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BITSAT 2009
Exam
BITSAT
Year
2009
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

BITSAT 2009 · Physics · Center of Mass, Momentum and Collision

A particle of mass 10 kg moving eastwards with a speed 5 ms^ -1...

A particle of mass 10kg10 \mathrm{ kg} moving eastwards with a speed 5ms−15 \mathrm{ ms}^{-1} collides with another particle of the same mass moving north-wards with the same speed 5ms−15 \mathrm{ ms}^{-1}. The two particles coalesce on collision. The new particle of mass 20kg20 \mathrm{ kg} will move in the north-east direction with velocity

Options

  1. A10ms−110 \mathrm{ ms}^{-1}
  2. B5ms−15 \mathrm{ ms}^{-1}
  3. C\quad(5 / \sqrt{2}) \mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm{ms}^{-1}
  4. Dnone of these

Correct answer

C. \quad(5 / \sqrt{2}) \mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm{ms}^{-1}

Step-by-step solution

Here i^mv+j^mv=2mV⃗\hat{i} m v+\hat{j} m v=2 m \vec{V} That is V→=v2(i^+j^)\overrightarrow{\mathrm{V}}=\frac{\mathrm{v}}{2}(\hat{\mathrm{i}}+\hat{\mathrm{j}}) Hence V=v2×2=v2⋅[V=\frac{v}{2} \times \sqrt{2}=\frac{v}{\sqrt{2}} \cdot \left[\right.Here \left.v=5 \mathrm{~ms}^{-1}\right]\text{ So, }V=V=\frac{5}{\sqrt{2}} \mathrm{~ms}^{-1}
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