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A large (4200 kg) truck collides with a small car (900 kg) that was originally at rest. The truck was mov? A large (4200 kg) truck collides with a small car (900 kg) that was originally at rest. The truck was moving at 34 m/s initially and 22 m/s immediately after the collision. a. What is the total momentum of the truck before the collision? b. What is the momentum of the truck after the collision? c. How fast is the car traveling after the collision?

By: Guest
Date: Unknown--
Response
0

Answer 1

Since c asks for velocity of the car, I assume elastic collision

=====> KE and Mo are both conserved

Mot0 = MV = 4200 * 34 m/s = 142,800 kg - m/s

Mot1 = MV = 4200 * 22 = 92,400 kg - m/s

del Mo = 142,800 - 92,400 = 50,400

Moc1 = 50,400

Moc1 = MV = 900 * V = 50,400

V = 56 m/s

Answer 2

Hello

You don't need to know if this is elastic or inelastic collision. You are only using the momentum equation, which applies to both.

Let Mt = mass of truck = 4200 kg

Mc = mass of car = 900 kg

Vti = Initial velocity of the truck = 34 m/s

Vci = Initial Velocity of the car = 0 m/s

Vtf = Final Velocity of the truck = 22 m/s

Vcf = Final Velocity of the car = ????

a) Initial Momentum of the truck (pti) = Mt*Vti

pti = 4200 kg * 34 m/s

pti = 142800 N-s

b) Final Momentum of the truck (ptf) = Mt *Vtf

ptf = 4200 kg * 22 m/s

ptf = 92400 N-s

c) Now for the fun part

Mt*Vti + Mc*Vci = Mt*Vtf + Mc*Vcf

Vci = 0

Mt*Vti = Mt*Vtf + Mc*Vcf

Mc*Vcf = Mt*Vti - Mt*Vtf

Mc*Vcf = Mt*(Vti - Vtf)

Vcf = (Mt/Mc)*(Vti - Vtf)

Vcf = (4200 kg/900 kg) * (34 m/s - 22 m/s)

Vcf = 56 m/s

Good Luck

[d] By: Guest
Date: Unknown---
Response
What is 1 + 100



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