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  • A moving mass of 8 kg collides elastically with a stationary mass of 2 kg.If KE be the initial kinetic energy of the moving mass, the kinetic energy left wit

A moving mass of 8 kg collides elastically with a stationary mass of 2 kg.If KE be the initial kinetic energy of the moving mass, the kinetic energy left with it after the collision will be :

Option: 1

0.4 KE


Option: 2

0.36 KE


Option: 3

0.30 KE


Option: 4

3.6 KE


Answers (1)

best_answer

Let v_1^{\prime}  be the final velocity of mass 8 \mathrm{~kg} and v_2^{\prime} be the final velocity of mass 2 \mathrm{~kg}.
Initially, mass of 2 \mathrm{~kg} is in rest, v_2=0
\begin{aligned} & \therefore \frac{1}{2} \times 8 \times v_1^2=1 \\ & \Rightarrow v_1=\frac{\sqrt{E}}{2} \end{aligned}
using formula,
\begin{aligned} v_2^{\prime} & =\frac{2 m_1 v_1}{m_1+m_2}-\frac{\left(m_1-m_2\right) v_2}{m_1+m_2} \\ \\& =\frac{2 \times 8 \times \sqrt{E}}{10 \times 2}=\frac{8}{10} \sqrt{E} \end{aligned}

v_1^{\prime}=\frac{\left(m_1-m_2\right) v_1}{m_1+m_2}+\frac{2 m_2 v_2}{m_1+m_2}

      =\frac{(8-2) \sqrt{B}}{2 \times 10}=\frac{3 \sqrt{E}}{10}

\text { Final k.e of } 8 \mathrm{~kg} \text { mass }=\frac{1}{2} \times 8 \times\left(\frac{3 \sqrt{6}}{10}\right)^2

                                                  =4 \times \frac{9 B}{100}

                                                  =0.36 \mathrm{E} .

Posted by

manish painkra

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