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Linear momentum question, need help

  • Thread starter 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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  • #2
PhanthomJay
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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?
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 ).
 
  • #3
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yeah i tried -9000 and i still got it wrong.
 
  • #4
PhanthomJay
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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).
 
  • #5
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Yeah i tried that too, and it said it was wrong
 
  • #6
PhanthomJay
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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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