = 20m/s [ downward]

ball is thrown from a height , h = 20m

velocity of ball to reach the floor , v² = u² + 2as

v² = 20² + 2 × 10 × 20

v = 20√2 m/s

so, velocity of ball before collision = 20√2 m/s

a/c to question,

the ball bounces to the same height from which it was thrown.

so, at a 20m height , velocity of ball , v= 0

we have to find velocity of ball after collision

so, use formula, v² = u² + 2as

0 = u² + 2(-10) × 20

v = 20m/s

In the case of a ball bouncing off a flat, stationary surface, the coefficient of restitution turns out to be:

coefficient of restitution = V/U

where V is velocity of object after collision.

U is velocity of object before collision.

hence, coefficient of restitution = 20/20√2 = 1/√2