Momentum of car and truck sytem.

[JJRKnights]

Homework Statement

A 1 175 kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the back of a 8 900 kg truck moving in the same direction at 20.0 m/s. The velocity of the car right after the collision is 18.0 m/s to the east.

(b) What is the change in mechanical energy of the car–truck system in the collision? (Use input values with an adequate number of significant figures to calculate this answer.)

m1 = 1175kg
m2 = 8900kg
v1i = 25m/s
v2i = 20m/s
vf = 20.583m/s

Homework Equations

K = 1/2mv^2
ME = U + K
U = mgh

The Attempt at a Solution

ΔK = (1/2m1v1f^2 + 1/2m2v2f^2) - (1/2m1v1i^2 + 1/2m2v2i^2)

am I on the right tracks?
if so how do i find v1f and v2f.
I know the relation:
20.583 = √(v1f^2 + v2f^2)
but there's 2 unknowns.

 

[anathema]

Yes, you are on the right track. To find v1f and v2f, you can use the conservation of momentum equation:

m1v1i + m2v2i = m1v1f + m2v2f

Substituting in the given values, we get:

(1175)(25) + (8900)(20) = (1175)(v1f) + (8900)(v2f)

Solving for v1f:

v1f = (1175)(25) + (8900)(20) - (8900)(v2f)
v1f = 29375 + 178000 - 8900v2f
v1f = 207375 - 8900v2f

Substituting this into the equation for mechanical energy change, we get:

ΔK = (1/2)(1175)(v1f)^2 + (1/2)(8900)(v2f)^2 - (1/2)(1175)(25)^2 - (1/2)(8900)(20)^2

Simplifying and substituting in the given values for vf, we get:

ΔK = (1/2)(1175)(20.583)^2 + (1/2)(8900)(20.583)^2 - (1/2)(1175)(25)^2 - (1/2)(8900)(20)^2

Solving for ΔK, we get:

ΔK = 153,159.22 J

Therefore, the change in mechanical energy of the car-truck system in the collision is approximately 153,159.22 J.