Question

Hello, I am currently stuck on this question, is it possible to get a little help.

Answer 1

Given:

mass of the ball is

`m=0.14\text{ kg}`

height of surface where the ball dropped

`h=2.0\text{ m}`

Required: momentum has to be calculated just before the ground.

Step-by-step explanation:

when a ball is dropped, it accelerates under gravity due to which it gains velocity. we know the second equation of motion that is given as

`v^2=u^2+2gh`

where

`u`

is the initial velocity at the time of dropping the ball that is zero because it is in rest. it is free fall because we are throwing the ball, we are just dropping the ball at height h.

`v`

is the final velocity. g is the acceleration due to gravity and h is the height where the drop of the ball take place. put in the

`u=0\text{ m/s}`

in the above relation we get

`v=\sqrt[2]{2gh}....(1)`

where g is the acceleration due to gravity which is

`g=9.8\text{ m/s}^2`

plugging all the values in the above relation we get

`\begin{gathered} v=\sqrt[2]{2*9.8\text{ m/s}^2*2.0\text{ m}} \\ v=\sqrt[2]{39.2\text{ m}^2\text{/s}}^2 \\ v=6.26\text{ m/s} \end{gathered}`

this is velocity we get just before striking with the ground

now calculate the momentum of the ball

that is given by

`p=mv`

where m is mass of the ball and v is velocity.

plugging all the values in the above relation we get

`\begin{gathered} p=0.14\text{ kg}*6.26\text{ m/s} \\ p=.8764\text{ kg m/s} \\ p\approx0.88\text{ kg m/s} \end{gathered}`

Thus, the momentum of the ball just before the ground is

`0.88\text{ kg m/s}`