Question

As observed in the laboratory system, a $6MeV$ proton is incident on a stationary $12C$ target. The velocity of center of mass of the system is (Take mass of proton to be $1amu$ )

• A. $2.6\times 10^{6}m{s}^{-1}$
• B. $6.2\times 10^{6}m{s}^{-1}$
• C. $10\times 10^{6}m{s}^{-1}$
• D. $10m{s}^{-1}$

Answer 1

Answer: (A)

$v_{cm}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$
$v_{cm}=\frac{\left(1\right)v_P+12\left(0\right)}{1+12}$
$v_{cm}=\frac{v_P}{13}$
$K.E.$ of proton,
$\frac{1}{2}m{v}_P^2=KE$
$\frac{1}{2}\left(1.67\times 10^{-27}\right)v_P^2=\left(6\times 10^6\right)\left(1.6\times 10^{-19}\right)$
$v_P^2=12\times 10^{14}$
$v_P=2\sqrt{3}\times 10^7$
$\therefore$ $v_{cm}=\frac{2\sqrt{3}}{13}\times 10^7$
$v_{cm}=2.6\times 10^{6}m{s}^{-1}$