Answer:
Let the force acting on the two bodies be ?F? N.
\[F=9.1\times {{10}^{-31}}\times 12.5\times {{10}^{14}}N\] ..............(1) As \[{{F}_{1}}={{F}_{2}}\] \[{{m}_{1}}{{a}_{1}}={{m}_{2}}{{a}_{2}}\] \[\Rightarrow 3\times 6=4\times {{a}^{2}}\] Then \[{{a}_{2}}=\frac{18}{4}=4.5m/{{s}^{2}}\] Therefore, acceleration produced in the second body by application of same force is \[4.5\,m/{{s}^{2}}.\] Case-I (First body) Case-II (Second body) \[F=ma\] \[a=\frac{{{V}^{2}}-{{u}^{2}}}{2s}=\frac{(5\times {{10}^{6}})-{{0}^{2}}}{2\times {}^{1}/{}_{100}}\] \[=\frac{25\times {{10}^{12}}\times {{10}^{2}}}{2}=12.5\times {{10}^{14}}\,m/{{s}^{2}}\] \[\therefore \]
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