Find the final velocity of the bobsled after the last bobsledder jumps in

In summary, 4 bob sledders are applying 25N each for the first 3 seconds with an empty sled, then one jumps in, leaving 3 of them pushing with 25 N each for 1 second etc… am I interpreting this correct? I think we need the sledders masses to find the velocity at the end?
  • #1
paulimerci
287
47
Homework Statement
A four man bob sled team each push on the bobsled with a force of 25N. The first bob sledder pushes for 3 secs, the second for 4 secs, the third for 5 secs and the fourth one for 6 secs. What is the total impulse the bobsledders exerted on the bobsled? supposing the bobsled has a mass of 45kg, what is the final velocity of the bobsled after the last bobsledder jumps in?
Relevant Equations
Impulse and change in momentum
For calculating the total impulse, I'm taking the individual impact force times the elapsed time and adding them all together to get the total impulse. Given F = 25N, ## \Delta t_1 =3 secs##, ##\Delta t_2 = 4secs##, ## \Delta t_3 = 5secs##, ## \Delta t_4 = 6secs##.
$$ Impulse 1 = F \Delta t_1 $$
$$ = 25 \cdot 3 $$
$$ = 75 Ns $$

$$ Impulse 2 = F\Delta t_2$$
$$ = 100 Ns $$

$$ Impulse 3 = 125 Ns$$
$$ Impulse 4 = 150Ns$$
Total impulse = Impulse 1+ Impulse 2+ Impulse 3 + Impulse 4
= 450Ns.
For the second part of the problem, I'm unsure whether I should consider the bobsled at rest or in motion when all three sledders are in it. And it looks way more complicated because individual masses of sledders are not provided.
And the answer to the above is not correct. Can anyone help me solve this problem?
 
Physics news on Phys.org
  • #2
Are 4 bob sledders are applying 25 N each for the first 3 seconds with an empty sled, then one jumps in, leaving 3 of them pushing with 25 N each for 1 second etc… am I interpreting this correct? I think we need the sledders masses to find the velocity at the end?
 
  • #3
erobz said:
Are 4 bob sledders are applying 25 N each for the first 3 seconds with an empty sled, then one jumps in, leaving 3 of them pushing with 25 N each for 1 second etc… am I interpreting this correct? I think we need the sledders masses to find the velocity at the end?
Yes that’s what it says in the question.
 
  • #4
We’ll the first phase of impulse is what?
 
  • #5
erobz said:
We’ll the first phase of impulse is what?
I don't understand. Are you asking about the first phase of impulse? I found that to be 75Ns. Was it wrong?
 
  • #6
paulimerci said:
I don't understand. Are you asking about the first phase of impulse? I found that to be 75Ns. Was it wrong?
If 4 people are pushing with 25N each for the first 3 seconds. Then yeah, I think you have the impulse wrong for that phase.
 
  • Like
Likes paulimerci
  • #7
erobz said:
If 4 people are pushing with 25N each for the first 3 seconds. Then yeah, I think you have the impulse wrong for that phase.
Thank you; you're correct. I got a total impulse of 1000 Ns. That's the answer on the answer sheet. I'm wondering how to solve the second part of the problem.
 
  • #8
paulimerci said:
Yes that’s what it says in the question.
erobz said:
If 4 people a pushing with 25N each for the first 3 seconds. Then yeah, I think you have the impulse wrong for that phase.
First, it doesn’t actually say that in post #1; secondly, it doesn't matter whether we add the impulses by phase or by bobsledder, the total should be the same; thirdly, it would have been simpler to add the four times and multiply the total by 25N.

Note that a bobsledder cannot keep pushing on the sled unless they keep up with it, so the last one's mass is irrelevant. Unfortunately, assuming each jumps in immediately after ceasing to push (rather than just keeping up with it), we do need the masses of the other three.
paulimerci said:
I got a total impulse of 1000 Ns.
Please explain how you get that. Doing it @erobz' way, by phases, I still get 100x3+75x1+50x1+25x1=450.
 
  • #9
haruspex said:
First, it doesn’t actually say that in post #1; secondly, it doesn't matter whether we add the impulses by phase or by bobsledder, the total should be the same; thirdly, it would have been simpler to add the four times and multiply the total by 25N.
I had a feeling it wouldn't matter how they are added, I was just trying to hold true to how bobsledders enter a sled.
haruspex said:
Note that a bobsledder cannot keep pushing on the sled unless they keep up with it, so the last one's mass is irrelevant. Unfortunately, assuming each jumps in immediately after ceasing to push (rather than just keeping up with it), we do need the masses of the other three.
Perhaps how bobsledder's sequentially board the sled in reality is the assumption that needs to go away to solve it. If they sequentially stop pushing and just run along with the sled until the last one has finished pushing and jump in all at once, then the problem can be solved.
haruspex said:
Please explain how you get that. Doing it @erobz' way, by phases, I still get 100x3+75x1+50x1+25x1=450.
@paulimerci So your first computation was fine after all. I should have checked the whole way. Sorry.
 
