Question

derive expression for
the velocities of two bodies
in terms of their initial Velocities before collision.
discuss the Special case also​

Answer 1

$\huge\mathfrak{\fcolorbox{red}{gray}{\green{A}\pink{N}\orange{S}\purple{W}\red{E}\blue{R}{!}}}$

Let two body of masses ${m}_{1}$ and ${m}_{2}$ moving with velocities ${u}_{1}$ and ${u}_{2}$ along the same straight line .

And consider that the two bodies collide and after collision ${v}_{1}$ and ${v}_{2}$ be the velocities of the masses .

________________________________________

Before collision :-

Momentum of mass ${m}_{1}$ = ${m}_{1}{u}_{1}$

Momentum of mass ${m}_{2}$ = ${m}_{2}{u}_{2}$

Total momentum before collision :-

${P}_{1}$ = ${m}_{1}{u}_{1}$ + ${m}_{2}{u}_{2}$

Kinetic energy of mass ${m}_{1}$ = $\frac{1}{2}{m}_{1}{{u}_{1}}^{2}$

Kinetic energy of mass ${m}_{2}$ = $\frac{1}{2}{m}_{2}{{u}_{2}}^{2}$

Thus,

Total Kinetic energy :-

K.E. = $\frac{1}{2}{m}_{1}{{u}_{1}}^{2}$ + $\frac{1}{2}{m}_{2}{{u}_{2}}^{2}$

___________________________________________

After collision :-

Momentum of mass ${m}_{1}$ = ${m}_{1}{v}_{1}$

Momentum of mass ${m}_{2}$ = ${m}_{2}{v}_{2}$

Total momentum after collision :-

${P}_{f}$ = ${m}_{1}{v}_{1}$ + ${m}_{2}{v}_{2}$

Kinetic energy of mass ${m}_{1}$ = $\frac{1}{2}{m}_{1}{{v}_{1}}^{2}$

Kinetic energy of mass ${m}_{2}$ = $\frac{1}{2}{m}_{2}{{u}_{2}}^{2}$

Total Kinetic energy :-

${K}_{f}$ = $\frac{1}{2}{m}_{1}{{u}_{1}}^{2}$ + $\frac{1}{2}{m}_{2}{{u}_{2}}^{2}$

__________________________________________

So according to the law of conservation of momentum,

${m}_{1}{u}_{1}$ + ${m}_{2}{u}_{2}$ = ${m}_{1}{v}_{1}$ + ${m}_{2}{v}_{2}$

=> ${m}_{1}({u}_{1} - {v}_{1})$ = ${m}_{2}({u}_{2} - {v}_{2})$ ------- (1)

__________________________________________

And according to the law of conservation of kinetic energy,

$\frac{1}{2}{m}_{1}{{u}_{1}}^{2}$ + $\frac{1}{2}{m}_{2}{{u}_{2}}^{2}$ = $\frac{1}{2}{m}_{1}{{v}_{1}}^{2}$ + $\frac{1}{2}{m}_{2}{{v}_{2}}^{2}$

=> ${m}_{1}({{u}_{1}}^{2} - {{v}_{1}}^{2})$ = ${m}_{1}({{u}_{2}}^{2} - {{v}_{2}}^{2})$

___________________________________________

Now dividing the equation,

$({u}_{1} + {v}_{1})$ = $({u}_{2} + {v}_{2})$

=> $({u}_{1} - {u}_{1})$ = $({v}_{1} - {v}_{2})$

Therefore, this is, relative velocity of approach is equal to relative velocity of separation.