Inelastic Collision HW Problem

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Homework Statement


Pete got into into a car accident and collided with another vehicle, which resulted in the death of the second vehicles passenger, David. Whether Pete was speeding or not determines if he goes to jail. The speed limit where he got into the accident was 45mph. How fast was pete going at the point of collision and will he go to jail?

Given Data :
Mass of David’s Car(m2)
1115lbs

Mass of Pete’s car (m1)
1755lbs

Distance the cars skidded
13m

Final velocity post skid
0mph

Time of skid
22sec

Initial velocity of David’s car
47mph

Speed limit
45mph

Homework Equations


$$ (m1v1) + (m2v2) = (m1 + m2)Vf $$


The Attempt at a Solution


First I converted from lbs and mph to kg and m/s for the masses and velocity since we use kg and m/s in my class, I dont know if this was neccesary.
1151 lbs = 505.8 kg
1755 lbs = 796.1 kg
47mph = 21.01 m/s

Then I used my equation
$$ (m1v1) + (m2v2) = (m1 + m2)Vf $$
$$ (796.1*v1) + (505.8*21.01) = (796.1 + 505.8)0 $$
$$ (796.1*v1) + (10626.858) = 0 $$
$$ 796.1*v1/796 = -10626.8/796 $$
$$ v1 = 13.35 m/s $$

I then converted that from m/s to mph and got 29.86mph, which would mean he would not be speeding and not go to jail.
Although I got all the way through, i feel im wrong since they gave me the data for time and distance and i never even used them, which makes me feel like im forgetting something. Could someone help me out or check it to see if i did skip a step?
 

Answers and Replies

  • #2
Nathanael
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Bit of a morbid physics problem ?:)

But anyway... the [itex]V_f[/itex] in your relevant equation is the speed immediately after the collision. This speed is not zero, otherwise the cars would not have skid.
 
  • #3
Bystander
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Then I used my equation
(m1v1)+(m2v2)=(m1+m2)Vf​
What you want here rather than Vf = 0, is the velocity of the two cars at the beginning of the 13m skid that took 22 seconds. You might want to double check the problem statement for values for time and distance (distance seems a bit short, and time seems like forever).
 
  • #4
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Yes it is morbid!! My physics teacher is a former funeral director untill he became a teacher!!
Back to the problem...based on what u two said I think that might mean I have to do
$$ V=D/T $$
$$ V=13/22 $$
$$ .59 m/s $$

Then Maybe I can plug that into this which makes sense
$$ (796.1∗v1)+(505.8∗21.01)=(796.1+505.8).59m/s $$
$$ (796.1∗v1)+(10626.858)= 768.121 $$
$$ (796.1∗v1) = -9858.737 $$
$$ (796.1∗v1)/796.1 = -9858.73/796.1 $$
$$ -12.38 m/s $$

I find it weird this is negative which makes me think Im wrong but Ill just ignore the negative and convert... which means $$ 12.38 m/s = 27.69mph $$... meaning he wasnt speeding

does this seem better? still feels a little strange
 
  • #5
Nathanael
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You used [itex]V=\frac{total.distance}{total.time}[/itex]

Total distance divided by total time just gives you the average velocity (over the time that they were skidding) which is not what you want.
 
  • #6
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V=13/22
Have you ever seen anything skid at constant speed and stop instantaneously?
 
  • #7
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Ok guess not... what if I try this one
$$ D=½(VF+VI)T $$
$$ 13=½(VF+21.01)22 $$
$$ VF = -19.83 m/s $$

Then I do this thing again
$$ (796.1∗v1)+(505.8∗21.01)=(796.1+505.8)-19.83 $$
$$ (796.1∗v1)+(10626.858)=-25816.677 $$
$$ (796.1∗v1)=-36443.533 $$
$$ (796.1∗v1)/796.1=-36443.533/796.1 $$
$$ V1 = -45.78 m/s $$

Still getting a negative but Ill ignore it again and that converts into 102.4 mph which is WAY over the speed limit..... I think i might of messed up again?
 
  • #8
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Let's find a value for deceleration of the cars from the total distance and time.
 
  • #9
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Ok...... so I did half of that above i think? Because i need to get average velocity to plug into an acceleration equation?
$$ V=D/T $$
$$ V=13/22 $$
$$ .59m/s $$

Then a step further for acceleration
$$ A=V/T $$
$$ A=.59/22 $$
$$ A = .027 m/s^2 $$

Im not sure where that would leave me to go next though
 
  • #10
Nathanael
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Ok guess not... what if I try this one
$$ D=½(VF+VI)T $$
$$ 13=½(VF+21.01)22 $$
Plugging in random formulas is never a good idea; you have to picture the situation.
In this new equation you wrote, we're looking at the time interval when they're skidding, (that's why "D" is the distance they skid and "T" is the time they skid for) so "VF" is not the same "VF" as in the other equation. So in this case, the initial velocity "VI" corresponds to the velocity right after the collision, and the final speed "VF" is the speed after they finish skidding.

P.S.
You are also assuming that the deceleration of the cars after the collision is constant. You need to make this assumption for the problem to be solvable, but it's good to be aware that you're making this assumption.


Edit:
Sorry I'm a little late; I didn't notice that bystander already replied
 
  • #11
Nathanael
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$$ A=V/T $$
$$ A=.59/22 $$
$$ A = .027 m/s^2 $$
Again, you're using the average velocity. Acceleration is not "average velocity per time" it is "change in velocity per time."
 
  • #12
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Ok the the thing you said about their being two different VF's helps me visualize a little better, i was thinking of it as one big problem when really theres the stuff after the collision when its skidding and then the stuff when it collides.

