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  4. An α - particle of mass m suffers one dimensional elastic collision with a nucleus of unknown mass. After the collision the α - particle is scattered directly backwards losing 75 % of its kinetic energy. Then the mass of the nucleus is

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Q. An $ \alpha - $ particle of mass $ m $ suffers one dimensional elastic collision with a nucleus of unknown mass. After the collision the $ \alpha - $ particle is scattered directly backwards losing $ 75\% $ of its kinetic energy. Then the mass of the nucleus is

Punjab PMETPunjab PMET 2010Nuclei

Solution:

$ \frac{1}{2} m_1 \,u_1^2 - \frac{1}{2} m_1\, v_1^2 = \frac{75}{100} \times \frac{1}{2} m_1\, u_1^2 $
$ \Rightarrow u_1^2 - v_1^2 = \frac{3}{4} u_1^2 $
$ \Rightarrow v_1 = \frac{1}{2} u_1 ...(i) $
Now $ v_1 = \frac{(m_2 - m_1)u_1}{(m_1 + m_2)} ...(ii) $
Thus, $ \frac{1}{2} u_1 = \frac{(m_2 - m_1)u_1}{(m_1 + m_2)} $
$ \Rightarrow m_2 = 3 m_1 = 3m $