Physics problems -- i.e., I'm stumped

[basketcase]

Okay, my kid is working on two Physics problems and that is outside my reservior of knowledge. Here they are:

1. Lilly, whose mass is 45.0 kg, is ice skating with a constant speed of 7.00 m/s when she hits a rough patch of ice with a coefficient of friction of 0.0800. How long will it be before Lilly coasts to a stop?

2. In a train yard, train cars are rolled down a long hill in order to link them up with other cars (parked on level ground at the bottom of an incline). A car with a mass of 4000. kg starts to roll from rest at the top of a hill 5.0 m high, and inclined at an angle of 5.0 degrees to the horizontal. The coefficient of rolling friction between the train and the track is 0.050. What velocity would the car have if it linked up with 3 identical cars sitting on flat ground at the bottom of the track?

Probably, the answers to this have motocycling applications in terms of (problem 1) how long will one slide on his arse if he falls off his bike? And (problem 2) how long can one roll powerless after running out of fuel?

:type

[BradfordBenn]

Okay, my kid is working on two Physics problems and that is outside my reservior of knowledge. Here they are:

:type

It has only been about 15 years since I took physics and I can not remember this stuff. username probably knows the answers to these...

[basketcase]

I could probably send her to a homework site, but it's a lot more fun around here.

Come to think of it, "more fun around here" might be why I'm clueless about physics... :huh

[Crow18]

1. Lilly, whose mass is 45.0 kg, is ice skating with a constant speed of 7.00 m/s when she hits a rough patch of ice with a coefficient of friction of 0.0800. How long will it be before Lilly coasts to a stop?

Speaking from my experience coaching beginners' hockey, if you weigh 45kg, you're skating 7 m/s and you hit a rough patch of ice, you divide half the mass by the velocity and then round up to arrive at the the number of stitches they'll put in your chin.

[cjack]

Okay, my kid is working on two Physics problems and that is outside my reservior of knowledge. Here they are:

1. Lilly, whose mass is 45.0 kg, is ice skating with a constant speed of 7.00 m/s when she hits a rough patch of ice with a coefficient of friction of 0.0800. How long will it be before Lilly coasts to a stop?

2. In a train yard, train cars are rolled down a long hill in order to link them up with other cars (parked on level ground at the bottom of an incline). A car with a mass of 4000. kg starts to roll from rest at the top of a hill 5.0 m high, and inclined at an angle of 5.0 degrees to the horizontal. The coefficient of rolling friction between the train and the track is 0.050. What velocity would the car have if it linked up with 3 identical cars sitting on flat ground at the bottom of the track?

Probably, the answers to this have motocycling applications in terms of (problem 1) how long will one slide on his arse if he falls off his bike? And (problem 2) how long can one roll powerless after running out of fuel?

:type

1. v sq = vo sq +2ax or x = -vo sq/2a when v=0. Need to find a. F=uN and F=ma where u = coeff of friction and N = normal force or m*g. Friction force -F=ma=uN and N=mg (mass times accel due to gravity which is 9.8 m/sec sq). So -F=ma=umg or a=-ug and the mass drops out. So x=Vo sq/2(ug)
Then x = 7 sq/2(ug) or 49/2(.08*9.8) or 31.25 meters.

2. I think this one is potential energy minus the work done by friction is equal to the kinetic energy at the level place. Then since the kinetic energy is equal to mv sq/2, and if energy is conserved then it is shared by the 3m (m is the mass of one car), then the velocity would become one half of the single car when it hooked up.
So I think this is mgy - uNx (where u is the coeff of friction, x is the distance of the track and N is the normal force, g is the accel of gravity) = mV1 sq /2 . Or mgy - umg(sin 85 deg)(5/sin 5 deg) = mV1 sq/2. Again the m drops out. You can plug in the values and solve for V1. Then mV1 sq = 4mV2 sq. So V2 sq = V1 sq/4. I get around 3.24 m/sec.

I think.

