JCECE Medical JCECE Medical Solved Paper-2010

  • question_answer
    A ball of mass 10 kg is moving with a velocity of 10 m/s. It strikes another ball of mass 5 kg, which is moving in the same direction with a velocity of 4 m/s. If the collision is elastic, then their velocities after collision will be respectively

    A)  12 m/s, 6 m/s   

    B)  12 m/s, 25 m/s

    C)  6 m/s, 12 m/s    

    D)  8 m/s, 20 m/s

    Correct Answer: C

    Solution :

     Since, collision is elastic momentum remains conserved, hence we have Momentum before collision = Momentum after collision Initially, \[p={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}=10\times 10+5\times 4=120\]?(i) Final, \[p={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}=10{{v}_{1}}+5{{v}_{2}}\] ?(ii) Equating Eqs. (i) and (ii), we get \[10{{v}_{1}}+5{{v}_{2}}=120\] \[\Rightarrow \] \[2{{v}_{1}}+{{v}_{2}}=24\] ?(iii) Since collision is elastic relative velocity remains unchanged in magnitude but is reversed in direction, \[e=1=\frac{{{v}_{2}}-{{v}_{1}}}{{{u}_{1}}-{{u}_{2}}}\] \[\Rightarrow \] \[10-4={{v}_{2}}-{{v}_{1}}\] ?(iv) \[\therefore \] \[{{v}_{2}}-{{v}_{1}}=6\] Solving Eqs. (iii) and (iv), we get \[{{v}_{1}}=6m/s,{{v}_{2}}=12m/s\]


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