Question

derive equation for a loss of kinetic energy in case of completely inelastic collision in One dimension

Answer 1

this the right answer

Answer 2

Explanation:

The loss of kinetic energy during a perfectly inelastic collision is converted to another type of energy such as sound, thermal, heat, light, and so on.

Let $\mathrm{KE}_{\mathrm{i}}$ represent the total kinetic energy before the collision and $\mathrm{KE}_{\mathrm{f}}$ represent the entire kinetic energy after the impact.

Before a collision, the total kinetic energy,

$\mathrm{KE}_{i}=\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}$..........(i)

After-collision total kinetic energy

$K_{f}=\frac{1}{2}\left(m_{1}+m_{2}\right) v^{2}$..........(ii)

Then the loss of kinetic energy is

$\begin{aligned}&\text { Loss of KE, } \Delta \mathrm{Q}=\mathrm{KE}_{\mathrm{f}}-\mathrm{KE}_{\mathrm{i}} \\&=\frac{1}{2}\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right) \mathrm{v}^{2}-\frac{1}{2} \mathrm{~m}_{1} \\&\mathrm{u}_{1}^{2}-\frac{1}{2} \mathrm{~m}_{2} \mathrm{u}_{2}^{2} \ldots \ldots \text { (3) }\end{aligned}$

Substituting equation

$\mathrm{v}=\frac{m_{1} u_{1}+m_{2} u_{2}}{\left(m_{1}+m_{2}\right)} in equation$(3), and on simplying (expand v by using the algebra)

$(a+b)^{2}=a^{2}+b^{2}+2 a b$, we get

Loss of KE, $\Delta \mathrm{Q}=\frac{1}{2}\left(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\right)\left(\mathrm{u}_{1}-\mathrm{u}_{2}\right)^{2}$