Answer to Question #105093 in Mechanics | Relativity for Carson

Question

A rubber ball is dropped from a height of 1.75 meters and hits the floor. The ball compressed and then reforms to spring upwards from the floor to the height of 1.14 meters. The time from the balls first contact of the floor until it leaves the floor was 7.35 x 10^-2 seconds. Calculate the balls acceleration while in contact with the floor

Answer 1

As per the given question,

Height from which ball was released$d_1 = 1.75m$

After compressed, it will reach to the height $d_2= 1.14m$

The time of impact $(t)=7.35 \times 10^{-2} km/hr$

acceleration of the ball (a)= ?

We know that,

Let velocity of the ball before the impact is u,

So, $u^2=0+2gd_1=2\times 9.8\times1.75=34.3$

$u=\sqrt{34.3}=5.85m/sec$

Let velocity after impact is v, so $0=v^2-2gd_2$

$v^2=2\times9.8\times1.14=22.344$

$v=\sqrt{22.344}=4.726m/sec$

change in momentum = force on the ball

So, acceleration $a=\dfrac{u-(-v)}{t}=\dfrac{u+v}{t}$

$a=\dfrac{5.85+4.726}{7.35\times 10^{-2}}=\dfrac{10.576}{7.35\times10^{-2}}=1.438\times 10^2 m/sec^2$