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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    A body is initially at rest on a smooth surface. A force F, whose time variation is shown in the adjacent figure acts on it for a duration of 4 s. The momentum of the ball at the end of the 4 s is (in N-s):

    A)  10                         

    B)         20

    C)  30                         

    D)         40

    E)  50

    Correct Answer: D

    Solution :

    From the definition of impulse, we have \[Fdt=dp\] Momentum is area under\[F-t\]graph. \[\therefore \] \[dp=\left( \frac{1}{2}\times base\times height \right)\]                                                 \[+\left( \frac{1}{2}base\times height \right)\] \[dp=\frac{1}{2}\times 2\times 20+\frac{1}{2}\times 2\times 20\]                 \[dp=40\]


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