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ipn [44]
3 years ago
13

Imagine that you replace the block in the video with a happy or sad ball identical to the one used as a pendulum, so that the sa

d ball strikes a sad ball and the happy ball strikes a happy ball. the target balls are free to move, and all the balls have the same mass. in the collision between the sad balls, how much of the balls' kinetic energy is dissipated?

Physics
2 answers:
BARSIC [14]3 years ago
5 0

Half of the kinetic energy is dissipated after the collision of the balls.

Further explanation:

Here, we have to calculate the amount of dissipation of the kinetic energy when sad ball collides.

In the video sad ball does not rebound back after collision.

So, this is the case of the perfect inelastic collision.

Here, we have given that the mass of the each ball is same.

Consider the mass of the each ball is \boxed{m}.  

Initially,

Sad ball of pendulum is at rest, so its velocity will be \boxed{0{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}.  

Assume the velocity of another sad ball is \boxed{u}.  

So, from the conservation of the linear momentum, total momentum before collision is equal to the total momentum after collision.

Therefore,

\begin{aligned}mu+0\cdotm&=\left( {m+m}\right)v\\mu&=2mv\\v&=\frac{u}{2}\\\end{aligned}

 

Now, calculate the total kinetic energy before collision,

\begin{aligned}{K_1}&=\frac{1}{2}m{u^2}+\frac{1}{2}m{\left(0\right)^2}\\&=\frac{1}{2}m{u^2}\\\end{aligned}

 

Now, calculate the total kinetic energy after collision,

\begin{aligned}{K_2}&=\frac{1}{2}m{v^2}+\frac{1}{2}m{v^2}\\&=\frac{1}{2}\left( {2m{v^2}}\right)\\\end{aligned}

 

Now, substitute the value of the v in above equation.

\begin{aligned}{K_2}&=\frac{1}{2}\left( {2m{{\left( {\frac{u}{2}} \right)}^2}}\right)\\&=\frac{1}{2}\left({\frac{{2m{u^2}}}{4}} \right)\\&=\frac{1}{2}\cdot\frac{1}{2}m{u^2}\\\end{aligned}

 

Now, replace \frac{1}{2}m{u^2} with {E_1} in the above equation,

\fbox{\begin\\{K_2}=\dfrac{{{K_1}}}{2}\end{minispace}}

 

Here, the total kinetic energy after collision is half of the total kinetic energy before collision.

So, half of the kinetic energy is dissipated.

Learn more:

1. Change in momentum due to collision: brainly.com/question/9484203

2. Calculate average kinetic energy: brainly.com/question/9078768

3. Motion under friction https: //brainly.com/question/9484203.

Answer detail:

Grade: Senior School

Subject: Physics

Chapter: Impulse and momentum

Keywords:

Sad ball strikes, replace, sad balls, happy balls, used as a pendulum. Strikes, same masses, kinetic energy, dissipated, happy ball strikes, target balls, kinetic energy, momentum, conservation of momentum.  

satela [25.4K]3 years ago
4 0
The sad ball does not rebound after it strikes the block. This means that the collision is inelastic. If two sad balls collide with each other, we can assume completely inelastic collision. Since momentum is conserved, the kinetic energy during the collision would be twice that of each of the ball's, half of the kinetic energy of each ball will be dissipated.
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Answer:

The force, exerted by Big Ben on the Empire State Building is 2.66972 × 10⁻⁷ N

Explanation:

The question relates to the force of gravity experienced between two bodies

The given parameters are;

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The distance between the two Big Ben and the Empire State Building, r = 5,000,000 meters

By Newton's Law of gravitation, we have;

F=G \times \dfrac{M_{1} \times M_{2}}{r^{2}}

Where;

F = The force exerted by Big Ben on the Empire State Building and vice versa

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

M₁, M₂, and r are the given parameters

By plugging in the values of the parameters and the constant into the equation for Newton's Law of gravitation, we have;

F=6.67430 \times 10^{-11} \times \dfrac{1 \times 10^8 \times 1 \times 10^9}{(5,000,000)^{2}} = 2.66972 \times 10^{-7}

The force, 'F', exerted by Big Ben on the Empire State Building is F = 2.66972 × 10⁻⁷ N.

3 0
3 years ago
yvonne van gennip of the netherlands ice skated 10.0 km with an average speed of 10.8 m/s. suppose vang ennip crosses the finish
Sophie [7]

By conservation of momentum, we will find that the mass is 4.97 kg.

So in the original system, we have two objects, the bouquet of flowers of mass M that is not moving and Yvonne, which has a mass of 63 kg and a speed of 10.8 m/s.

Then the total momentum of this system is:

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Remember the conservation of momentum, thus, the final momentum must be equal to the above one.

In the final situation, Yvonne and the bouquet move together with a speed of 10.01 m/s,

Then the final momentum is:

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And that must be equal to the initial momentum, then we have the equation:

(63kg + M)*(10.01 m/s) = 680.4 kg*m/s.

630.63kg*m/s + M*10.01 m/s = 680.4 kg*m/s.

M*10.01 m/s = 680.4 kg*m/s - 630.63kg*m/s = 49.77 kg*m/s

M = (49.77 kg*m/s)/(10.01 m/s) = 4.97 kg

So the mass of the bouquet is 4.97 kg

If you want to learn more about momentum, you can read:

brainly.com/question/19636349

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2 years ago
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Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
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A student wants to create a 6.0V DC battery from a 1.5V DC battery. Can this be done using a transformer alone
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Answer:

Therefore, we need an invert, and a rectifier, along with the transformer to do the job.

Explanation:

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Hence, the transformer only take AC voltage as an input, and converts it to another AC voltage. So, the output voltage of a transformer is also AC voltage.

So, in order to convert a 6 V DC to 1.5 V DC we need an inverter to convert 6 V DC to AC, then a step down transformer to convert it to 1.5 V AC, and finally a rectifier to convert 1.5 V AC to 1.5 V DC.

<u>Therefore, we need an invert, and a rectifier, along with the transformer to do the job.</u>

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tensa zangetsu [6.8K]

Answer:

While the use of the type of transformer in a rectifier depends on the voltage requirement or to meet desired operating conditions, a step-down transformer is used mainly to reduce the voltage. It is used to bring the high AC voltage level to a reasonable value or the desired output voltage.

Explanation:

Hope it helps

Correct me if Im wrong

3 0
2 years ago
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