Physics Numerical Problems No. 3.7, Unit-3: Motion and Forces HSSC-I (class 11) Physics (Punjab Curriculum Textbook Board, Lahore) With Verified Answers

An electron (m = 9.1 ｘ 10⁻³¹ kg) traveling at 2 ｘ 10⁷ m s⁻¹ undergoes a head on collision with a hydrogen atom (m = 1.67 ｘ 10⁻²⁷kg), which is initially at rest. Assume the collision to be perfectly elastic and motion to be a straight line, find the velocity of hydrogen atom.

(Ans: 2.2 ｘ 10⁴ m s⁻¹)

Given:

Mass of electron = m₁ = 9.1 ｘ 10⁻³¹ kg

Initial velocity of electron = v₁ = 2 ｘ 10⁷ m s⁻¹

Mass of Hydrogen = m₂ = 1.67 ｘ 10⁻²⁷kg

Initial velocity of Hydrogen = v₂ = 0 m s⁻¹ (at rest)

To Find:

Final velocity of Hydrogen = v₂' = ?

Solution:

The formula to find Final velocity v₂' of Hydrogen of mass m₂ is:

v₂' =

$\frac {(m₂-m₁)v₂}{m₁+m₂}$ +

$\frac {2m₁v₁}{m₁+m₂}$

by putting values

v₂' =

$\frac {(1.67 ｘ 10⁻²⁷kg - 9.1 ｘ 10⁻³¹ kg) (0 m s⁻¹)}{9.1 ｘ 10⁻³¹ kg + 1.67 ｘ 10⁻²⁷kg}$ +

$\frac {2 ｘ 9.1 ｘ 10⁻³¹ kg x (2 ｘ 10⁷ m s⁻¹) }{9.1 ｘ 10⁻³¹ kg+ 1.67 ｘ 10⁻²⁷kg}$

v₂' = 0 +

$\frac {2 ｘ 9.1 ｘ 10⁻³¹ kg x (2 ｘ 10⁷ m s⁻¹) }{9.1 ｘ 10⁻³¹ kg + 16700 ｘ 10⁻³¹kg}$

v₂' =

$\frac {36.4ｘ 10⁻²⁴ kg m s⁻¹}{16709.1 ｘ 10⁻³¹ kg }$

v₂' = 0.002178 ｘ 10⁷ m s⁻¹

v₂' = 2.178 ｘ 10⁴ m s⁻¹ -------------Ans.

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