JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Head on Inelastic Collision

(1) Velocity after collision : Let two bodies A and B collide inelastically and coefficient of restitution is e.

Where                        

\[e=\frac{{{v}_{2}}-{{v}_{1}}}{{{u}_{1}}-{{u}_{2}}}=\frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}}\]

\[\Rightarrow \] \[{{v}_{2}}-{{v}_{1}}=e({{u}_{1}}-{{u}_{2}})\]      \ \[{{v}_{2}}-{{v}_{1}}=e({{u}_{1}}-{{u}_{2}})\]                                       ...(i)

From the law of conservation of linear momentum     

\[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\]                         ...(ii)

By solving (i) and (ii) we get

\[{{v}_{1}}=\left( \frac{{{m}_{1}}-e{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{1}}+\left( \frac{(1+e)\,{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{2}}\] Similarly 

\[{{v}_{2}}=\left[ \frac{(1+e)\,{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right]\,{{u}_{1}}+\left( \frac{{{m}_{2}}-e\,{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{2}}\]

By substituting e = 1, we get the value of \[{{v}_{1}}\] and \[{{v}_{2}}\] for perfectly elastic head on collision.

(2) Ratio of velocities after inelastic collision : A sphere of mass m moving with velocity u hits inelastically with another stationary sphere of same mass.

\[\therefore \] \[e=\frac{{{v}_{2}}-{{v}_{1}}}{{{u}_{1}}-{{u}_{2}}}=\frac{{{v}_{2}}-{{v}_{1}}}{u-0}\]

\[\Rightarrow \] \[{{v}_{2}}-{{v}_{1}}=eu\]                                                        ...(i)

By conservation of momentum :

Momentum before collision = Momentum after collision

\[mu=m{{v}_{1}}+m{{v}_{2}}\]

\[\Rightarrow \] \[{{v}_{1}}+{{v}_{2}}=u\]                                                             ...(ii)

Solving equation (i) and (ii) we get \[{{v}_{1}}=\frac{u}{2}(1-e)\]

and \[{{v}_{2}}=\frac{u}{2}(1+e)\]

\[\therefore \] \[\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{1-e}{1+e}\]

(3) Loss in kinetic energy     

Loss in K.E. (DK) = Total initial kinetic energy                  

- Total final kinetic energy

= \[\left( \frac{1}{2}{{m}_{1}}u_{1}^{2}+\frac{1}{2}{{m}_{2}}u_{2}^{2} \right)-\left( \frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2} \right)\]

Substituting the value of \[{{v}_{1}}\] and \[{{v}_{2}}\] from the above expressions

Loss (DK) = \[\frac{1}{2}\left( \frac{{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\,(1-{{e}^{2}})\,{{({{u}_{1}}-{{u}_{2}})}^{2}}\]

By substituting e = 1 we get \[\Delta K=0\] i.e. for perfectly elastic collision, loss of kinetic energy will be zero or kinetic energy remains same before and after the collision.