question_answer 65)

                  A ball of mass m, moving with a speed \[2\,{{\upsilon }_{0}}\], collides inelastically \[(e>0)\] with an identical ball at rest. Show that                 (a) For head-on collision, both the balls move forward.                 (b) For a general collision, the angle between the two velocities of scattered balls is less than \[{{90}^{o}}\].                

Answer:

                  ((i) Let \[{{\upsilon }_{1}}\] and \[{{\upsilon }_{2}}\], be the velocity of two balls after collision.                 Initial momentum \[=m\times 2{{\upsilon }_{0}}=2m{{\upsilon }_{0}}\]                 Final momentum \[=m{{\upsilon }_{1}}+m{{\upsilon }_{2}}\]                 \[=m({{\upsilon }_{1}}+{{\upsilon }_{2}})\]                 According to the law of conservation of linear momentum \[2\,m\,{{\upsilon }_{0}}+m({{\upsilon }_{1}}+{{\upsilon }_{2}})\]                 or            \[2{{\upsilon }_{0}}={{\upsilon }_{1}}+\,{{\upsilon }_{2}}\]                         ? (i)                 Now,  \[\text{e = }\frac{\text{relative}\,\,\text{velocity}\,\,\text{of}\,\text{separation}}{\text{relative}\,\,\text{velocity}\,\,\text{of}\,\text{approach}}\]                 \[=\frac{{{\upsilon }_{2}}-{{\upsilon }_{1}}}{2{{\upsilon }_{0}}}\]                 or \[{{\upsilon }_{2}}-\,{{\upsilon }_{1}}=2e{{v}_{0}}\]                   ?. (ii)                 Using Eqn. (i), we get                 \[2{{\upsilon }_{0}}\,-{{\upsilon }_{1}}-\,{{\upsilon }_{1}}=2e{{\upsilon }_{0}}\] or \[{{\upsilon }_{1}}=\,{{\upsilon }_{0}}\,(1-e)\]                 and \[{{\upsilon }_{2}}={{\upsilon }_{0}}(1-e)\,+\,2e{{\upsilon }_{0}}\,={{\upsilon }_{0}}(1+\,e)\]                 Since e is positive, therefore, \[{{\vec{\upsilon }}_{1}}\] and \[{{\vec{\upsilon }}_{2}}\] have the same directions as that of the direction as that of the direction of \[{{\upsilon }_{0}}\].                 (b) In general collision, K.E. before collision is greater than die K.E. after collision                 Since, \[K.E.=\,\frac{{{p}^{2}}}{2m},\] where, \[p=m\upsilon \]                 \[\therefore \]\[\,\frac{{{p}^{2}}}{2m}>\,\frac{p_{1}^{2}}{2m}+\,\frac{p_{2}^{2}}{2m}\]                 or\[{{p}^{2}}>\,p_{1}^{2}+\,p_{2}^{2}\]                                    ?. (i)                      Let \[\theta \] be the angle between \[{{\vec{p}}_{1}}\] and \[{{\vec{p}}_{2}}\], then resultant of \[\vec{p}\] and \[{{\vec{p}}_{2}}=\,\,p_{1}^{2}+p_{2}^{2}+\,2{{p}_{1}}{{p}_{2}}\,\cos \theta \]                 \[=\,p_{1}^{2}+\,p_{2}^{2}\] if \[\theta \,\,={{90}^{o}}\]                 Thus, \[\theta \] must be less than \[{{90}^{o}}\].