question_answer 77)

                  Two identical steel cubes (masses 50 g, side 1 cm) collide head-on face to face with a speed of 10 cm/s each. Find the maximum compression of each. Young?s modulus for steel \[=Y=2\times {{10}^{11}}N/{{m}^{2}}\].                

Answer:

                  \[Y=\,\frac{F/A}{\Delta L/L}=\,\frac{FL}{A\Delta L}\]                 or \[F=\,\frac{YA\Delta L}{L}\]                 Let K be the spring constant or compression constant.                 \[\therefore \]\[K=\,\frac{F}{\Delta L}\,=\frac{YL}{L}\]                 \[=\,\frac{2\times \,{{10}^{11}}\times 1\,\times {{10}^{-4}}}{{{10}^{-2}}}\]                 \[=2\times {{10}^{7}}N{{m}^{1}}\]                 Kinetic energy of two cubes \[=2\times \,\frac{1}{2}\,m{{\upsilon }^{2}}\]\[=m{{\upsilon }^{2}}\]                 \[=50\times {{10}^{3}}\times {{10}^{2}}=5\times {{10}^{4}}J\]                 If \[\Delta L\] be the compression of each cube, then potential energy of two cubes at the time of collision.                 P.E. \[=\,2\,\times \,\frac{1}{2}\,k\,{{(\Delta L)}^{2}}\,=\,k\,{{(\Delta L)}^{2}}\]                 K.E. f cubes is converted into P.E.                 \[\therefore \]\[k\,{{(\Delta L)}^{2}}=\,5\times \,{{10}^{-4}}\]                 or \[\Delta L=\,\sqrt{\frac{5\,\times \,{{10}^{-4}}}{2\times {{10}^{7}}}}=5\,\times \,{{10}^{-6}}\,m\]