[Belquar]

1. v sq = vo sq +2ax or x = -vo sq/2a when v=0. Need to find a. F=uN and F=ma where u = coeff of friction and N = normal force or m*g. Friction force -F=ma=uN and N=mg (mass times accel due to gravity which is 9.8 m/sec sq). So -F=ma=umg or a=-ug and the mass drops out. So x=Vo sq/2(ug)
Then x = 7 sq/2(ug) or 49/2(.08*9.8) or 31.25 meters.

2. I think this one is potential energy minus the work done by friction is equal to the kinetic energy at the level place. Then since the kinetic energy is equal to mv sq/2, and if energy is conserved then it is shared by the 3m (m is the mass of one car), then the velocity would become one half of the single car when it hooked up.
So I think this is mgy - uNx (where u is the coeff of friction, x is the distance of the track and N is the normal force, g is the accel of gravity) = mV1 sq /2 . Or mgy - umg(sin 85 deg)(5/sin 5 deg) = mV1 sq/2. Again the m drops out. You can plug in the values and solve for V1. Then mV1 sq = 4mV2 sq. So V2 sq = V1 sq/4. I get around 3.24 m/sec.

I think.

:doh :deal :thumb :cry :banghead :clap :sick :hungover :drink :bluduh :uhoh :snore :help :ha

[rinty]

Pretty good for an electrical guy, Jack.

Rinty

[username]

for problem 1, cjack is correct. he gets an A for getting the right answer, and he gets an A+ for showing his work. that's nice! :thumb

i'm not so sure on problem 2, i think it's a trickier question. it is possible that the goal of the exercise was to solve it in the manner that cjack did, but in doing so, it's wrong, and unrepresentative of the real world. if that is the case, the teacher/textbook author deserves an F.

when the single car comes down the hill and collides with the other three, and they stick together, this is an example of a perfectly inelastic collision. in this case, momentum must be conserved. it's the law. system KINETIC energy does not have to be conserved, which is why cjack's solution is incorrect. (sorry cjack.)

the car rolling down the hill is denoted by the subscript 1
the parked cars are denoted by the subscript 2.
momentum is called "p" by physicists.
the subscript "i" is for the initial part of the problem (one car rolling down the hill, the other three at rest) and the subscript "f" is for the final part of the problem, all four cars rolling together as one big mass.

(m1*V1)i + (m2*V2)i = (m1 + m2)*Vf

note that in the initial case, the three cars are at rest, so their velocity, V2i is zero.

the equation reduces to

m1*v1 = (m1 + m2)*Vf

we know that the cars waiting at the bottom of the hill are each equal to the mass of the first car, so

m2 = 3*m1

therefore the final mass is 4*m1

so solving for Vf we see

Vf = m1*V1/(4*m1)

the m1s cancel and the final velocity is one fourth the initial.

i calculated the initial velocity of the first car to be 6.46 m/sec. (this agrees with cjack's calculation as well.) so i get

Vf = 1.62 m/sec

one can fall into the trap of arguing that my solution does not obey the law of conservation of energy. that is incorrect, and here is why: in real life, when the items collide, a certain amount of energy is transmitted to the items. this is why your bike gets smashed up when you hit something. so the energy to bend/deform the object is part of the total energy, along with the kinetic energy. (before anyone argues, *everything* bends, even big tough rail cars.)

the amount of energy that is transmitted is determined by the conservation of momentum. this is how shock absorbers and crumple zone ares designed! it's way cool. first, you assume the system obeys conservaiton of momentum, and solve the problem as i did. note that if you attempt to balance the initial and final KINETIC energy in my solution, they do not balance. what that means is that the "missing" kinetic energy got turned into some other energy. (ultimately heat, but i will spare you that discussion.)

the difference in kinetic energy between the initial and final cases was 62,726 joules. that energy MUST be absorbed by the system. in this case, i'd expect most of it to be lost in the form of bending the joining mechanism between the two cars. so you can look at that design, and put that much energy into, and see how far it'd deflect, and then boom, you know if the mechanism will fail.

all of this applies to the design of things like the armor in your jacket, and the way your bike crumples when you smash into something bigger/faster than you.

i hope this helps.

[cjack]

When I make a mistake, I always learn something. I'm thinking now that this problem set was intended to remind us just what you wrote. That momentum is always conserved and energy is not except in totally elastic collisions. So good for the teacher in that case. A+ for you.