A body of mass m moving with an initial velocity u collides inelastically with a body of mass M initially at rest. If the collision is completely inelastic, then find (i) final velocity of combined system. (ii) loss in kinetic energy during collision.

(i) As the collision is completely inelastic therefore the two bodies stick together and the sticked bodies of mass  'm+M' move with final velocity let us say 'v'  which we have to find out 

If the external force on system is zero then by law of conservation of momentum of system , we have ,

 initial momentum of system = final momentum of system 

⇒mu +M×0 = (m+M) v 

⇒(m+M) v = mu

⇒v = mu ⁄ (m+M) 

Thus the above value of v is the required final velocity 

(ii)  initial kinetic energy of system =

      ½mu² +½ M×0² = ½mu²

Final kinetic energy of system  = ½(m+M) v²

Now putting the value of v= mu⁄(m+M) in final KE

       ∴final KE  =½(m+M) × m² u²⁄ (m+M) ²

                           =½m² u² ⁄ (m+M) 

∴ loss of kinetic energy = initial KE – final KE 

                     =½mu² – ½ m²u² ⁄(m+M) 

                     = {½mu²×(m+M) –½m²u²} ⁄ (m+M) 

                    ={½m²u² +½mMu² – ½m²u²}  ⁄ (m+M)

                  =½mMu² ⁄ (m+M)  

 ∴loss of kinetic energy =½mMu² ⁄ (m+M) 

(i) As the collision is completely inelastic therefore the two bodies stick together and the sticked bodies of mass  'm+M' move with final velocity let us say 'v'  which we have to find out 

If the external force on system is zero then by law of conservation of momentum of system , we have ,

 initial momentum of system = final momentum of system 

⇒mu +M×0 = (m+M) v 

⇒(m+M) v = mu

⇒v = mu ⁄ (m+M) 

Thus the above value of v is the required final velocity 

(ii)  initial kinetic energy of system =

      ½mu² +½ M×0² = ½mu²

Final kinetic energy of system  = ½(m+M) v²

Now putting the value of v= mu⁄(m+M) in final KE

       ∴final KE  =½(m+M) × m² u²⁄ (m+M) ²

                           =½m² u² ⁄ (m+M) 

∴ loss of kinetic energy = initial KE – final KE 

                     =½mu² – ½ m²u² ⁄(m+M) 

                     = {½mu²×(m+M) –½m²u²} ⁄ (m+M) 

                    ={½m²u² +½mMu² – ½m²u²}  ⁄ (m+M)

                  =½mMu² ⁄ (m+M)  

 ∴loss of kinetic energy =½mMu² ⁄ (m+M)