question_answer

A block of mass 1.2 kg moving at a speed of 20 cm/s collides head-on with a similar block kept at rest. The coefficient of restitution is 3/5. Find the loss of Identic energy during the collision.

Answer:

                Here, \[{{m}_{1}}=1.2kg\]¸\[{{u}_{1}}=20cm/s\] \[{{m}_{2}}=1.2kg\],            \[{{u}_{2}}=0\] If \[{{\upsilon }_{1}}\] and \[{{\upsilon }_{2}}\] are velocities of the two blocks after collision, then according to the principle of conservation of momentum, \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{\upsilon }_{1}}+{{m}_{2}}{{\upsilon }_{2}}\] \[1.2\times 20+0=1.2{{\upsilon }_{1}}+1.2{{\upsilon }_{2}}\]                                                                        ..... (i) velocity of approach       = \[{{u}_{1}}-{{u}_{2}}=20\,cm/s\] velocity of separation \[={{\upsilon }_{2}}-{{\upsilon }_{1}}\] By definition, \[e=\frac{{{\upsilon }_{2}}-{{\upsilon }_{1}}}{{{u}_{1}}-{{u}_{2}}}\] \[\frac{3}{5}=\frac{{{\upsilon }_{2}}-{{\upsilon }_{1}}}{{{u}_{1}}-{{u}_{2}}}\]     \[\therefore \]\[{{\upsilon }_{2}}-{{\upsilon }_{1}}=\frac{20\times 3}{5}=12\]                                               .?(ii) From (i) and (ii), \[{{\upsilon }_{1}}=4cm/s,\]\[{{\upsilon }_{2}}=16\,cm/s,\] Loss in K.E. \[=\frac{1}{2}{{m}_{1}}u_{1}^{2}-\frac{1}{2}\left( {{m}_{1}} \right)\upsilon _{1}^{2}-\frac{1}{2}{{m}_{2}}\upsilon _{2}^{2}=\frac{1}{2}\times 1.2{{\left( \frac{20}{100} \right)}^{2}}-\frac{1}{2}\left( 1.2 \right){{\left( \frac{4}{100} \right)}^{2}}-\frac{1}{2}\times 1.2{{\left( \frac{16}{100} \right)}^{2}}\] \[=2.4\times {{10}^{-2}}-0.096\times {{10}^{-2}}-1.536\times {{10}^{-2}}\] Loss in K.E. = \[0.768\times {{10}^{-2}}=7.7\times {{10}^{-3}}J\]