Question

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 sad 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?

Answer 1

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.

Answer 2

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: https://brainly.com/question/9484203

2. Calculate average kinetic energy: https://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.