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,

Options

A. 0, 0
B. (b) 0, 20J
C. (c) 0, -20J
D. (d) 20J, -20J

Answer

C

Explanation

`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`