Last edited:
  • #10
haruspex said:
First, it doesn’t actually say that in post #1; secondly, it doesn't matter whether we add the impulses by phase or by bobsledder, the total should be the same; thirdly, it would have been simpler to add the four times and multiply the total by 25N.

Note that a bobsledder cannot keep pushing on the sled unless they keep up with it, so the last one's mass is irrelevant. Unfortunately, assuming each jumps in immediately after ceasing to push (rather than just keeping up with it), we do need the masses of the other three.

Please explain how you get that. Doing it @erobz' way, by phases, I still get 100x3+75x1+50x1+25x1=450.
Please don't laugh at what I described. Okay, I believe the first phase consists of four sledders pushing the bobsled with a force of 25 N for three seconds, and impulse 1 is (25x 4x3=300Ns). In the second phase, I think the first one jumped in, so there are only 3 pushing the bob for 4 seconds, and impulse 2 is (25x3x4 = 300Ns). In the third phase the second one jumped in, so there are two pushing the bob for 5 sec, and impulse 3 is (25x2x5=250Ns). Finally, only one person pushes the sled for six seconds, and the fourth impulse is 150 Ns. Total impulse =1000Ns. I know it is not specified in the question. So which one is right?
 
  • Like
Likes renormalize
  • #11
paulimerci said:
Please don't laugh at what I described. Okay, I believe the first phase consists of four sledders pushing the bobsled with a force of 25 N for three seconds, and impulse 1 is (25x 4x3=300Ns). In the second phase, I think the first one jumped in, so there are only 3 pushing the bob for 4 seconds, and impulse 2 is (25x3x4 = 300Ns). In the third phase the second one jumped in, so there are two pushing the bob for 5 sec, and impulse 3 is (25x2x5=250Ns). Finally, only one person pushes the sled for six seconds, and the fourth impulse is 150 Ns. Total impulse =1000Ns. I know it is not specified in the question. So which one is right?
Maybe how they are supposedly pushing is a little bit ambiguous...
 
  • #12
paulimerci said:
Please don't laugh at what I described. Okay, I believe the first phase consists of four sledders pushing the bobsled with a force of 25 N for three seconds, and impulse 1 is (25x 4x3=300Ns). In the second phase, I think the first one jumped in, so there are only 3 pushing the bob for 4 seconds, and impulse 2 is (25x3x4 = 300Ns). In the third phase the second one jumped in, so there are two pushing the bob for 5 sec, and impulse 3 is (25x2x5=250Ns). Finally, only one person pushes the sled for six seconds, and the fourth impulse is 150 Ns. Total impulse =1000Ns. I know it is not specified in the question. So which one is right?
In that scheme, the second pushed for 3+4=7secs, the third for 3+4+5=12 secs, the fourth for 3+4+5+6=18 secs.
That conflicts with post #1. Are you sure you quoted the problem correctly?
 
  • #13
haruspex said:
In that scheme, the second pushed for 3+4=7secs, the third for 3+4+5=12 secs, the fourth for 3+4+5+6=18 secs.
That conflicts with post #1. Are you sure you quoted the problem correctly?
Yes, that's what the question says.
 
  • Wow
Likes erobz
  • #14
It's kind of an unrealistic interpretation:



but that must be the desired interpretation, given the answer.
 
  • Like
Likes nasu
  • #15
It looks like this problem is unsolvable.
 

Attachments

  • Screenshot 2023-02-06 at 5.02.10 PM.png
    Screenshot 2023-02-06 at 5.02.10 PM.png
    20 KB · Views: 71
  • #16
paulimerci said:
It looks like this problem is unsolvable.
They assume the bobsledders all jump in at once at the very end ( so they are just running beside it after they push), and push the way you last calculated...however unrealistic that may be ( notice in the video the are all in the sled after about 5 seconds).
 
Last edited:
  • Like
Likes paulimerci
  • #17
erobz said:

If that's the case, there is no need for individual masses?
 
  • #18
No, you won't need the masses of the bobsledders because the mass of the sled isn't changing in each pushing phase.
 
  • #19
erobz said:
No, you won't need the masses of the bobsledders because the mass of the sled isn't changing in each pushing phase.
Okay, should I use conservation of momentum? or with the final impulse that I calculated in post #10?
 
  • #20
paulimerci said:
Okay, should I use conservation of momentum? or with the final impulse that I calculated in post #10?
Use Impulse/Momentum. ##\Delta p = \int F dt ##. The Momentum is changing.

