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Home / Doubts and Solutions/JEE/Physics

A cone is falling at a speed of vo strikes and penetrates a block.The acceleration of the cone after the impact is - a=g-cx2 where c is a positive constant and x is penetration distance.If the maximum penetration distance is xm.Then find the value of xm

Asked by anujrudrakar25 19th August 2018, 11:26 PM
Answered by Expert
Answer:
The required equation of motion to solve this question is " V2 = u2 - 2×a×S " .
where v is final velocity which is zero in this case, u is initial velocity which is given as vo,  
a is retardation  and S is the the stopping distance which is given as xm .
 
then we have ,  0 = vo2 - 2×a×xm     or  begin mathsize 12px style x subscript m space end subscript equals space fraction numerator v subscript 0 superscript 2 over denominator 2 a end fraction space equals space fraction numerator v subscript 0 superscript 2 over denominator 2 cross times open parentheses g minus c x subscript m superscript 2 close parentheses end fraction end style  .................(1)
Eqn.(1) is cubic that can be solved only by numerical method by substituting the values of vo, g and c
Answered by Expert 20th August 2018, 11:31 AM
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