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A spacecraft of mass $100 \mathrm{~kg}$ breaks into two when its velocity is $10^4 \mathrm{~m} \mathrm{~s}^{-1}$. After the break, a mass of $10 \mathrm{~kg}$ of the spacecraft is left stationary. The velocity of the remaining part is
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Verified Answer
The correct answer is:
$11.11 \times 10^3 \mathrm{~m} \mathrm{~s}^{-1}$
From conservation of momentum
$M v=m_1 v_1+m_2 v_2$
Here, $M=100 \mathrm{~kg}, v=10^4 \mathrm{~m} \mathrm{~s}^{-1}$
$\begin{array}{ll} & m_2=90 \mathrm{~kg}, v_2=? \\ \therefore & 100 \times 10^4=10 \times 0+90 \times v_2 \\ \text { or } & v_2=\frac{100 \times 10^4}{90}=11.11 \times 10^3 \mathrm{~m} \mathrm{~s}^{-1}\end{array}$
$M v=m_1 v_1+m_2 v_2$
Here, $M=100 \mathrm{~kg}, v=10^4 \mathrm{~m} \mathrm{~s}^{-1}$
$\begin{array}{ll} & m_2=90 \mathrm{~kg}, v_2=? \\ \therefore & 100 \times 10^4=10 \times 0+90 \times v_2 \\ \text { or } & v_2=\frac{100 \times 10^4}{90}=11.11 \times 10^3 \mathrm{~m} \mathrm{~s}^{-1}\end{array}$
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