Question

A ping pong ball of mass 0.0027 kg and radius 0.020 m is completely submerged 0.3 m in water. How fast is it moving when it emerges from the water?

Answer 1

The value speed is
`v = 8.192 \ m/s`

Step-by-step explanation:

From the question we are told that

The mass of the ping pong ball is
`m = 0.0027 \ kg`

The radius is
`r = 0.020 \ m`

The depth of the ping pong inside the water is
`s = 0.3 \ m`

Gnerally the force with which the ball will emerge will from the water is mathematically represented as

`F = B - m g`

Here B is the Buoyant force on the ball which is mathematically represented as

`B = \rho_w * V_b * g`

Here
`V_b` is the volume of the ball which is mathematically represented as

`V_b = (4)/(3) * \pi * r^3`

=>
`V_b = (4)/(3) * 3.142 * 0.020 ^3`

=>
`V_b = 3.35*10^(-5) \ m^3`

`\rho_w` is the density of water with value
`\rho_w = 1000 \ kg/m^3`

So

`B = 1000 * 3.35*10^(-5) * 9.8`

=>
`B = 0.3284 \ N`

So

`F = 0.3284 - 0.0027 * 9.8`

=>
`F = 0.3284 - 0.02646`

=>
`F = 0.3020 \ N`

Generally force is mathematically represented as

`F = m * a`

So

`0.3020 = 0.0027 * a`

=>
`a = 111.84 \ m/s^2`

Generally from kinematic equation

`v^2 = u^2 + 2as`

Here u is the velocity of the ping pong at depth of 0.3 m and the value is zero given that at that point there was no motion

So

`v^2 = 0^2 + 2 * 111. 84 * 0.3`

=>
`v = 8.192 \ m/s`