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>
First, find the GCF. For these two numbers, it is 36. (Because 36 times 1 is 36 and 36 times 2 is 72)
Divide both sides by this, which gives you a
2:1 ratio
From point A, draw a line segment at an angle to the given line, and about the same length. The exact length is not important. Set the compasses on A, and set its width to a bit less than one fifth of the length of the new line. Step the compasses along the line, marking off 5 arcs. Label the last one C. With the compasses' width set to CB, draw an arc from A just below it. With the compasses' width set to AC, draw an arc from B crossing the one drawn in step 4. This intersection is point D. Draw a line from D to B. Using the same compasses' width as used to step along AC, step the compasses from D along DB making 4 new arcs across the line. Draw lines between the corresponding points along AC and DB. Done. The lines divide the given line segment AB in to 5 congruent parts.
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