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Centre of Mass, Linear Momentum, Collision — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsCentre of Mass, Linear Momentum, Collision5 Marks
Question
A neutron initially at rest, decays into a proton, an electron and an antineutrino. The ejected electron has a momentum of $1.4 \times 10^{-26}kg-m/s$ and the antineutrino $6.4 \times 10^{-27}kg-m/s$. Find the recoil speed of the proton:
  1. If the electron and the antineutrino are ejected along the same direction.
  2. If they are ejected along perpendicular directions.
  3. Mass of the proton $= 1.67 \times 10^{-27}kg$.
✓

Answer

Mass of proton $= 1.67 \times 10^{-27}$ Let $'V_p\ '$ be the velocity of proton.
Given, momentum of electron $= 1.4 \times 10^{-26}kg m/ \sec$. Given, momentum of antineutrino $= 6.4 \times 10^{-27}kg m/ \sec$
  1. The electron the antineutrino are ejected in the same direction. As the total momentum is conserved the proton should be ejected in the opposite direction.
$1.67 \times 10^{-27} \times V_p = 1.4 \times 10^{-26} + 6.4 \times 10^{-27} = 20.4 \times 10^{-27}$
$\Rightarrow\text{V}_{\text{p}}=\Big(\frac{20.4}{1.67}\Big)=12.2\text{m}/\text{sec}$ in the opposite direction.
  1. The electron antineutrino are ejected $\perp$ to each other.
Total momentum of electron and antineutrino,
$=\sqrt{(14)^2+(6.4)^2}\times10^{-27}\text{kg}\ \ \text{m}/\text{s}=15.4\times10^{-27}\text{kg}\ \ \text{m}/\text{sec}$
Since $, 1.67 \times 10^{-27}, V_p = 15.4 \times 10^{-27}kg m/s$
So $, V_p = 9.2m/s$

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