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Two identical `5kg` blocks are moving with same speed of `2ms^-1` towards each other along a frictionless horizontal surface. The two blocks collide, stick together, and come to rest. Consider the two blocks as a system. The works done by external and internal forces are, respectively,A. (a) 0, 0B. (b) 0, 20JC. (c) 0, -20JD. (d) 20J, -20J |
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Answer» Correct Answer - C `F_(ext)=0`, therefore, `W_(ext)=vecF_(ext)*vecs=0` By work-energy theorem: `W=K_F-K_I` `W_(ext)+W_(i nt)=0-(1/2mv^2+1/2mv^2)` `W_(i nt)=-mv^2=-5xx2^2=-20J` |
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