Ok so let me try to stop messing up on the velocity. If i need change in velocity I think this is how I would do the equation
$$ A = Vf-Vi/T $$

Im stuck here though.......... I think the Vf is 0 but where do I get the Vi from...... because all I have is initial velocity
All I could think of is trying this
$$ D=½(Vf+Vi)T $$
$$ 13=½(0+Vi)22 $$
$$ Vi = 1.18 m/s $$

Wait maybe Im getting something...... could I then go here
$$ A = Vf-Vi/T $$
$$ A = 0-1.18/22 $$
$$ A = -.054 m/s^2 $$

which I assume would go positive? Sorry Im trying not to plug in equations... im just using this reply box as like my constant stream of thinking as i contemplate your responses.. I would still need to figure out what to do with acceleration if correct
 
  • #13
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Wait I dont think it goes positive actually... since its a negative acceleration as the car slows.. i am just scared of negative numbers i think haha
 
  • #14
Bystander
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And that deceleration acts for 22 seconds for what change in velocity?
 
  • #15
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And that deceleration acts for 22 seconds for what change in velocity?
I think
$$ A = ΔV/T $$
$$ -0.54 = ΔV/22 $$
$$ ΔV = -11.88 m/s $$

You said this is my change in velocity.... does that mean I cant put this in my VF all the way at the original equation? If so what would I do with this/where would I go now since I thought we needed VF.
 
  • #16
Nathanael
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I thought it was -0.054 not -0.54, so that gives you 1.188 m/s, but really it should be 1.18 because 0.054 is not the exact value for the acceleration, you rounded it.
The exact value of the acceleration was 1.18/22, so (1.18/22)*22=1.18
There's a good lesson on how premature rounding can affect your answer :oldtongue:

Anyway in post #12 you said Vi=1.18 m/s
What does this "Vi" represent? It was the velocity when they first started skidding, right? Which means it was the velocity right after the collision, right?
 
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  • #17
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A=−.054m/s 2
This is good.
−0.54=ΔV/22
This is getting sloppy transcribing numbers.
ΔV=−11.88m/s
Correct your arithmetic error, and this is the "initial" velocity of the wreckage following the collision, from which you do the calculation of Pete's velocity from the conservation of momentum.
 
  • #18
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First off, yeah I definitely made an error (Sorry :() the actual answer would be
$$ Vi = 1.18m/s $$

Now let me just write were in the post-collision world here so I remember. Next I think I have to get back on track and find the VF after the collision if I have the initial velocity
So I think Im now allowed to do this?
$$ D=½(Vf+Vi)T $$
$$ 13=½(Vf+1.18)22 $$
$$ Vf = .00182m/s $$

seems low to me kind of... should i keep moving forward or did i make an error
 
  • #19
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I kind of feel like Im getting lost in the problem now.... we are so far down (my fault since I just cant seem to comprehend it)
I'm gonna restate this.....
$$ (m1v1)+(m2v2)=(m1+m2)Vf $$

$$(796.1∗v1)+(505.8∗21.01)=(796.1+505.8)Vf $$

I really now just need help figuring out if the stuff I did in like my last two posts can help me get the right VF so i can find V1

I am operating under the assumption this is still the original equation I should be using
 
  • #20
Nathanael
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If you assume that the acceleration after the collision is constant, then you can say [itex]V_{avg}=\frac{V_f+V_i}{2}[/itex]. Since you know that the final velocity is zero, (they come to a stop) then the initial velocity (assuming constant acc.) is twice the average velocity. Now the average velocity is just the total distance per total time. You wrote it in post #9, it is 0.59 m/s, therefore the initial velocity (a.k.a. the velocity immediately after the collision) is 1.18 m/s

This would be the velocity you use for the conservation of momentum equation.
 
  • #21
Nathanael
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The whole challenge of the problem was to use the information about how far the cars skid, and how long it took, to find the speeds of the cars immediately after the collision (and then use conservation of momentum to find the speed of the car). To do this, you had to assume the deceleration was constant.

It got a little bit muddled, but hopefully you see the big picture now.
 
  • #22
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Ok, after reading that maybe I understand a little better not perfectly though. Is this how you wanted me to proceed?

$$ (796.1∗v1)+(505.8∗21.01)=(796.1+505.8)1.18 $$
$$ (796.1*V1)+(10610.05)=1536.242 $$
$$ 796.1Vl = -9073.808 $$
$$ 796.1Vl/796.1 = -9073.808/796.1 $$
$$ Vi = -11.40 m/s $$

Thats 25.5 mph..... I hope i did this right but I may of even messed myself up even more. I think it all hinged on whether I was supposed to put 1.18 where I just did in this equation
 
Last edited:
  • #23
Nathanael
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I didn't go through the calculation myself but it looks good.

You should get in to the habit of not plugging in the numbers until after you've manipulated the equations. It makes it easier for others and for yourself to read and spot errors. Plus, it can give insight in to what is going on.

Do you happen to know what part you're not perfectly understanding?
 
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  • #24
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Thanks. Ill try to format it better next time... I guess the only part I dont understand is why if were calling 1.18 m/s a velocity initial value... are we plugging it into Vf on the equation? I assume this has something to do with there being two different worlds (post collision and directly at point of collision)
 
  • #25
Nathanael
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Yeah. I wouldn't write the "Vf" in the conservation of momentum equation as "Vf" because it's not truly the "final velocity," I would write maybe [itex]V_{ac}[/itex] for "velocity after collision" (or something).

That's why it's good to understand formulas instead of remember them; if you just remember them, small things like that can throw you off.

Good luck with your physics class Dave!
 
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