Energy Transfer in Collision of 200000N & 400000N Cars

  • Thread starter Thread starter Tubs
  • Start date Start date
  • Tags Tags
    Energy
AI Thread Summary
In a collision between a 200,000 N car traveling at 3 m/s and a stationary 400,000 N car, the kinetic energy (KE) before and after the collision must be calculated to determine energy transfer. The initial KE of the moving car is 0.5 * mass * velocity^2, which needs to be computed. After the collision, the combined mass of the two cars will move together, requiring momentum conservation to find the final velocity. The difference between the initial KE and the final KE reveals the energy lost to other forms during the collision. This analysis highlights that energy is not conserved in the system due to the inelastic nature of the collision.
Tubs
Messages
20
Reaction score
0
This question is making me go in circles, any tips?

A car weighing 200000 N traveling 3 m/s slams into a stationary car weighing 400000 N. The collision connects the two cars. How much energy is transferred from kinetic energy to other forms of energy in the collision?
 
Physics news on Phys.org
In this question, the energy of the system (ie ,the two cars) is not conserved. You have to find the KE of the cars before and after collision and the difference would give you the KE lost to other forms.
Can you find this difference? (Hint: Conservation of momentum)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top