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?

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:

(1)

Where:

- Mass of the 4.8-g particle, measured in kilograms.

- Mass of the 7.4-g particle, measured in kilograms.

- Initial speed of the 4.8-g particle, measured in meters per second.

- 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 , , and , then the final speed of the system is:

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

(2)

Where:

, - 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:

(3)

If we get that  , , , and , the internal energy associated with the system is:

And the initial energy of both particles is:

Lastly, the percentage of the original kinetic energy that is convertible to internal energy is: (, )

(4)

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

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.