Question

As observed in the laboratory system, a $6 \, MeV$ proton is incident on a stationary $12 \, C$ 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. $6.2\times 10^{6}ms^{- 1}$
• C. $10\times 10^{6}ms^{- 1}$
• D. $10ms^{- 1}$

Answer 1

Answer: (A)

$v_{c m}=\frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
$v_{c m}=\frac{\left(1\right) v_{P} + 12 \left(0\right)}{1 + 12}$
$v_{c m}=\frac{v_{P}}{13}$
$K.E.$ of proton,
$\frac{1}{2}mv_{P}^{2}=KE$
$\frac{1}{2}\left(1.67 \times 1 0^{- 27}\right)v_{P}^{2}=\left(6 \times 1 0^{6}\right)\left(1.6 \times 1 0^{- 19}\right)$
$v_{P}^{2}=12\times 10^{14}$
$v_{P}=2\sqrt{3}\times 10^{7}$
$\therefore $ $v_{c m}=\frac{2 \sqrt{3}}{13}\times 10^{7}$
$v_{c m}=2.6\times 10^{6} \, m \, s^{- 1}$