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shepuryov [24]
3 years ago
12

A 2 kg ball going 10 m/s follows a path that is perpendicular to the surface of a wall. It impacts the wall and looses 20 % of i

ts momentum in the collision. What is the balls impulse?
Physics
1 answer:
Ostrovityanka [42]3 years ago
4 0

Answer:

If two objects make a head on collision, they can bounce and move along the same direction they approached from (i.e. only a single dimension). However, if two objects make a glancing collision, they'll move off in two dimensions after the collision (like a glancing collision between two billiard balls).

For a collision where objects will be moving in 2 dimensions (e.g. x and y), the momentum will be conserved in each direction independently (as long as there's no external impulse in that direction).

In other words, the total momentum in the x direction will be the same before and after the collision.

\Large \Sigma p_{xi}=\Sigma p_{xf}Σp

xi

​

=Σp

xf

​

\Sigma, p, start subscript, x, i, end subscript, equals, \Sigma, p, start subscript, x, f, end subscript

Also, the total momentum in the y direction will be the same before and after the collision.

\Large \Sigma p_{yi}=\Sigma p_{yf}Σp

yi

​

=Σp

yf

​

\Sigma, p, start subscript, y, i, end subscript, equals, \Sigma, p, start subscript, y, f, end subscript

In solving 2 dimensional collision problems, a good approach usually follows a general procedure:

Identify all the bodies in the system. Assign clear symbols to each and draw a simple diagram if necessary.

Write down all the values you know and decide exactly what you need to find out to solve the problem.

Select a coordinate system. If many of the forces and velocities fall along a particular direction, it is advisable to use this direction as your x or y axis to simplify calculation; even if it makes your axes not parallel to the page in your diagram.

Explanation:

i think this what your aswer is let me know if you got  it right if not i fix it or i will look in my answer book

please rate me in say thanks

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Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the di
vekshin1

Here is the full question

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the disk of the galaxy, about how far (on average) would it be to the nearest civilization?

(Hint: Start by finding the area of the Milky Way's disk, assuming that it is circular and 100,000 light-years in diameter. Then find the average area per civilization, and use the distance across this area to estimate the distance between civilizations.)

Answer:

1000 light-years (ly)

Explanation:

If we go by the hint; The area of the disk can be expressed as:

A = \pi (\frac{D}{2})^2

where D = 100, 000 ly

Let's divide the Area by the number of civilization; if we do that ; we will be able to get 'n' disk that is randomly distributed; so ;

d= \frac{A}{N} =\frac{\pi (\frac{D}{2})^2 }{10, 000}

The distance between each disk is further calculated by finding the radius of the density which is shown as follows:

d = \pi r^2 e

r^2_e= \frac{d}{\pi}

r_e = \sqrt{\frac{d}{\pi} }

replacing d = \frac{\pi (\frac{D}{2})^2 }{10, 000} in the equation above; we have:

r_e = \sqrt{\frac{\frac{\pi (\frac{D}{2})^2 }{10, 000}}{\pi} }

r_e = \sqrt{\frac{(\frac{D}{2})^2 }{10, 000}}

r_e = \sqrt{\frac{(\frac{100,000}{2})^2 }{10, 000}}

r_e = 500 ly

The distance (s) between each civilization = 2(r_e)

= 2 (500 ly)

= 1000 light-years (ly)

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Answer:

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Explanation:

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