Question

An alpha-particle of mass m suffers 1-dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing, 64% of its initial kinetic energy. The mass of the nucleus is ______.

Options

• 2 m

• 3.5 m

• 1.5 m

• 4 m 

Answer 1

An alpha-particle of mass m suffers 1-dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing, 64% of its initial kinetic energy. The mass of the nucleus is 4 m.

Explanation:

Using conservation of momentum,

mv₀ = mv₂ - mv₁

${\displaystyle \frac{1}{2}{\text{mv}}_1^2=0.36\times \frac{1}{2}{\text{mv}}_0^2}$

⇒ v₁ = 0.6 v₀

The collision is elastic. So,

${\displaystyle \frac{1}{2}{\text{Mv}}_2^2=0.64\times \frac{1}{2}{\text{mv}}_0^2}$    [∴ M = mass of nucleus]

⇒ v₂ = ${\displaystyle \sqrt{\frac{\text{m}}{\text{M}}}\times 0.8{\text{v}}_0}$

mv₀ = ${\displaystyle \sqrt{\text{mM}}\times 0.8{\text{v}}_0-\text{m}\times 0.6{\text{v}}_0}$

⇒ 1.6 m = 0.8${\displaystyle \sqrt{\text{mM}}}$ 

⇒ 4m² = mM

∴ M = 4m