InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Two identical `5 kg` blocks are moving with same speed of `2 m//s` towards each other along a friction less horizontal surface . The two blocks collide , stick together and come to rest. Consider the two blocks as a system . The work done by the external and ijnternal force are respectively: |
|
Answer» Here, `m_(1)=m_(2)=5kg,` `v_(1)=2m//s, v_(2)=-2m//s` Final velocity `v=0` As no external forces are involves, therefore, `vec(F)_(ext)=0, W_(ext)=vec(F)_(ext).vec(s)=Zero` According to work energy theorem, Total work done `=` Change is KE `W_("ext")+W_("int")=` final KE - inital KE `=0-((1)/(2)m_(1)v_(1)^(2)+(1)/(2)m_(2)v_(2)^(2))` `0+W_(i nt)=-(1)/(2)[5(2)^(2)+5(-2)^(2)]=-20J` `:. W_("int")=-20J` Negative sign show that internal forces of action and reaction act on the two blocks in a direction opposite to their motion. |
|