a ball of mass 4 kg moving on a smooth horizontal surface makes an elastic collision with another ball of mass m at rest in the line of motion of first ball. if after collision first ball moves in the same direction with one fourth of its velocity before collision then mass of second ball is(A)4kg(B)4.4kg(C)2.4kg(D)2kg
Answers
m₁ = mass of first ball = 4 kg
v₁ = velocity of first ball before collision = v
v'₁ = velocity of first ball after collision = v/4
m₂ = mass of second ball = m
v₂ = velocity of second ball before collision = 0
v'₂ = velocity of second ball after collision = ?
using conservation of momentum
m₁ v₁ + m₂ v₂ = m₁ v'₁ + m₂ v'₂
inserting the above values
4 v + m (0) = 4 (v/4) + m v'₂
m v'₂ = 3 v
v'₂ = 3v/m eq-1
using conservation of kinetic energy for the elastic collision
m₁ v²₁ + m₂ v²₂ = m₁ v²'₁ + m₂ v²'₂
4 v² + m (0)² = 4 (v/4)² + m v²'₂
4 v² = (0.25) v² + m v²'₂
(3.75) v² = m v²'₂
using eq-1
(3.75) v² = m (3v/m)²
(3.75) v² = m (9 v²)/m²
(3.75) = (9)/m
m = 9/3.75
m = 2.4 kg
so the correct choice is "C"
Collision is elastic therefore e=1