Manipal Medical Manipal Medical Solved Paper-2010

  • question_answer
    A ball falls from 20 m height on floor and rebounds to 5 m. Time of contact is 0.02 s. Find acceleration during impact.

    A)  \[1200\text{ }m/{{s}^{2}}\]      

    B)  \[1000\text{ }m/{{s}^{2}}\]

    C)  \[2000\text{ }m/{{s}^{2}}\]      

    D)  \[1500\text{ }m/{{s}^{2}}\]

    Correct Answer: D

    Solution :

     According to Newtons second law of motion force acting on a body is equal to the rate of change of momentum during impact. \[F=\frac{\Delta p}{\Delta t}\] Also,     \[F=ma\] \[\therefore \] \[ma=\frac{{{p}_{2}}-{{p}_{1}}}{\Delta t}\] Or \[a=\frac{m{{v}_{2}}-(-m{{v}_{1}})}{m\Delta t}\] Or \[a=\frac{{{v}_{2}}+{{v}_{1}}}{\Delta t}\] \[\therefore \] \[a=\frac{\sqrt{2\times 10\times 20}+\sqrt{2\times 10\times 2}}{0.02}\] Or \[a=\frac{20+10}{0.02}\] \[=1500m/{{s}^{2}}\]


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