Change in internal energy of an inelastic collision.

AI Thread Summary
In an inelastic collision between a 2200 kg car and a 4500 kg truck, both vehicles stick together post-collision, with the car's pre-collision velocity at <35, 0, 0> m/s and the truck's at <-18, 0, 27> m/s. The attempt to calculate the change in internal energy involved using the kinetic energy before and after the collision, but the initial calculations were incorrect due to the use of vector notation. Kinetic energy should be treated as a scalar, requiring the magnitudes of the velocities to be calculated first. The correct approach involves recalculating the kinetic energies using these magnitudes to find the change in internal energy accurately. This adjustment is essential for obtaining the correct answer in the energy change calculation.
flyboy9
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Homework Statement



A car of mass 2200 kg collides with a truck of mass 4500 kg, and just after the collision the car and truck slide along, stuck together, with no rotation. The car's velocity just before the collision was < 35, 0, 0 > m/s, and the truck's velocity just before the collision was < -18, 0, 27 > m/s.

Homework Equations


Pf=Pi+F*time
Ef=Ei+Q+W

The Attempt at a Solution


I attempted to use the change in kinetic energy of the system to solve of the change in internal energy.

Ef=.5*m*v2
Ef=.5*6700*<-.597,0,18.134>2 <-- final velocity of the system was calculated in an earlier step and found to be correct
Ef,sys=<1194,0,1101620>

Ei,car=.5*2200*<35,0,0>2
=<1347500,0,0>
Ei,truck=.5*4500*<-18,0,27>2
=<72900,0,1640250
EI,sys=Ei,car+Ei,truck
= <1420400,0,1640250>

I then found the change in energy by Ef,sys-Ei,sys and then getting the magnitude to solve get change in internal energy. This however did not get me the correct answer. I am stuck and not sure what to try.
 
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flyboy9 said:
Ef=.5*m*v2
Ef=.5*6700*<-.597,0,18.134>2 <-- final velocity of the system was calculated in an earlier step and found to be correct
Ef,sys=<1194,0,1101620>

Ei,car=.5*2200*<35,0,0>2
=<1347500,0,0>
Ei,truck=.5*4500*<-18,0,27>2
=<72900,0,1640250
EI,sys=Ei,car+Ei,truck
= <1420400,0,1640250>
Looks like you're using vector notation to express the energy. KE is a scalar, not a vector--it has no components. Use the full speed when calculating KE.
 
so would I calculate the magnitudes of the velocities first?
 
flyboy9 said:
so would I calculate the magnitudes of the velocities first?
Yes.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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