Trains colliding and latching together

  • Thread starter Thread starter mrshappy0
  • Start date Start date
  • Tags Tags
    Trains
AI Thread Summary
The discussion revolves around a physics problem involving two train cars colliding and latching together. The first train car has a mass of 550 kg and a speed of 7.2 m/s, while the final speed after the collision is 4.7 m/s. The conservation of momentum equation is applied, but there is confusion regarding the initial conditions of the second train car's motion. One participant initially assumed the second car was stationary, leading to an incorrect solution, while another confirmed that the momentum equation used was correct but yielded a mass of 292.55 kg for the second car. The key takeaway is the importance of accurately determining the initial conditions of both train cars to solve the problem correctly.
mrshappy0
Messages
97
Reaction score
0

Homework Statement


A train car with mass m1 = 550 kg is moving to the right with a speed of v1 = 7.2 m/s and collides with a second train car. The two cars latch together during the collision and then move off to the right at vf = 4.7 m/s.2)What is the mass of the second train car?

Homework Equations




momentum before = momentum after.

kinetic energy before does not equal kinetic after.

The Attempt at a Solution



okay, I assumed the second car is not moving and solved for v final using this:

m1v1+m2v2=(m1+m2)v final.

My answer was wrong. Any ideas. My guess is that I shouldn't assume the second train car isn't moving but I don't know another way to solve it in that case.
 
Physics news on Phys.org
How do u know that your answer is incorrect i attempted your problem your equation m1v1+m2v2=(m1+m2)vf appears correct to me it gives 550*7.2+0=(550+m2)4.7 and I got an answer of 292.55 kg.
 
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