Question

A proton moving with a velocity of $1.25\times {10}^5$m / s collides with a stationary helium atom. The fraction of K.E transferred from proton to helium is :

• A. $0.64$
• B. $0.36$
• C. $0.48$
• D. $1$

Answer 1

The correct option is A $0.64$

Let mass of proton be $m$.

Its velocity before collision $u_1=1.25\times {10}^{5}m/s$

Thus initial kinetic energy of proton $E_i=\frac{1}{2}m{u}_1^2$

Mass of helium atom $M=4m$

Its velocity before collision $u_2=0$

So, velocity of helium atom after collision is given by $v_2=\frac{(m_2-e{m}_1)u_2+(1+e)m_1{u}_1}{m_1+m_2}$

where $m_1=m$, $m_2=M=4m$ and $e=1$

$\therefore$ $v_2=\frac{(4m-m)\left(0\right)+(1+1)m{u}_1}{m+4m}=0.4u_1$

Thus kinetic energy of helium after collision $E_f=\frac{1}{2}M{v}_2^2$

$E_f=\frac{1}{2}\left(4m\right)\left(0.4u_1^2\right)=0.64\times \frac{1}{2}m{u}_1^2=0.64E_i$

Thus fraction of K.E transferred from proton to helium $\frac{E_f}{E_i}=0.64$