Question

A ball is dropped ona horizontal surface from height h .if it rebounds upto upylto h/2after first collision then coefficient of restitution between ball and surface

Answer 1

Initial speed of ball just before it collide with the floor from height h is given by

$v_i = \sqrt{2gh}$

now after it collide with the floor with rebound height is h/2 is given by

$v_f = \sqrt{2g\frac{h}{2}}$

now we have to use the formula of coefficient of restitution

$v_f = e*v_i$

$\sqrt{gh} = e\sqrt{2gh}$

$1 = e*\sqrt2$

$e = \frac{1}{\sqrt2}$

$e = 0.707$

so the coefficient of restitution is 0.707 in this collision

Answer 2

According to the  data given, the initial speed of the ball be

                                              u = √ 2 g h

which is before the collision.

And will change after collision as

                                            v = √ 2 g h/ 2.

Thereby as we know that coefficient of restitution is the factor which changes the initial velocity, we have the formula as v = e x u where e is the coefficient of restitution.

Thereby,

                                    √ 2 g h/ 2 = e √ 2 g h

                                      => √  g h = e x √ 2 g h

                                      => e = 1 / √ 2

                                      => e = 0.707.