hey guys just wondering if anyone can help me...
Three identical freight cars have the velocities indicated.
Assuming that car B is first hit by car A, determine the velocity of each car after all collisions have taken place if all three cars get automatically coupled.
Velocity of car A is 7.5km/h
velocity of car B is 0km/h
velocity of car C is 6Km/h
car A is traveling to the right and car C to the left, car B is stationary in between.
i understand what the question is asking me to do but i don't understand how to do it without knowing the respective weights of the cars, it states the carts are identical but still surely there weights and distance between them need to be known?!?!
can anyone help me?
cheers
distance is of little importance since you use the conservation of momentum. Put 1 kg for each cars mass, or use m and you will find that mass if negligible as well should the conditions of quetsion account for idential cars and if your question is designed properly.
Since A and B collide first, what is the momentum of A+B after impact? Then do the same for ( A+B ) and C. 
Edit: It seems that condensing the parentheses above gave me this joyful monster: (A+
Edit: It seems that condensing the parentheses above gave me this joyful monster: (A+
cheers for your help guys I've followed it up and attempted it but it still doesn't look right to me but here's what i have:
assuming the cars collide in an inelastic collision the following equation can be used:
(ma+va(initial))+(mb+vb(initial))=(m1+m2)v(final)
removing the mass leaves us with an equation that says the combined initial velocities divided by two equals the final velocity. therefore the final velocity of (A+ will equal 3.25km/h.
when the combined cars (A+ collide with C (A+ will have double the mass of C therefore when this is put into the equation the overall speed of the three cars is 0.25km/h
is this correct? could someone please check and give me another hint as to where i might be going wrong, if i am indeed going wrong at all!?
assuming the cars collide in an inelastic collision the following equation can be used:
(ma+va(initial))+(mb+vb(initial))=(m1+m2)v(final)
removing the mass leaves us with an equation that says the combined initial velocities divided by two equals the final velocity. therefore the final velocity of (A+
will equal 3.25km/h.when the combined cars (A+
collide with C (A+ will have double the mass of C therefore when this is put into the equation the overall speed of the three cars is 0.25km/h is this correct? could someone please check and give me another hint as to where i might be going wrong, if i am indeed going wrong at all!?
Intuitively that sounds right, but aren't you supposed to tell us? 
so long as I'm correct in it being inelastic collisions rather than elastic collisions then I'm sure that my answers correct! the only other doubt i have comes from the fact that the question asks at what speed EACH of the cars are traveling at and to me that sounds like they all have different speeds, but hey i guess ill just keep my fingers crossed!
Let us know how it shakes out!
The conservation of momentium means that the total momentum when you start and when you end will be the same. Momentium is M*V
So, you have MaVa, MbVb, and McVc when you start.
You have (Ma+Mb+Mc)Vf when you are done.
So, how do you relate those terms?
The momentum equation would be:
MaVa + MbVb + McVc = (Ma+Mb+Mc)Vf
MaVa is the momentum of car a, and so on.
What this means is that the final momentum will be the combination of the momentums of each individual momentum, since the collision is inelastic.
When it is over, there will be three units of mass stuck together moving at some Vf.
Remember that velocity is a vector, so the direction counts
Assume mass = 1.
The middle car has no momentum, but its mass slows things down.
1*7.5 + 1*0 + 1*(-6) = 3Vf
Vf = 0.5
So, you have MaVa, MbVb, and McVc when you start.
You have (Ma+Mb+Mc)Vf when you are done.
So, how do you relate those terms?
The momentum equation would be:
MaVa + MbVb + McVc = (Ma+Mb+Mc)Vf
MaVa is the momentum of car a, and so on.
What this means is that the final momentum will be the combination of the momentums of each individual momentum, since the collision is inelastic.
When it is over, there will be three units of mass stuck together moving at some Vf.
Remember that velocity is a vector, so the direction counts
Assume mass = 1.
The middle car has no momentum, but its mass slows things down.
1*7.5 + 1*0 + 1*(-6) = 3Vf
Vf = 0.5
now i feel like an idiot, just went through my workings again everything was correct but i managed to make 7.5/2=3.25 not 3.75! cheers for your help guys!