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Inelastic Collisions

Final Speed

If two objects with different masses and speeds are stuck together after the collision, the underlying process is called inelastic scattering, because deformations of both objects take place. Because both objects are connected after the collision, the speed of both objects has to be equal. Now we can formulate the momentum conservation law according to $$m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'$$ Adding the condition $v'_1 = v'_2 = v'$ results in $$m_1v_1 + m_2v_2 = (m_1+m_2)v'$$ In the last step, this formula has to be solved for $v'$, in order to obtain the final seed of both objects, and the following result is obtained.
Inelastic Collision


If two objects with the masses $m_1$ and $m_2$ and the speeds $v_1$ and $v_2$ collid totally inelastic, they move on with the speed of their center of mass: $$v' = \frac{m_1v_1+m_2v_2}{m_1+m_2}$$

This very important formula states, that both objects travel at the speed of the center of mass before the collision took place.
Example

A car with a mass of $1\,\mathrm{t}$ travels inside the city with a speed of $v = 50\,\mathrm{km/h}$ and crashes into a car with the same mass that stands at the entrance of a house. After the crash, both cars stuck with each other. In this case, $v_2$ is equal to $0$ and $v_1 = 50\,\mathrm{km/h}$. In addition, $m_1$ is equal to $m_2$. The final speed is then given as $$v' = \frac{m_1v_1}{2m_1} = \frac{1}{2}v_1 = 25\,\frac{\mathrm{km}}{\mathrm{h}}$$ In this special case, the final speed of both cars is equal to half of the speed of the moving car.

Special Cases

We can now take a look at some special cases: 1. $v_2 = 0$, $m_1 = m_2$: This condition means that the second object is in rest and both masses are identical. In this case, the resulting speed after the collision is always half of the speed of the first object. 2. $v_2 = -v1$, $m_1 = m_2$: Again, both masses are equal, but now both objects collide with the same speed in opposite directions. After the collision, both objects will be at rest because the resulting velocity gets $0$.

Energy Conservation

Now we can take a look at the energy before and after the collision. In the beginning, the total energy of both objects is given as $$E_\mathrm{kin} = \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$ The kinetic energy of both objects after the collision is given as $$E_\mathrm{kin} = \frac{1}{2}(m_1+m_2)v'^2 + \Delta E$$ where $\Delta E$ is the energy that is transferred into the objects by deformation. Solving this equation for $\Delta E$ and inserting the derived formula for $v'^2$, the difference in the kinetic energies is given as $$\boxed{\Delta E = \frac{1}{2}\frac{m_1\cdot m_2}{m_1 + m_2}(v_1-v_2)^2}$$ This equation is equal to the kinetic energy of the center of mass and shows that in the case of an inelastic collision the kinetic energy is not conserved. However, the total energy of a system cannot change. Therefore, this energy difference has to be transferred into the deformation of both objects and into the creation of heat. As one can see, the difference in the kinetic energy and therefore the deformation increases with the difference in the speeds prior to the collision.
Example

One truck with a mass of $30\,\mathrm{t}$ and a speed of $120\,\mathrm{km/h}$ hits a car with a mass $1\,\mathrm{t}$ which travels with a speed of $100\,\mathrm{km/h}$. The deformation energy is then given as $$\Delta E = 1807\,\mathrm{J}$$

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