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AIIMS 201925 May
Exam
AIIMS
Year
2019
Date
25 May
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

AIIMS 2019 · 25 May · Physics · Center of Mass, Momentum and Collision

A spacecraft of mass 100 ~kg breaks into two when its velocity is...

A spacecraft of mass 100 kg100 \mathrm{~kg} breaks into two when its velocity is 104 m s−110^4 \mathrm{~m} \mathrm{~s}^{-1}. After the break, a mass of 10 kg10 \mathrm{~kg} of the spacecraft is left stationary. The velocity of the remaining part is

Options

  1. A103 m s−110^3 \mathrm{~m} \mathrm{~s}^{-1}
  2. B11.11×103 m s−111.11 \times 10^3 \mathrm{~m} \mathrm{~s}^{-1}
  3. C11.11×102 m s−111.11 \times 10^2 \mathrm{~m} \mathrm{~s}^{-1}
  4. D104 m s−110^4 \mathrm{~m} \mathrm{~s}^{-1}

Correct answer

B. 11.11×103 m s−111.11 \times 10^3 \mathrm{~m} \mathrm{~s}^{-1}

Step-by-step solution

From conservation of momentum Mv=m1v1+m2v2M v=m_1 v_1+m_2 v_2 Here, M=100kg,v=104ms−1M=100 \mathrm{ kg}, v=10^4 \mathrm{ m} \mathrm{ s}^{-1}m2=90kg,v2=?∴100×104=10×0+90×v2 or v2=100×10490=11.11×103ms−1\begin{array}{ll} & m_2=90 \mathrm{ kg}, v_2=? \\ \therefore & 100 \times 10^4=10 \times 0+90 \times v_2 \\ \text{ or } & v_2=\frac{100 \times 10^4}{90}=11.11 \times 10^3 \mathrm{ m} \mathrm{ s}^{-1}\end{array}$$
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