Answer:
The velocity of the second car after the collision, v₂ = 11 m/s
The second car bounces back from its original direction.
Step-by-step explanation:
Given data,
The mass of the car going east, m₁ = 1200 kg
The velocity of the car going east, u₁ = 20 m/s
The mass of the second car, m₂ = 1000 kg
The velocity of the second car, u₂ = -25 m/s
The first car bounces back at v₁ = -10 m/s
According to the law of conservation of momentum
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
v₂ = (m₁u₁ + m₂u₂ - m₁v₁) / m₂
Substituting the given values,
v₂ = (1200 x 20 - 1000 x 25 + 1200 x 10)/1000
v₂ = 11 m/s
Hence, the velocity of the second car after the collision, v₂ = 11 m/s
The second car bounces back from its original direction.