JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Self Evaluation Test - Laws of Motion

  • question_answer
    Two bodies of masses 1 kg and 2 kg moving with same velocities are stopped by the same force. Then the ratio of their stopping distances is

    A) \[1:2\]

    B) \[2:1\]  

    C) (c)\[\sqrt{2}:1\]

    D) \[1:\sqrt{2}\]

    Correct Answer: A

    Solution :

    [a] \[S=\frac{{{V}^{2}}-{{U}^{2}}}{2a}\text{ and }a=F/m,\text{ so}\] \[\frac{{{S}_{1}}}{{{S}_{2}}}=\frac{\left( \frac{0-{{U}^{2}}}{2F/m} \right)}{\left( \frac{0-{{U}^{2}}}{2F/2m} \right)}=1.2\]


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