Physics, asked by saikar, 6 months ago

A ball falls from a certain height on to a hard floor and bounces with 25 percent of its initial velocity after impinging on the floor the coefficient of restitution is​

Answers

Answered by parulsehgal06
1

Answer:

The coefficient of restitution = e = 1/4v².

Explanation:

Let v be the velocity with which ball strikes the floor.

and v' be the velocity of ball with which bounces back with 25 percent of initial velocity.

              v' = (25/100)v

              v' = (1/4)v

 By law of conservation of energy,

   When ball strikes the floor,

              mgh = (1/2)mv²

                  gh = (1/2)v²

                     v² = 2gh

                     v = √2gh   ---------(i)

      When ball bounces back

             mgh = (1/2)mv'²

                gh = (1/2)v'²

                  v'² = 2gh

                  v' = √2gh --------------(ii)

      from (i) and (ii) we can write

                  v = v'

       By the formula coefficient of restitution,

             e = (relative final velocity)/(relative initial velocity)

             e = v'/v

             e = [(1/4)v]/v

              e = 1/4v²

    Hence, the coefficient of restitution = e = 1/4v².

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Answered by swethassynergy
0

The coefficient of restitution is \frac{1}{4}.

Explanation:

Given:

A ball falls from a certain height on to a hard floor.

Ball bounces with 25 percent of its initial velocity after impinging on the floor.

To Find:

The coefficient of restitution.

Solution:

Let m is the mass of ball and it falls from a certain height k on to a hard floor.

Let p be the velocity with which ball strikes the floor.

Applying law of energy conservation, before the ball collision the floor.

Kinetic Energy just before the collision will be loss in Potential Energy.

mgk= \frac{1}{2} mp^{2}

2mgk=  mp^{2}

p^{2}=2gk

p=\sqrt{2gk}

Let q is the velocity of ball with which bounces back with 25 percent of initial velocity,

As given,ball bounces with 25 percent of its initial velocity after impinging on the floor.

q=\frac{25}{100} p

   =\frac{1}{4} p  

   =\frac{1}{4} (\sqrt{2gk} )

The coefficient of restitution, e=\frac{ velocity \ after\  bounce\ back }{  velocity \ before\  collision }

                                                     =\frac{\frac{1}{4} \sqrt{2gk} }{  \sqrt{2gk}    }

                                                     =\frac{1}{4}

Thus,the coefficient of restitution is \frac{1}{4}.

#SPJ3

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