question_answer

Two balls A and B of masses 0.3 kg and 0.2 kg respectively are moving along positive X-axis and negative X-axis with velocities 2.0 m/s. They collide and thereafter move in the directions opposite to their original directions. Find the velocities of A and B after collision. Also, calculate total K.E. of the balls before and after collision.                                

Answer:

                Here, \[{{\text{m}}_{\text{1}}}=0.\text{3kg}\], \[{{\text{m}}_{\text{2}}}=0.\text{3kg}\] \[{{u}_{1}}=2.0\,m/s\] , \[{{u}_{2}}=-2.0\,m/s\] Let \[{{\upsilon }_{1}}=vel.\] of A after collision ; along negative X-axis,  Fig.4(HT).1.  \[{{\upsilon }_{2}}=vel.\] of B after collision; along positive X-axis According to the law of conservation of linear momentum,  \[{{m}_{1}}{{\upsilon }_{1}}+{{m}_{2}}{{\upsilon }_{2}}={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\] \[0.3\,{{\upsilon }_{1}}+0.2{{\upsilon }_{2}}=0.3\times 0.2+0.2\left( -2.0 \right)\] \[0.3\,{{\upsilon }_{1}}+0.2{{\upsilon }_{2}}=0.2\]                                               ..    (i) Relative velocity of separation after collision = Relative velocity of approach before collision \[{{\upsilon }_{2}}-{{\upsilon }_{1}}={{u}_{1}}-{{u}_{2}}=2.0-\left( -2.0 \right)=4.0\]                      ... (ii)   \[{{\upsilon }_{2}}=4+{{\upsilon }_{1}}\]                                                                 Put in (i), \[0.3{{\upsilon }_{1}}+0.2\left( 4+{{\upsilon }_{1}} \right)=0.2\] \[0.5{{\upsilon }_{1}}=0.2-0.8=-0.6\] \[{{\upsilon }_{1}}=-\frac{0.6}{0.5}=-1.2m{{s}^{-1}}\] \[\therefore \] Ball A moves with velocity 1.2 m/s along negative X-axis. From (ii),  \[{{\upsilon }_{2}}=4-1.2=2.8m/s\] \[\therefore \] Ball B moves with velocity 2.8 m/s along + X-axis. Total KE of balls before collision  \[=\frac{1}{2}{{m}_{1}}u_{1}^{2}=\frac{1}{2}{{m}_{2}}u_{2}^{2}=\frac{1}{2}[0.3{{\left( 2.0 \right)}^{2}}+0.2{{\left( -2.0 \right)}^{2}}]=1.0J\] Total KE of balls after collision \[=\frac{1}{2}{{m}_{1}}\upsilon _{1}^{2}+\frac{1}{2}{{m}_{2}}\upsilon _{2}^{2}=\frac{1}{2}[0.3{{\left( -1.2 \right)}^{2}}+0.2{{\left( 2.8 \right)}^{2}}]=1.0J\] Thus KE of balls is conserved during the elastic collision.