Manipal Medical Manipal Medical Solved Paper-2008

  • question_answer
    A body of mass 2 kg has an initial velocity of 3 m/s along OE and it is subjected to a force of 4 N in a direction perpendicular to OE. The distance of body from 0 after 4 s will be

    A)  12m           

    B)  20m

    C)  8m              

    D)  48m

    Correct Answer: B

    Solution :

     The acceleration of the body perpendicular to \[OE\]is \[a=\frac{F}{m}=\frac{4}{2}=2m/{{s}^{2}}\] Displacement along OE, \[{{s}_{1}}=vt=3\times 4=12\,m\] Displacement perpendicular to OE \[{{s}_{2}}=\frac{1}{2}a{{t}^{2}}\] \[=\frac{1}{2}\times 2\times {{(4)}^{2}}=16m\] The resultant displacement \[s=\sqrt{s_{1}^{2}+s_{2}^{2}}\] \[=\sqrt{144+256}\] \[=\sqrt{400}\] \[=20m\]


You need to login to perform this action.
You will be redirected in 3 sec spinner