A ball dropped from a building of height 12 m falls on a slab of 1 m height from the ground and makes a perfect elastic collision.
Later the ball falls on a wooden table of height 0.5 m, makes inelastic collision and falls on the ground. If the coefficient of restitution between the ball and the table is 0.5, then the velocity of the ball while touching the ground is about
(Acceleration due to gravity, `g=10 ms^(-2)`)

Options

A. `15.5 ms^(-1)`
B. `14.5 ms^(-1)`
C. `9.2 ms^(-1)`
D. `8.2 ms^(-1)`

Answer

D

Explanation

According to the question,

When ball is droped from point P, and collides at point Q, from third equation of the motion,
`v_(Q)^(2)=u^(2)+2gh=0+2 times 10 times (12-1)=220`
`therefore` Velocity at a point Q, `V_(Q)=sqrt(220)m//s`
In perfect elastic collision at point Q, there is no loss of kinetic energy.
Hence, the height gain by ball after collision,
`rArr " "1/2mv_(Q)^(2)=mgh^(.)`
`rArr" "1/2(sqrt(220))^(2)=10 times h^(.)`
`rArr " "h^(.)=11m`
Let velocity of the ball before inelastic collision `=v_(s)`
From energy conservation law,
`rArr " "1/2mv_(s)^(2)=mg(11+0.5)`
`rArr " "v_(s)^(2)=2 times 10 times 11.5`
`rArr " "v_(s)=sqrt(230)m//s`
`because` coefficient of restitution, `" "e=0.5="velocity after collision (v)"/("velcoity before collision "(v_(s)))`
`rArr 0.5=v/sqrt(230) rArr v=0.5sqrt(230)m//s`
Again from energy conservation law,
`" "1/2mv^(2)=mgh..`
`rArr " "1/2(0.5sqrt(230))^(2)=10 times h`
`rArr h..=(0.5 times 230 times 0.5)/(20)=2.875 m`
Hence, the total height from point T,
`" "h=h..+0.5=2.875+0.5=3.375m`
From energy conservation law,
`" "1/2mv_(T)^(2)=mgh`
Velocity of ball while touching the ground,
`" "v_(T)=sqrt(2gh)`
`" "=sqrt(2 times 10 times 3375)`
`" "=8.215 m//s`