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Q.
A proton moving with a velocity of $1.25\times 10^{5} \, m \, s^{- 1}$ collides with a stationary helium atom. The velocity of the proton after the collision is
NTA AbhyasNTA Abhyas 2020
Solution:
From momentum conservation
$\text{m}_{p}u_{1}=\text{m}_{p}v_{1}+\text{m}_{He}v_{2}$
$\Rightarrow u_{1}=v_{1}+4v_{2}$ ...(i)
For elastic collision $e=1$
$e=\frac{v_{2} - v_{1}}{u_{1}}$
$\Rightarrow u_{1}=v_{2}-v_{1}$ ...(ii)
From equation (i) and equation (ii), we get
$v_{2}=\frac{2}{5}u_{1}\text{ and }v_{1}=-\frac{3}{5}u_{1}$
$\therefore v_{1}=-\frac{3}{5}\times \text{1.25}\times 10^{5}ms^{- 1}=-\text{0.75}\times 10^{5}ms^{- 1}$
