Answer:
The velocity of the car after the collision is -5.36 m/s
Step-by-step explanation:
An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies remains the same.
let the following:
m₁ = mass of the car = 1400 kg
m₂ = mass of the truck = 3200 kg
u₁ = velocity of the car before collision = 13.7 m/s
u₂ = velocity of the truck before collision = 0 m/s
v₁ = velocity of the car after collision
v₂ = velocity of the truck after collision
v₁ = [ u₁ * (m₁ - m₂) + u₂ * 2m₂ ]/ (m₁ + m₂)
= [ 13.7 * (1400 - 3200) + 0 * 2 * 3200 ]/ (1400 + 3200)
= - 5.36 m/s
So, <u>the velocity of the car after the collision is -5.36 m/s</u>
Do you have a picture so that ik what the track looks like
If you divide 2380 by 4, it will give you 10% of the whole:
1380/4=345
Then you can times 345 by 6 To find the 60%:
345*6=2070
PS Please check my calculations with a calculator as I worked these out in my head.
This is just about substitution!! 9(5)-3(5), 3 • 5 = 15
9 • 5 = 45
so Now you have, 45 - 15 = 30
