A2578-kg van runs into the back of a 825-kg compact car at rest. they move off together at 8.5 m/s. assuming the friction with the road is negligible, calculate the initial speed of the van.

The initial speed of the car = 11.22 m/s

Explanation:

Law of conservation of energy: It states that when two bodied collide in a closed system, the sum of momentum before collision is equal to the sum of momentum after collision.

Note: A close system is one that is free from from external forces. E.g Frictional force.

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

m₁u₁ + m₂u₂ = (m₁ + m₂)V Equation 1

Where m₁ = mass of the van, m₂ = mass of the car, u₁ = initial velocity of the van, u₂ = initial velocity of the car, V = Common velocity.

Given: m₁ = 2578 kg, m₂ = 825 kg, u₂ = 0 ( the car was at rest), V= 8.5 m/s

Substituting these values into equation 1, and solving for u₁

2578(u₁) + (825 × 0) = (2578 + 825)8.5

2578u₁ = 28925.5

Dividing both side of the equation by the coefficient of u₁

2578u₁/2578 = 28925.5/2578

u₁ = 11.22 m/s

The initial speed of the car = 11.22 m/s

kinetic

i'm so sorry if this is wrong i'm only in 6th grade xd

prolly a beacuse the momentuem the car has increases while going down hill