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JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2019
- Hard
- Numerical
An alpha-particle of mass \(m\) suffers \(1-\) dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing, \(64\%\) of its initial kinetic energy. The mass of the nucleus is .......... \(\mathrm{m}\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(4\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
\left( i \right)\,\,\,\,m{v_0} = - m{v_1} + m{v_2}\\
\left( {ii} \right)\,\,{v_0} = {v_1} + {v_2}\\
\,\,\,\,\,\,\,\,\,\frac{{2m{v_0}}}{{m + M}} = {v_2}\\
K{E_f} = \frac{1}{2}mv_1^2 = \frac{1}{2}m{\left( {\frac{{M - m}}{{M + m}}} \right)^2}v_0^2 = \frac{{36}}{{100}} \times \frac{1}{2}mv_0^2\\
\Rightarrow \,M = 4\,m.
\end{array}\)

\left( i \right)\,\,\,\,m{v_0} = - m{v_1} + m{v_2}\\
\left( {ii} \right)\,\,{v_0} = {v_1} + {v_2}\\
\,\,\,\,\,\,\,\,\,\frac{{2m{v_0}}}{{m + M}} = {v_2}\\
K{E_f} = \frac{1}{2}mv_1^2 = \frac{1}{2}m{\left( {\frac{{M - m}}{{M + m}}} \right)^2}v_0^2 = \frac{{36}}{{100}} \times \frac{1}{2}mv_0^2\\
\Rightarrow \,M = 4\,m.
\end{array}\)

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