JEE Main & Advanced Physics NLM, Friction, Circular Motion JEE PYQ - NLM Friction Circular Motion

  • question_answer
    A projectile moving vertically upwards with a velocity of 200 \[\text{m}{{\text{s}}^{-1}}\]breaks into two equal parts at a height of 490 m. One part starts moving vertically upwards with a velocity of 400 \[\text{m}{{\text{s}}^{-1}}\].                    [JEE ONLINE 12-05-2012]
    How much time it will take, after the break up with the other part to hit the ground?

    A) \[2\sqrt{10}s\]

    B) 5 s

    C) 10 s

    D) \[\sqrt{10}s\]

    Correct Answer: C

    Solution :

    Momentum before explosion
    = Momentum after explosion
    \[m\times 200\hat{j}=\frac{m}{2}\times 400\hat{j}+\frac{m}{2}v\]
    \[=\frac{m}{2}\left( 400\hat{j}+v \right)\]\[\Rightarrow \]\[400\hat{j}-400\hat{j}=v\]
    \[\therefore \]\[v=0\]
    i.e., the velocity of the other part of the mass, v = 0
    Let time taken to reach the earth by this part be t Applying formula, \[h=ut+\frac{1}{2}g{{t}^{2}}\]
    \[490=0+\frac{1}{2}\times 9.8\times {{t}^{2}}\]\[\Rightarrow \]\[{{t}^{2}}=\frac{980}{9.8}=100\]
    \[\therefore \] \[t=\sqrt{100}=10\sec \]

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