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

Efficiency is an action of doing something. for example when you take a test. good efficiency means you did well but bad effeciency or no effiency means you did bad.

Answer from: Quest

ι вєℓιєνє ιт ιѕ тнє fιяѕт ¢нσι¢є(α.)

нє ѕнσυℓ∂ ѕмιℓє ρσℓιтℓу αи∂ ѕρєαк тσ тнє ρєяѕσи ωнσ ιѕ ιитєяяυριиg αfтєя тнє ∂ιѕ¢υѕѕισи

нσρє ι ¢συℓ∂ нєℓρ.

Another question on Physics

Physics, 22.06.2019 00:00

Suppose that recording station n is located 200 km north of the epicenter of an earthquake, and recording station s is located 100 km south of the epicenter of the same earthquake. which of the following statements would be true?

Answers: 2

Answer

Physics, 22.06.2019 06:20

Oxygen undergoes an isentropic expansion from 120 degree c and 280 kpa absolute to 140 kpa absolute. determine the final temperature of the oxygen and the amount of work per kg of o_2 produced in the process if it is adiabatic.

Answers: 1

Answer

Physics, 23.06.2019 00:50

Aplayer uses a hockey stick to push a puck at a constant velocity across the ice. the weight of the puck is 170 n. the coefficient of friction is 0.0600. with what force must the player push the puck so that his force just equals the frictional force? 28.3 n 0.999 n 0.102 n 0.0104 n

Answers: 1

Answer

Physics, 23.06.2019 02:00

How can you measure the amplitude of a longitudinal wave?

Answers: 2

Answer

You know the right answer?

A 4.8-g particle is moving toward a stationary 7.4-g particle at 3.0 m/s. What percentage of the ori...

Questions