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 ______.

विकल्प

• 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

`1/2"mv"_1^2=0.36xx1/2"mv"_0^2`

⇒ v₁ = 0.6 v₀

The collision is elastic. So,

`1/2"Mv"_2^2=0.64xx1/2"mv"_0^2`    [∴ M = mass of nucleus]

⇒ v₂ = `sqrt("m"/"M")xx0.8"v"_0`

mv₀ = `sqrt"mM"xx0.8"v"_0-"m"xx0.6"v"_0`

⇒ 1.6 m = 0.8`sqrt"mM"` 

⇒ 4m² = mM

∴ M = 4m