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Gwar [14]
2 years ago
7

Elastic and Inelastic Collisions​

Physics
1 answer:
Igoryamba2 years ago
4 0

\large{\color{gold}{\bigstar }}\:  \: { \underline{\mathfrak{Together \: We \: Go \: Far}}}  \:  \:   \color{gold}\bigstar

\rule{70mm}{2.9pt}

<h3>Elastic and Inelastic Collisions</h3>

A collision is an interaction between two objects resulting in the exchange of impulse and momentum. The time of impact is usually small; the impulse provided by external forces like friction during this time is negligible. When two or more bodies collide, the momentum of the system is therefore approximately conserved. In an isolated system, the total momentum before the collision is equal to the total momentum of the system after collision.

total momentum before collision = total momentum after collision

There are two types of collision and are categorized according to whether the system's total kinetic energy changes. It may or may not be conserved depending on the type of collision. It may lose during collisions when (1) it is converted to heat or other forms like binding energy, sound, light (if there is spark), etc. and (2) it is spent in producing deformation or damage, such as when two cars collide.

-  -  -  -  -  -  -  -  -  -  -  - -   \: -

The two types of collisions are:-

1. Elastic collision ☞ states that the system's total kinetic energy does not change and colliding objects bounce off after collision. (no kinetic energy is loss, no damage, no heat)

Examples:

  • Motion of atoms and molecules
  • Hitting billiard balls
  • When a soccer player kicks a ball since the player's foot and the ball do indeed remain completely separate after collision.

2. Inelastic collision ☞ states that the system's total kinetic energy changes (i.e., converted to some other form of energy). Collision is said to be perfectly inelastic if after collision the objects stick together and move as one mass with one velocity.

Example:

  • Celestial bodies collide, like two asteroids, they fuse together to form a larger body.

=  =  =  =  =  =  =  =  =  =  =  =  =  =  = = = =

Additional Information:-

☞ brainly.com/question/24616147

☞ brainly.com/question/12941951

\rule{70mm}{2.9pt}

~ \large {\color{iron}{(:}} \:  \bold {MrHarold  \: \color{iron}{ :)}}

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A horizontal pipe of inner diameter 2.2 cm carries water with a density of 1000.0 kg/m3 flowing at a rate of 1.5 kg/s. If the pi
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The speed of the water in the wider part will be 1.194 m/sec. Speed is a time-based quantity. Its SI unit is m/sec.

<h3> What is speed?</h3>

Speed is defined as the rate of change of the distance or the height attained.

The given data in the problem is;

The initial diameter is,\rm d_1 = 2.2 \ cm

initial radius,

r_1 = \frac{d_1}{2} \\\\ r_1 = \frac{2.2}{2} \\\\ r_1 = 1.1\ cm

The initial crossection area;

\rm A_1 = \pi r_1^2 \\\\ \rm A_1 = 3.14 \times  (1.1\times 10^{-2})^2 \\\\ \rm A_1 =3.8 \times 10^{-4} \ m^2

The final crossection area;

\rm A_2 = \pi r_2^2 \\\\ \rm A_2 = 3.14 \times ( 2 \times 10^{-2})^2 \\\\ \rm A_2 = 12.56 \ m^2

The initial flow rate is;

R = density ×velocity ×area

\rm R = \rho A V \\\\ 1.5 = 1000 \times V_1 \times 3.8 \times 10^{-4} \\\\ V_1  = 3.947 \ m/sec

The speed of the water in the wider part will be;

From the continuity equation;

\rm A_1 V_1 = A_2V_2  \\\\\ 3.8 \times 10^{-4} \times 3.947 = 12.56 \times 10^{-4} \times V_2 \\\\ V_2= 1.194 \ m/sec

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Two blocks in contact with each other are pushed to the right across a rough horizontal surface by the two forces shown. If the
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I assume the blocks are pushed together at constant speed, and it's not so important but I'll also assume it's the smaller block being pushed up against the larger one. (The opposite arrangement works out much the same way.)

Consider the forces acting on either block. Let the direction in which the blocks are being pushed by the positive direction.

The 2.0-kg block feels

• the downward pull of its own weight, (2.0 kg) <em>g</em>

• the upward normal force of the surface, magnitude <em>n₁</em>

• kinetic friction, mag. <em>f₁</em> = 0.30<em>n₁</em>, pointing in the negative horizontal direction

• the contact force of the larger block, mag. <em>c₁</em>, also pointing in the negative horizontal direction

• the applied force, mag. <em>F</em>, pointing in the positive horizontal direction

Meanwhile the 3.0-kg block feels

• its own weight, (3.0 kg) <em>g</em>, pointing downward

• normal force, mag. <em>n₂</em>, pointing upward

• kinetic friction, mag. <em>f₂</em> = 0.30<em>n₂</em>, pointing in the negative horizontal direction

• contact force from the smaller block, mag. <em>c₂</em>, pointing in the <u>positive</u> horizontal direction (this is the force that is causing the larger block to move)

Notice the contact forces form an action-reaction pair, so that <em>c₁</em> = <em>c₂</em>, so we only need to find one of these, and we can get it right away from the net forces acting on the 3.0-kg block in the vertical and horizontal directions:

• net vertical force:

<em>n₂</em> - (3.0 kg) <em>g</em> = 0   ==>   <em>n₂</em> = (3.0 kg) <em>g</em>   ==>   <em>f₂</em> = 0.30 (3.0 kg) <em>g</em>

• net horizontal force:

<em>c₂</em> - <em>f₂</em> = 0   ==>   <em>c₂</em> = 0.30 (3.0 kg) <em>g</em> ≈ 8.8 N

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5. Suppose a skydiver (mass = 75kg) is falling toward the Earth. When the skydiver is 100m above the Earth he is moving at 60m/s
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1) Gravitational potential energy: 73 500 J

2) Kinetic energy: 135 000 J

3) Mechanical energy: 208 500 J

Explanation:

1)

The gravitational potential energy of an object is the energy possessed by the object due to its position in the Earth's gravitational field, and it is given by

U=mgh

where

m is the mass of the object

g=9.8 m/s^2 is the acceleration of gravity

h is the height of the object relative to the ground

For the skydiver in this problem,

m = 75 kg is his mass

h = 100 m is his height above the Earth

Substituting, we find

U=(75)(9.8)(100)=73500 J

2)

The kinetic energy of a body is the energy due to its motion, and it is given by

K=\frac{1}{2}mv^2

where

m is the mass of the body

v is its speed

For the skydiver in this problem,

m = 75 kg is his mass

v = 60 m/s is his speed

Substituting, we find the kinetic energy

K=\frac{1}{2}(75)(60)^2=135000 J

3)

The mechanical energy of an object is the sum of the gravitational potential energy and the kinetic energy of the object:

E=U+K

where

U is the gravitational potential energy

K is the kinetic energy

For the skydiver in this problem,

U = 73,500 J is the gravitational potential energy

K = 135,000 J is the kinetic energy

Substituting, we find

E=73,500 + 135,000 = 208,500 J

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