Answer
given,
mass of body = 4 Kg
travelling at speed = 3 m/s
exploded into two fragments of equal mass i.e. = 2 Kg
energy after collision increases by 18 J
original kinetic energy
[tex]KE = \dfrac{1}{2}MV^2[/tex]
[tex]KE = \dfrac{1}{2}\times 4 \times 3^2[/tex]
KE = 18 J
energy increases after collision = 18 J
total energy = 36 J
now,
[tex]\dfrac{1}{2}m_1v_1^2 + \dfrac{1}{2}m_2v_2^2 = 36[/tex]
[tex]\dfrac{1}{2}v_1^2 + \dfrac{1}{2}v_2^2 = 36[/tex]
[tex]v_1^2+v_2^2 = 72[/tex]
using conservation of momentum
m₁v₁ + m₂v₂ = MV
2 v₁ + 2 v₂ = 4 x 3
v₁ + v₂ = 6-------------(1)
squaring both side in the above equation
v₁² + v₂² + 2v₁v₂ = 36
72 + 2v₁v₂ = 36
v₁v₂ = -18-------(2)
putting value of equation (1) in (2)
v₁ (-18 - v₁) = 6
-18 v₁ - v₁² = 6
v₁² + 18 v₁ + 6 = 0
on solving the above equation we get
v₁ = -0.34 m/s
putting value of v₁ in equation (1)
v₂ = 6 + 0.34
v₂ = 6.34 m/s
