Physics, 12.11.2020 19:10 darricus

A 4.8-g particle is moving toward a stationary 7.4-g particle at 3.0 m/s. What percentage of the original kinetic energy is convertible to internal energy?

Answers: 2

Show answers

Answers

Answer from: juliestaten5984

60.185 percent of the original kinetic energy is convertible to internal energy.

Explanation:

Let suppose that collision between both particles is entirely inellastic. If there is no external forces exerted on any of the particles, then we can apply the Principle of Linear Momentum Conservation. That is:

$m_{A}\cdot v_{A,o} + m_{B}\cdot v_{B,o} = (m_{A}+m_{B})\cdot v$

$v = \frac{m_{A}\cdot v_{A,o}+v_{B}\cdot v_{B,o}}{m_{A}+m_{B}}$ (1)

Where:

$m_{A}$ - Mass of the 4.8-g particle, measured in kilograms.

$m_{B}$ - Mass of the 7.4-g particle, measured in kilograms.

$v_{A,o}$ - Initial speed of the 4.8-g particle, measured in meters per second.

$v_{B,o}$ - Initial speed of the 7.4-g particle, measured in meters per second.

- Final speed of the collided particles, measured in meters per second.

If we know that $m_{A} = 4.8\times 10^{-3}\,kg$ , $m_{B} = 7.4\times 10^{-3}\,kg$ , $v_{A,o} = 3\,\frac{m}{s}$ and $v_{B,o} = 0\,\frac{m}{s}$ , then the final speed of the system is:

$v = \frac{(4.8\times 10^{-3}\,kg)\cdot \left(3\,\frac{m}{s} \right)+(7.4\times 10^{-3}\,kg)\cdot \left(0\,\frac{m}{s} \right)}{4.8\times 10^{-3}\,kg+7.4\times 10^{-3}\,kg}$

$v = 1.180\,\frac{m}{s}$

During the collision part of the initial energy is dissipated in the form of heat, which is related to the internal energy ( $\Delta U$ ), measured in joules. According to the Principle of Energy Conservation, we have the following model:

$\Delta U = K_{A}+K_{B}-K$ (2)

Where:

$K_{A}$ , $K_{B}$ - Initial translational kinetic energies of each particle, measured in joules.

- Final translational kinetic energy of the collided particles, measured in joules.

By applying the definition of translational kinetic energy, we expand and simplify the equation above:

$\Delta U = \frac{1}{2}\cdot m_{A}\cdot v_{A,o}^{2}+\frac{1}{2}\cdot m_{B}\cdot v_{B,o}^{2} -\frac{1}{2}\cdot (m_{A}+m_{B})\cdot v^{2}$ (3)

If we get that $m_{A} = 4.8\times 10^{-3}\,kg$ , $m_{B} = 7.4\times 10^{-3}\,kg$ , $v_{A,o} = 3\,\frac{m}{s}$ , $v_{B,o} = 0\,\frac{m}{s}$ and $v = 1.180\,\frac{m}{s}$ , the internal energy associated with the system is:

$\Delta U = \frac{1}{2}\cdot (4.8\times 10^{-3}\,kg)\cdot \left(3\,\frac{m}{s} \right)^{2}+ \frac{1}{2}\cdot (7.4\times 10^{-3}\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-\frac{1}{2}\cdot (4.8\times 10^{-3}\,kg+7.4\times 10^{-3}\,kg)\cdot \left(1.180\,\frac{m}{s} \right)^{2}$

$\Delta U = 0.013\,J$

And the initial energy of both particles is:

$E_{o} = \frac{1}{2}\cdot (4.8\times 10^{-3}\,kg)\cdot \left(3\,\frac{m}{s}\right)^{2}+\frac{1}{2}\cdot (7.4\times 10^{-3}\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}$

$E_{o} = 0.0216\,J$

Lastly, the percentage of the original kinetic energy that is convertible to internal energy is: ( $\Delta U = 0.013\,J$ , $E_{o} = 0.0216\,J$ )

$\%e = \frac{\Delta U}{E_{o}}\times 100\,\%$ (4)

$\%e = \frac{0.013\,J}{0.0216\,J}\times 100\,\%$

$\%e = 60.185\,\%$

60.185 percent of the original kinetic energy is convertible to internal energy.

Answer from: Quest

the skeletons of most mammals reach a certain size and then stop growing. however, many animals, including some mammals, keep growing throughout their lives. kangaroos, for example, just keep growing and growing until they die. also, the longer an animal lives the greater the chance it might catch some deadly disease.

Answer from: Quest

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form