Question

An elastic ball at rest with coefficient of
restitution 1/2 is
dropped from a height h on a
smooth floor. The total distance covered by the ball is

option is
5/3 h
5 h
3 h
4/3h


plzzz help me ​

Answer 1

Answer:

5h/4

(the option does not match so either option or my solution needs checking)

Explanation:

We know that

coefficient of restitution = velocity after collision/velocity before collision

Let the velocity of the ball just before the collision is v

Then the velocity of the ball just after the collision

= velocity before collision × coefficient of restutution

= $v\times\frac{1}{2}$

= $\frac{v}{2}$

When the ball is dropped from height h, using Newton's third equation of motion

$v^2=0^2+2gh$

or, $v=\sqrt{2gh}$

Again if the distance covered by the ball after the collision is h' then from the third equation of motion

$0^2=(\frac{v}{2} )^2-2gh'$

or, $\frac{ v^2}{4}=2gh'$

or, $h' = \frac{2gh}{8g}$

or, $h'=\frac{h}{4}$

∵ the ball is dropped from height h and then again goes to height h/4

∴ total distance covered = $h+\frac{h}{4} =\frac{5h}{4}$

Answer 2

Ans.

5h/4

Explanation:

We know that

coefficient of restitution = velocity after collision/velocity before collision

Let the velocity of the ball just before the collision is v

Then the velocity of the ball just after the collision

= velocity before collision × coefficient of restutution

= v\times\frac{1}{2}v×21

= \frac{v}{2}2v

When the ball is dropped from height h, using Newton's third equation of motion

v^2=0^2+2ghv2=02+2gh

or, v=\sqrt{2gh}v=2gh

Again if the distance covered by the ball after the collision is h' then from the third equation of motion

0^2=(\frac{v}{2} )^2-2gh'02=(2v)2−2gh′

or, \frac{ v^2}{4}=2gh'4v2=2gh′

or, h' = \frac{2gh}{8g}h′=8g2gh

or, h'=\frac{h}{4}h′=4h

∵ the ball is dropped from height h and then again goes to height h/4

∴ total distance covered = h+\frac{h}{4} =\frac{5h}{4}h+4h=45h