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DIA [1.3K]
3 years ago
9

A small ball of mass m is aligned above a larger ball of mass M = 0.63kg (with a slight separation) and the two are dropped simu

ltaneously from a height of 1.8m. If the larger ball rebounds elastically from the floor and the small ball rebounds elastically from the larger ball what value of m results in the larger ball stopping when it collides with the small ball?
Physics
1 answer:
natta225 [31]3 years ago
8 0
 <span>a) M is the big one 
m is the little one 

v is the speed of each of them when they impact 

V = speed of mii after collision 

Conservation of momentum 

Mv – mv = mV 

Conservation of energy 

1/2Mv^2 + 1/2mv^2 = 1/2 mV^2 

This pair simplify to give 

M = 3m 

V = 2v 

So m = 0.21kg 

and h = 4 . 2.7 = 10.8 m</span>Source(s):<span>Old teacher</span>
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Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

The distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

T^2 \propto R^3

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.

By using Kepler's third law, this can be written as,

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Substituting the values, we get the value of constant k for mars.

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k = 3.92 \times 10^{-29}

The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.

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Hence we can conclude that the distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

To know more about Kepler's third law, follow the link given below.

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