Answer:
14062.5 J
Explanation:
From the law of conservation of momentum,
Total momentum before collision = Total momentum after collision.
V = (m₁u₁ + m₂u₂)/(m₁+m₂).................1
Where V = common velocity after collision
Given: m₁ = m₂ = 25000 kg, u₁ = 2.5 m/s, u₂ = 1 m/s
Substitute into equation 1
V = [25000(2.5) + 25000(1)]/(25000+25000)
V = (62500+25000)/50000
V = 87500/50000
V = 1.75 m/s.
Note: The collision is an inelastic collision as such there is lost in kinetic energy of the system.
Total Kinetic energy before collision = kinetic energy of the first train car + kinetic energy of the second train car
E₁ = 1/2m₁u₁² + 1/2m₂u₂²........................ Equation 2
Where E₁ = Total kinetic energy of the body before collision, m₁ and m₂ = mass of the first train car and second train car respectively. u₁ and u₂ = initial velocity of the first train car and second train car respectively.
Given: m₁ = m₂ = 25000 kg, u₁ = 2.5 m/s, u₂ = 1 m/s
Substitute into equation 2
E₁ = 1/2(25000)(2.5)² + 1/2(25000)(1.0)²
E₁ = 12500(6.25) + 12500
E₁ = 78125+12500
E₁ = 90625 J.
Also
E₂ = 1/2V²(m₁+m₂)....................... Equation 3
Where E₂ = total kinetic energy of the system after collision, V = common velocity, m₁ and m₂ = mass of the first and second train car respectively.
Given: V = 1.75 m/s, m₁ = m₂ = 25000 kg
Substitute into equation 3
E₂ = 1/2(1.75)²(25000+25000)
E₂ = 1/2(3.0625)(50000)
E₂ = (3.0625)(25000)
E₂ = 76562.5 J.
Lost in kinetic Energy of the system = E₁ - E₂ = 90625 - 76562.5
Lost in kinetic energy of the system = 14062.5 J
