Distance traveled during a head on collision

[Kevin Krock]

Homework Statement

During the head on collision of a 1500kg car moving 20 m/s and a 2400 kg truck moving 12.5 m/s does the car move farther than, a shorter distance than, the same distance as the truck, or is this indeterminate with the given information.

Homework Equations

m1(v1)+m2(v2)=(m1+v1)vf=0
average acceleration of car = 20/t
average acceleration of truck = 12.5/t

The Attempt at a Solution

the final velocity of both cars is 0 m/s. So if the duration of the collision is .5 seconds, then the average acceleration of the car is 40 m/s^2 and the truck is 24 m/s^2. the area under the graph -40x+20 (the velocity of the car) is 5 m. the area under the graph -24x+12.5 (velocity of truck) is 3.25 m. So this would mean the car travels a greater distance. My gut says that the car moves the same distance as the truck but that is not an adequate explanation for my professor. How do I calculate this??

 

[gneill]

Why do you conclude that the final velocities of both vehicles is zero? Are there enough details provided to determine the post-collision situation?

 

[Kevin Krock]

Why do you conclude that the final velocities of both vehicles is zero? Are there enough details provided to determine the post-collision situation?

the vehicle's final velocities are zero because 1500*(20)+2400*(-12.5) = (1500+2400)*0

This is ALL of the information in the problem and my professor grades extremely hard on explanations, so I'm trying to make sure that I can say the correct thing, here.

 

[Wasay]

the vehicle's final velocities are zero because 1500*(20)+2400*(-12.5) = (1500+2400)*0

This is ALL of the information in the problem and my professor grades extremely hard on explanations, so I'm trying to make sure that I can say the correct thing, here.

Homework Statement

During the head on collision of a 1500kg car moving 20 m/s and a 2400 kg truck moving 12.5 m/s does the car move farther than, a shorter distance than, the same distance as the truck, or is this indeterminate with the given information.

Homework Equations

m1(v1)+m2(v2)=(m1+v1)vf=0
average acceleration of car = 20/t
average acceleration of truck = 12.5/t

The Attempt at a Solution

the final velocity of both cars is 0 m/s. So if the duration of the collision is .5 seconds, then the average acceleration of the car is 40 m/s^2 and the truck is 24 m/s^2. the area under the graph -40x+20 (the velocity of the car) is 5 m. the area under the graph -24x+12.5 (velocity of truck) is 3.25 m. So this would mean the car travels a greater distance. My gut says that the car moves the same distance as the truck but that is not an adequate explanation for my professor. How do I calculate this??

Look when uh find the momentum of car and truck they both are same,regardless of car having less mass than truck the car has enough velocity to equalise the momentum with truck so if two different bodies with different mass but same momentum collide both bodies travel same distance or uh could say come at rest cancel each other's momentum. And for final velocity take it as zero

 

[phinds]

1500*(20)+2400*(-12.5) = (1500+2400)*0

That is an astounding piece of math. You might want to rethink the way you have expressed it.

 

[gneill]

the vehicle's final velocities are zero because 1500*(20)+2400*(-12.5) = (1500+2400)*0

This is ALL of the information in the problem and my professor grades extremely hard on explanations, so I'm trying to make sure that I can say the correct thing, here.

How do you know that the collision is perfectly inelastic? It wasn't mentioned in the problem statement. Nor were any details given about the structure of the vehicles regarding rigidity (safety crumple zones, etc.).

 

[Kevin Krock]

I know that the question says nothing about the elasticity of the collision but head on car collisions are generally inelastic so I took that as an assumption. This is the first homework assignment regarding forces and momentum and we really haven't gone over any of the equations except f=ma. Basically everything I've done has come from personal research. My professor makes us figure things out on our own and then grades homework like its an upper level physics course. So I am aware of all of the conditions I just am trying to think like an experienced physicist, even though I'm not, so that I can try to get some decent homework grades. If it is indeterminate I have to be able to THOROUGHLY explain why.

 

[NickAtNight]

explain why.

