BVP Medical BVP Medical Solved Paper-2006

  • question_answer
    A metal ball of mass 2 kg moving with a velocity of 36 km/h has a head on collision with a stationary ball of mass 3 kg. If after the collision, the two balls move together, the loss in kinetic energy due to collision is :

    A)  140 J                                    

    B)  100 J

    C)  60 J                                       

    D)  40 J

    Correct Answer: C

    Solution :

                    Key Idea: The first key idea is that final velocity v is given by the law of conservation of momentum. Initial momentum = Final momentum \[\therefore \]  \[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}=({{m}_{1}}+{{m}_{2}})v\] Given:   \[{{v}_{1}}=36km/h=36\times \frac{5}{18}=10m/s\]                 \[{{v}_{2}}=0\]                 \[{{m}_{1}}=2kg,\]                 \[{{m}_{2}}=3kg\] \[\therefore \]  \[v=\frac{{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}}{{{m}_{1}}+{{m}_{2}}}\]                 \[=\frac{2\times 10+3\times 0}{2+3}\] or            \[v=\frac{20}{5}=5m/s\] Loss in kinetic energy                 \[=\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}-\frac{1}{2}({{m}_{1}}+{{m}_{2}}){{v}^{2}}\]                 \[=\frac{1}{2}\times 2\times {{(10)}^{2}}+0-\frac{1}{2}(2+3)\times {{(4)}^{2}}\]                 \[=100-40=60J\]


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