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KVPY Sample Paper KVPY Stream-SX Model Paper-2

  • question_answer
    Two identical trucks each of mass M (excluding sacks of rice) move on national highways with speeds \[{{v}_{1}}\] and \[{{v}_{2}}\] towards each other. When they meet each other, a sack of rice of mass \[m\]is thrown from one track to the other and an identical sack of rice is thrown from the second to the first. Calculate their velocities \[{{v}_{1}}'\] and \[{{v}_{2}}'\] after the exchange of sacks, given m = 50 kg, M=200kg, \[\left| \overrightarrow{\,{{v}_{1}}}\, \right|=50m/s\] and \[\left| \overrightarrow{\,{{v}_{2}}}\, \right|=200m/s\] Friction of the road may be neglected-

    A) \[0,-\,150\text{ }m/s\]

    B) \[150\text{ }m/s,-\,50\text{ }m/s\]

    C) 100 m/s, 150 m/s

    D) 100 m/s, 50 m/s

    Correct Answer: A

    Solution :

    [A]
    Let the first truck move to the right and the second to the left. Let us take the rightward direction as positive. In the horizontal direction, friction being absent, no external force is acting on the system. Hence momentum is conserved.
    The momentum carried away by a sack thrown from the first is \[m{{v}_{1}}.\] the momentum brought in the second sack to the first is \[m(-{{v}_{2}}).\]
    \[\therefore \] By the law of conservation of momentum of (truck +sacks of rice)
    \[(M+m)\,{{v}_{1}}-m{{v}_{1}}+m\,(-{{v}_{2}})=(M+m)\,{{v}_{1}}'\]
    Where \[{{v}_{1}}'\] is the new velocity of truck 1.
    \[\therefore M{{v}_{1}}-m{{v}_{2}}=(M+m)\,{{v}_{1}}'\]
    \[\therefore \,\,{{v}_{1}}'=\frac{M}{M+m}{{v}_{1}}-\frac{m}{M+m}{{v}_{2}}\]
    \[=\frac{200}{250}.50-\frac{50}{250}.200=0\]
    Similarly considering conservation of momentum of the second (truck \[\div \] sacks of rice), we get for the second truck.
    \[{{v}_{2}}'=\left[ \frac{M}{M+m}{{v}_{2}}-\frac{m}{M+m}.{{v}_{1}} \right]\]
    \[=-\,\left[ \frac{200}{250}\times 200-\frac{50}{250}\times 50 \right]=-\,150\,\,m/s.\]


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