So could not each vehicle rebound with exactly the same velocity they collide with but in the opposite direction?

Why are you assuming the two objects become one object?

 

[NickAtNight]

As in:
m1(v1i)+m2(v2i)=m1(v1f)+m2(v2f)

 

[Kevin Krock]

As in:
m1(v1i)+m2(v2i)=m1(v1f)+m2(v2f)

so this would mean that the car would travel farther than the truck? or is it indeterminate because we don't know from the question whether the collision is elastic or inelastic?

 

[NickAtNight]

or is it indeterminate because we don't know from the question whether the collision is elastic or inelastic?

Very good. If it is fully elastic, what is the answer?

 

[gneill]

I know that the question says nothing about the elasticity of the collision but head on car collisions are generally inelastic so I took that as an assumption. This is the first homework assignment regarding forces and momentum and we really haven't gone over any of the equations except f=ma. Basically everything I've done has come from personal research. My professor makes us figure things out on our own and then grades homework like its an upper level physics course. So I am aware of all of the conditions I just am trying to think like an experienced physicist, even though I'm not, so that I can try to get some decent homework grades. If it is indeterminate I have to be able to THOROUGHLY explain why.

You'll want to lay out all your assumptions and justify them, then draw conclusions from there. If you're going with an inelastic collision as a model, state why and how well you think it will approximate real life. Definitely mention the initial total momentum and what you can conclude from that if the collision is inelastic.

How much of a vehicle has to be moving to say that the whole vehicle is moving? Most cars these days have lots of glass and plastic bits that shatter and scatter in a collision They are also built with crumple zones to absorb and dissipate the energy of a collision. Trucks tend to be less forgiving in their structure.

 

[Kevin Krock]

Very good. If it is fully elastic, what is the answer?

i would say that the car would travel farther than the truck if it was fully elastic because it has a greater velocity

 

[gneill]

i would say that the car would travel farther than the truck if it was fully elastic because it has a greater velocity

What will stop either vehicle? And does motion several minutes after the collision still count?

 

[NickAtNight]

i would say that the car would travel farther than the truck if it was fully elastic because it has a greater velocity

Good.

Do you not have an equation for the conservation of the energy?

Perhaps you should write out and apply this equation?

 

[NickAtNight]

What will stop either vehicle? And does motion several minutes after the collision still count?

Good point. It is distance not velocity that they asked for.

 

[Kevin Krock]

We haven't discussed frictional forces very much or how to calculate them so I know friction will stop both vehicles but I don't know how far either will go, I just know that the car will travel farther, having less mass and higher velocity.

 

[Kevin Krock]

If the collision is perfectly inelastic, the car will travel the same distance as the truck, because two bodies with equal momentum that undergo inelastic collision will accelerate to zero velocity at the point that the collision occurred. If the collision is fully elastic, then the kinetic energy before the collision will equal the kinetic energy after the collision; KE = .5*m*v^2 . This means that the car will have the same velocity before and after the collision pointed in opposite directions, thus traveling a greater distance than the truck during the collision.this is my explanation.

 

[NickAtNight]

this is my explanation.

Hmm, keep working on it.

Did you write out your Kinetic Energy equation yet?

How many variables do you have?
How many knowns?
How many unknowns?
How many equations do you have?

 

[NickAtNight]

We haven't discussed frictional forces very much or how to calculate them so I know friction will stop both vehicles but I don't know how far either will go, I just know that the car will travel farther, having less mass and higher velocity.

We you given any information to calculate frictional forces with?

Which vehicle has Michelin All-weather tires?

Is one of the roads iced up?

One of the vehicles may have a higher velocity, but without any information on the forces slowing the vehicles, how can you calculate any distance?

 

[NickAtNight]

What is the initial kinetic energy of the car?

What is the initial kinetic energy of the truck?

 

[Kevin Krock]

Well I did a bit more editing to my explanation. I have been on this problem for about six hours now. I must relax now. Thanks for your help though!

 

[haruspex]

That is an astounding piece of math. You might want to rethink the way you have expressed it.

