Question

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

• A. m
• B. 2m
• C. 3m
• D. $\frac{3}{2}m$

Answer 1

Answer: (C)

$\frac{1}{2}{m}_1{u}_1^2-\frac{1}{2}{m}_1{v}_1^2=\frac{75}{100}\times \frac{1}{2}{m}_1{u}_1^2$
$\Rightarrow u_1^2-v_1^2=\frac{3}{4}{u}_1^2$
$\Rightarrow v_1=\frac{1}{2}{u}_1...(i)$
Now $v_1=\frac{(m_2-m_1)u_1}{(m_1+m_2)}...(ii)$
Thus, $\frac{1}{2}{u}_1=\frac{(m_2-m_1)u_1}{(m_1+m_2)}$
$\Rightarrow m_2=3m_1=3m$