Calculating the Change in Mechanical Energy of a Car-Truck Collision

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Homework Help Overview

The discussion revolves around a car-truck collision problem involving the calculation of the change in mechanical energy of the system. The scenario includes a 1,260 kg car and an 8,000 kg truck, with specific initial and final velocities provided for both vehicles.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of conservation of momentum to find the truck's velocity after the collision and explore the calculation of the change in mechanical energy using kinetic energy formulas.
  • There are varying results for the change in mechanical energy, with some participants questioning the significance of figures and the correctness of their calculations.
  • Concerns are raised about the signs of the energy change and the potential for errors in the calculations.

Discussion Status

The discussion is ongoing, with multiple participants sharing their calculations and results. There is no clear consensus on the correct change in mechanical energy, as different values have been proposed and some participants express uncertainty about their methods.

Contextual Notes

Participants mention the possibility of significant figure issues affecting their results, particularly regarding the mass of the truck. There is also an indication that the problem may involve careful attention to signs in the calculations.

lefthand
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A 1 260 kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the back of a 8 000 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.

(a) What is the velocity of the truck right after the collision?
( i got 21.10 as my vt)...using this:m(c) . v(c) + m(t) . v(t) = m(c) . V (c) + m(t) . V(t)

m/s 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.)
(for part be i was using the formula: (1/2)m1(v1^2) + (1/2)m2(v2^2)-(1/2)m1(v1^2) + (1/2)m2(vt)^2, and i got my answer as 8790, then it said my sign was wrong, and made it to -8790, and its still wrong)...does anyone know how to do this problem?
 
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lefthand said:
A 1 260 kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the back of a 8 000 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.

(a) What is the velocity of the truck right after the collision?
( i got 21.10 as my vt)...using this:m(c) . v(c) + m(t) . v(t) = m(c) . V (c) + m(t) . V(t)

m/s 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.)
(for part be i was using the formula: (1/2)m1(v1^2) + (1/2)m2(v2^2)-(1/2)m1(v1^2) + (1/2)m2(vt)^2, and i got my answer as 8790, then it said my sign was wrong, and made it to -8790, and its still wrong)...does anyone know how to do this problem?
It might be a significant figure thing...one could probably argue that there is only 1 significant figure in the truck's mass of 8000 kg...so try -9000 joules? Otherwise , I get your same negative answer of -8790 J as the mechanical energy change ( a loss in mechanical energy ).
 
yeah i tried -9000 and i still got it wrong.
 
I did the math again...your final speed of the truck agrees with my number (21.1 m/s east), but for the energy change I get - 69000 J. Watch plus minus signs and carefully define terms(vc_car_initial, vt truck_initial , vc_car_final, vt_truck_final).
 
Yeah i tried that too, and it said it was wrong
 
Maybe we did the math wrong..what number do you get using the change in KE as

(1/2)mcar(vcarfinal^2) + (1/2)mtruck(vtruckfinal]^2) - (1/2)mcar(vcarinitial^2) - (1/2)mtruck(vtruckinitial^2)

I get different numbers every time I try so maybe the maths is no good or the book answer is no good.
 
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