CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    The recoil velocity of a 4.0 kg rifle that shoots a 0.050 kg bullet at a speed of\[280\text{ }m{{s}^{-1}}\]is:

    A)  \[+3.5\text{ }m{{s}^{-1}}\]      

    B)         \[-3.5\text{ }m{{s}^{-1}}\]

    C)  \[-\sqrt{(3.5)}\text{ }m{{s}^{-1}}\] 

    D)         \[+\sqrt{(3.5)}\text{ }m{{s}^{-1}}\]

    E)  \[\text{+7 }m{{s}^{-1}}\]

    Correct Answer: B

    Solution :

    \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}=0\] \[\Rightarrow \]               \[{{u}_{1}}=-\frac{{{m}_{2}}{{u}_{2}}}{{{m}_{1}}}\]                 \[=\frac{0.050\times 280}{4}=-3.5\,m{{s}^{-1}}\]


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