You are just going to use the total impulse you calculated in #10, equated to the change in momentum of the sled just before they all jump in.
 
Last edited:
  • #21
erobz said:
use the total impulse you calculated in #10
Except that, as I wrote in post #12, the interpretation needed for that calculation simply does not fit the description in post #1.
 
  • #22
haruspex said:
Except that, as I wrote in post #12, the interpretation needed for that calculation simply does not fit the description in post #1.
Okay, for finding the individual impulse phases with the times you provided, I got the total impulse as 1875Ns.
 
  • #23
erobz said:
Use Impulse/Momentum. ##\Delta p = \int F dt ##. The Momentum is changing.

You are just going to use the total impulse you calculated in #10, equated to the change in momentum of the sled just before they all jump in.
I've changed the times, as @haruspex pointed out.
$$ I_T = mv_f - mv_i $$
$$1875 = 45 \cdot v_f$$
$$ v_f = 41m/s$$
 
  • #24
paulimerci said:
I've changed the times, as @haruspex pointed out.
$$ I_T = mv_f - mv_i $$
$$1875 = 45 \cdot v_f$$
$$ v_f = 41m/s$$
Does the answer look okay?
 
  • #25
haruspex said:
Except that, as I wrote in post #12, the interpretation needed for that calculation simply does not fit the description in post #1.
I agree, but it fits the desired outcome of the question setter, based on the answers given in post 15. So what are we to do.

The interpretation is ambiguous, and it does not align with the start of an actual bobsled run.
 
  • #26
paulimerci said:
Does the answer look okay?
Where did 1875 [Ns] come from? I thought you said you got 1000 [Ns].
 
  • #27
erobz said:
Where did 1875 [Ns] come from?
With the time provided in post #12, I changed the individual phases with their respective times (3, 7, 12, and 18) and got that answer for total impulse.
 
  • #28
paulimerci said:
With the time provided in post #12, I changed the individual phases with their respective times (3, 7, 12, and 18) and got that answer for total impulse.
That way is 1 force at a time.

25*( 3+7+12+18)=1000

Incidentally I don’t like this method because “if” they were jumping in the sled sequentially this wouldn’t describe the momentum change accurately for each phase. That being said in this problem they aren’t jumping in the sled till the end.
 
  • #29
erobz said:
The interpretation is ambiguous
Not to me. Either 1000Ns is wrong or the problem statement is wrong.
 
  • Like
Likes SammyS
  • #30
erobz said:
That way is 1 force at a time.

25*( 3+7+12+18)=1000

Incidentally I don’t like this method because “if” they were jumping in the sled sequentially this wouldn’t describe the momentum change accurately for each phase. That being said in this problem they aren’t jumping in the sled till the end.
Oh, I see. Thank you! That makes sense.
 
  • #31
haruspex said:
Not to me. Either 1000Ns is wrong or the problem statement is wrong.
Yeah, the problem statement doesn’t match the solution. But what at @paulimerci did in #10 given the circumstances is a reasonable approach…that’s what’s important IMO.

@palimerci please feel free to complain to the instructor about the better interpretation not jiving with the solution. The impulse should be 450 [Ns] IMHO. I assume @haruspex agrees?
 
Last edited:
  • Like
Likes paulimerci
  • #32
erobz said:
Yeah, the problem statement doesn’t match the solution. But what at @paulimerci did in #10 given the circumstances is a reasonable approach…that’s what’s important IMO.

@palimerci please feel free to complain to the instructor about the better interpretation not jiving with the solution. The impulse should be 450 [Ns] IMHO.
 
Last edited:
  • #33
paulimerci said:
Thank you, @erobz. It appears that I provided a reasonable interpretation this time.
Sure, will do!
 
  • #34
paulimerci said:
Thank you, @erobz. It appears that I provided a reasonable interpretation this time.
You see what is being said right? To get the answer they want, it doesn’t align with the wording of the OP. It’s like they were saying the second guy pushes for 6 more seconds than the first, etc… when they actually said the second guy pushes for 6 seconds…The 1000 [Ns] impulse is the result of a very poor/completely wrong interpretation.
 
Last edited:
  • #35
erobz said:
You see what is being said right? To get the answer they want, it doesn’t align with the wording of the OP. It’s like they were saying the second guy pushes for 6 more seconds than the first, etc… when they actually said the second guy pushes for 6 seconds…The 1000 [Ns] impulse is the result of a very poor/completely wrong interpretation.
I see that the question is poorly stated. Is it wrong for the second guy to push longer than the first? The time taken by each sledder can be different, right? I understand the answers doesn't make sense.
 
Back
Top