Question

PLEASE HELP!!!
A 14 kg rock starting from rest free falls through a distance of 5.0 m with no air resistance. Find the momentum change of the rock caused by its fall and the resulting change in the magnitude of earths velocity. earth mass is 6.0 * 10^24 kg. show your work assuming the rock earth system is closed.
PLEASE HELP!!!!

Answer 1

mass of the rock = 14 kg

distance that rock will move down = 5.0 m

now for motion under gravity we know that

$d = v_i t + \frac{1}{2} at^2$

here we have

$5 = 0 + \frac{1}{2}(10)t^2$

by solving above equation we have

$t = 1 s$

now change in momentum of rock is related to the force on it by equation below

$\Delta P = F \Delta t$

$\Delta P = (mg)\Delta t$

$\Delta P = (14 kg)(10 m/s^2)(1)$

$\Delta P = 140 kg m/s$

Now we know that same amount of force will act on the Earth also

So Earth will also have same momentum change

now we have

$\Delta P_{earth} = \Delta P_{rock}$

$m_{earth}(\Delta v) = 140$

$6.0 \times 10^{24} kg(\Delta v) = 140$

$\Delta v = \frac{140}{6 \times 10^{24}}$

$\Delta v = 2.33 \times 10^{-23} m/s$

Answer 2

The momentum change of the rock caused by its fall will be $2.33\times10^{-23}$m/sec and the magnitude of the earth velocity will be 140 kgm/s

When the rock starts from rest and free falls through a certain distance causes a change in the momentum of the rock.

How to find the change in the momentum of the body when the body is free falls?

Given,

Mass of the rock = 14 kg

Distance traveled by the rock under the influence of gravity =5.0m

Mass of earth =$6.0\times10^{24}$

Assumptions: the rock earth system is closed.

To find  

The momentum change of the rock    

Megnitude of earth velocity

   

  $S=ut +1/2{at^2}$  

 If the body freely falls initial velocity will be zero.

 so ; u = initial velocity = 0

 $5=0+\dfrac{1}{2} at^{2}$

  a= acceleration due to gravity ; g =10 m/s^2

 $t =\dfrac{5 \times2}{10}$

  t= 1 second

Change in the momentum of the rock will be,

ΔP= FΔT

As we know F= mg

ΔP = (mg ) ΔT

ΔP= (14$\times10$)

 ΔP=140 kg-m/s²

The change in the momentum of the rock is 140kg-m/s²

As we know that change in the momentum has conserved

change in the momentum of earth = change in the momentum of the  rock

$\rm ( \Delta P )_{earth} = ( \Delta P )_{ rock }$                   $\rm ( \Delta P ) _{rock }$$= 140 \;\rm kg m/s^2$

$\rm ( M) _{earth} \Delta V = ( \Delta P )_{rock }$    

$6.0214\times10^{24}\times$ΔV$= 140\rm \; kg m/s^2$

ΔV =$2.33\times10^{-23}$m /sec

The change in the megnitude of the earth velocicty  is  $2.33\times10^{-23}$ m /sec .

To know more about the change in momentum refers to

https://brainly.com/question/904448