Looks fine to me.

 

[haruspex]

Why do you conclude that the final velocities of both vehicles is zero? Are there enough details provided to determine the post-collision situation?

It says 'during the collision'. I think we can take that to mean up to the point where their mass centres are at closest approach. Since the total momentum is zero, that would mean when both vehicles are stationary.

 

[haruspex]

During the head on collision of a 1500kg car moving 20 m/s and a 2400 kg truck moving 12.5 m/s does the car move farther than, a shorter distance than, the same distance as the truck, or is this indeterminate with the given information.

You need to consider what is meant by the distance a vehicle travels during a collision. Not all parts of it will be moving at the same speed, nor travel the same distance. What might be a suitable definition?

 

[phinds]

Looks fine to me.

I said the way it is expressed, not the conclusion. The laws of arithmetic support the final conclusion of zero but not the figures as shown.

1500*(20)+2400*(-12.5) = (1500+2400)*0

 

[jbriggs444]

Concerns about the nature of frictional forces or about the elasticity of the collision may be avoided by noting that there is a particular unchanging relationship between the velocity of the car and the velocity of the truck. Momentum is not just the same after the collision as it was before the collision. It is also conserved throughout the duration of the collision.

Haruspex' hint in post#25 might help you to phrase that relationship precisely.

 

[haruspex]

I said the way it is expressed, not the conclusion. The laws of arithmetic support the final conclusion of zero but not the figures as shown.

1500*(20)+2400*(-12.5) = (1500+2400)*0

Still not seeing the issue. You'll need to be less subtle.

 

[jbriggs444]

Still not seeing the issue. You'll need to be less subtle.

Interpreted as mathematics, it could be seen as evaluating 1500 * 20 + 2400 * -12.5 and arriving at an intermediate result of (1500+2400)*0. While correct, it does not immediately follow -- it is not an intermediate result. That is the way I too looked at it at first. It was jarring.

Interpreted as physics, it can be seen as evaluating the momentum of the two bodies and equating that to the momentum of the combined body broken down as combined mass times combined velocity. From that viewpoint, it is a perfectly reasonable and meaningful equation.

Is that the 'gist of the difficulty being batted back and forth?

 

[phinds]

Still not seeing the issue. You'll need to be less subtle.

1500*(20)+2400*(-12.5) = (1500+2400)*0

Makes it look like he is equating 20 - 12.5 with 0

 

[haruspex]

1500*(20)+2400*(-12.5) = (1500+2400)*0

Makes it look like he is equating 20 - 12.5 with 0

Sorry, I still can't see how you can read it that way. Are you suggesting some interpretation of the asterisks other than multiplication?

 

[phinds]

Sorry, I still can't see how you can read it that way. Are you suggesting some interpretation of the asterisks other than multiplication?

See post #29 for someone who had an immediate response identical to mine.

 

[haruspex]

See post #29 for someone who had an immediate response identical to mine.

Well, I can't see how that fits with what you wrote in post #30, and jbriggs' "physics" interpretation is how I read it. Mr. Krock took the equation m₁v_1i+m₂v_2i=(m₁+m₂)v_f, plugged in the known values and found that v_f must be 0. He then posted that fact in the form of a solved equation. I'm sure it was not intended to represent the logic by which he arrived at the result. As a way of illustrating where he was up to, I have no problem with it.

 

[phinds]

Well, I can't see how that fits with what you wrote in post #30,

You're right. Post #30 was an incorrect way to express my dismay at his intermediate result.

I'm sure it was not intended to represent the logic by which he arrived at the result.

Well, I'm not a mind reader. I was commenting on what he posted, not what he meant (although actually, I addressed that as well).

As a way of illustrating where he was up to, I have no problem with it.

Well, I do. Let's agree to disagree.

 

[haruspex]

You're right. Post #30 was an incorrect way to express my dismay at his intermediate result.

Well, I'm not a mind reader. I was commenting on what he posted, not what he meant (although actually, I addressed that as well).

Well, I do. Let's agree to disagree.

Ok, all good.