Question

a partical is dropped on the earth from height r and bounce back to a height r/2 the coefficient of restitution is​

Answer 1

nitial velocity=(2gh)^1/2

so, (v)=(e^n)((2gh)^1/2

thus, as it is uniform motion so,

2gH=v^2

2gH=(e^2n)2gh

thus,   H=(e^2n)h

so, option (a) is correct!

Answer 2

The coefficient of restitution is $\frac{1}{\sqrt{2} }$

• Coefficient of restitution is the ratio of relative velocity of separation to the relative velocity of approach after collision.
• It is denoted by e.
• Since it is a ratio, it has no units and dimensions.
• It is given by

            $e = \frac{v}{u} = \sqrt{\frac{h_{2} }{h_{1} } }$

• Here,

         $h_{1} = r, h_{2} = r/2$

• Therefore,

           $e = \sqrt{\frac{r/2}{r} } \\ = \frac{1}{\sqrt{2} }$