Momentum collision in two dimensions

[menco]

Homework Statement

A light passenger vehicle weighing 1470 N collides with a train engine weighing 1.23 x 10^5
N, which was being moved from one rail siding to another. The train engine and the vehicle were entangled after the accident and from your measurements you have been able to determine they skidded 15 m before finally coming to rest at an angle of 68 degress to the crossing. The co-efficient of friction between the vehicle’s tyres and wet road surface
are 0.25. The eyewitness reports highlight the passenger vehicle drove straight through the stop sign and ignored the warning horn blasts of the train and continued onto the railway level crossing without due care. You need to ascertain the vehicle’s actual entrance speed to the crossing and whether the driver has exceeded the speed limit of 60 km/hr?

Homework Equations

FIMPACT=FFRICTION
F = μ * m * g
W = f * d
KE = 1/2mv^2
x component of momentum: (m1 + m2) * Vf * cos θ = m1 * v1i
y component of momentum: (m1 + m2) * Vf * sin θ = m2 * v2i
1470N = 150kg
1.23x10^5 = 12551kg
total mass = 12701kg

The Attempt at a Solution

I made the train (m1) travel in the x direction and the car (m2) is the y direction.

Ff = 0.25 * (12701) * 9.8 = 31117.45 N

W = f*d = 31117.45 * 15 = 466761.75J

KE = 466761.75 = 1/2 mv^2
v = 8.57 m/s

y component (car) = (12701)(8.57)(sin68) = 150 v2
v2 = 672.81 m/s
= 2422km/h

Now obviously that's a lot and i think it has to do with the fact that the car only weights 150kg so I think they made a mistake in the question. If I change the initial mass to 14700N I get a final result of 74.4 m/s (267.84 km/h) which also seems like a lot for a car.

 

[gneill]

Yeah, it looks like they mucked up the question somehow. I also don't see how they can justify the implication that the steel wheels of the train will have the same coefficient of friction as that of the rubber tires of the car. I'd think that the steel wheels with their sharp rim projections would bite into the ground/asphalt and change the scenario drastically.

 

[menco]

solved thanks can someone please delete?

 

[menco]

There was a change to the question so now the vehicle weighs 1470kg, Now I still get velocity of 8.57m/s after the vehicles collide.

Using this to find the initial speed of the vehicle in the x direction

(14021)(8.57sin68) = 1470 * u_car
u_car = 75.8 m/s
= 273 km/h

Now the problem is were were told the vehicle should have an initial velocity less than 130km/h. So I am not sure where I have gone wrong to me it looks like I have calculated it all correctly

 

[gneill]

Is it possible that the "angle to the crossing" is meant to be measured with respect to the road rather than the track? Thus the term "railroad crossing" refers to the road crossing the railroad, rather than the other way around...

 

[menco]

I think your right I tried it again last night changing the way I was looking at the angles and got the result I was expecting. Thanks

 

[a_skier]

did you get something like 30.852 km/h for your final answer? I tried this out and got that. Although one thing I didn't understand was what the angle was given for. Because how is that relevant to the speed of the car upon entering the crossing? Unless you know the length of the train engine you wouldn't be able to discern if it hit the fore or aft part of the train engine anyways right?

Also (I am a total noob to physics), but why did you talk about the x and y components of momentum? (not hating or anything I genuinely don't understand so it confused me haha. And what did Vf stand for?

Sorry for the dumb questions I'm just discovering my fascination with physics but have no formal teaching other than Alg II in school. HAHAHA