People Jumping Off a Car (Momentum Problem)

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In summary: I wouldn't use energy here, just momentum.In summary, Jumping off 1 by 1 increases the velocity of the vehicle more than jumping off all at once.
  • #1
Winter_Dew
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Hi

I haven't study physics since the one time I took AP Physics 1 three years ago, but my friend presented me this problem. There's a vehicle with passengers on top of it. Does the vehicle have higher velocity after all the passengers jumped off in the same direction and force at the same time or after the passengers jump off one by one (No external forces)? He is taking physics right now and told me the answer was the latter, but I'm not sure I 100% agree with his explanation. This seems like it would be a problem already mentioned, but I couldn't find the thread for it, so can someone explain this to me?
 
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  • #2
Winter_Dew said:
Does the vehicle have higher velocity after all the passengers jumped off in the same direction ...
Which direction?

Winter_Dew said:
...and force
Same force over the same time?

Winter_Dew said:
(No external forces)?
Just coasting with no propulsion?
 
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  • #3
Winter_Dew said:
This seems like it would be a problem already mentioned, but I couldn't find the thread for it, so can someone explain this to me?

try looking at the following thread;

Jumping while on a moving bus | Physics Forums - The Fusion of ...
https://www.physicsforums.com/threads/jumping-while-on-a-moving-bus.179123/Aug 2, 2007 ... If you are on a moving bus (at a constant velocity) and you are standing in the aisle then you jump directly upwards, would you move backwards or forwards relative to ... The combined linear momentum of you and the bus in the direction of travel does not change as you jump straight
 
  • #4
I interpreted this as essentially a rocket problem. Each person jumps off the back of the bus to speed it up. What happens if they all jump at once or one after the other?

One approach is to consider that each jumper releases a fixed amount of energy into the system.
 
  • #5
A.T. said:
Which direction?

Some force over the same time?Just coasting with no propulsion?

Lets say opposite direction to the way the vehicle is moving, same time, no propulsion just coasting.
 
  • #6
drvrm said:
try looking at the following thread;

Jumping while on a moving bus | Physics Forums - The Fusion of ...
https://www.physicsforums.com/threads/jumping-while-on-a-moving-bus.179123/Aug 2, 2007 ... If you are on a moving bus (at a constant velocity) and you are standing in the aisle then you jump directly upwards, would you move backwards or forwards relative to ... The combined linear momentum of you and the bus in the direction of travel does not change as you jump straight
[

I understand these concepts. I just want a proper explanation to why jumping off 1 by 1 would increase the velocity of the vehicle more than everyone jumping off at once (in the direction opposite to the bus's direction vector)
 
  • #7
PeroK said:
I interpreted this as essentially a rocket problem. Each person jumps off the back of the bus to speed it up. What happens if they all jump at once or one after the other?

One approach is to consider that each jumper releases a fixed amount of energy into the system.

So if each jumper releases a fixed amount of energy, why does people jumping off 1 by 1 increase the vehicles speed? He mentioned how with each jump, the force in each jump is being put onto a decreasing mass because people are jumping off (final mass being the vehicle). But I'm wondering why would that increase the vehicle's speed more than if everyone jumped off and put the entirety of their jumping forces at once onto the vehicle?
 
  • #8
Winter_Dew said:
Lets say opposite direction to the way the vehicle is moving, same time, no propulsion just coasting.
Same force over the same time means that each jump adds the same impulse to the rest (car + remaining passengers). To make math very simple assume that the empty car has the same mass, as each person.
 
  • #9
Winter_Dew said:
So if each jumper releases a fixed amount of energy, why does people jumping off 1 by 1 increase the vehicles speed? He mentioned how with each jump, the force in each jump is being put onto a decreasing mass because people are jumping off (final mass being the vehicle). But I'm wondering why would that increase the vehicle's speed more than if everyone jumped off and put the entirety of their jumping forces at once onto the vehicle?

You have to do the calculations. If they all jump off together they all have the same energy. But, if they jump off one at a time, perhaps they have less energy and the bus gets more? Or, the other way round.

You have to do the maths, as they say.
 
  • #10
PeroK said:
You have to do the calculations. If they all jump off together they all have the same energy. But, if they jump off one at a time, perhaps they have less energy and the bus gets more?

I wouldn't use energy here, just momentum. As stated, each jump applies the same impulse p. Then for two passengers of mass m, on a car of mass M we have:

Jump together:
delta v = 2 p/M

Jump with delay:
delta v = p/(M+m) + p/M

So delta v of the car is greater for jumping together
 
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  • #11
A.T. said:
So delta v of the car is greater for jumping together
I got the same answer and it makes sense (now). I was reluctant to post my answer 'cos I thought I must have made a slip. This must have consequences in the management of space flight. A big, short burn must be better than a long, low level burn.
 
  • #12
A.T. said:
I wouldn't use energy here, just momentum. As stated, each jump applies the same impulse p. Then for two passenger of mass m, on a car of mass M we have:

Jump together:
delta v = 2 p/M

Jump with delay:
delta v = p/(M+m) + p/M

So delta v of the car is greater for jumping together

Why assume a common impulse? Rather than common energy burn?

In the extreme case where you jumped off something very light, how much momentum could you generate?
 
  • #13
PeroK said:
In the extreme case where you jumped off something very light, how much momentum could you generate?
That would be a different matter. The first jumper would give the second jumper v/2 velocity and the second number would impart almost the whole of his v. Two jumpers together would only impart v.
There must be a M:m ratio where the two strategies give the same resultant velocity. Go on go on go on - you know you want to.
 
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  • #14
sophiecentaur said:
That would be a different matter. The first jumper would give the second jumper v/2 velocity and the second number would impart almost the whole of his v. Two numbers together would only impart v.
There must be a M:m ratio where the two strategies give the same resultant velocity. Go on go on go on - you know you want to.
What if each jumper was a coiled spring? That would give a constant energy increase to the system. Not a common impulse.
 
  • #15
PeroK said:
Why assume a common impulse?
As I understand the OP, all jumps apply the same force over the same duration.
 
  • #16
A.T. said:
As I understand the OP, all jumps apply the same force over the same duration.
It's not explicitly stated one way or the other. In general, as the vehicle gets lighter, to maintain constant impulse you need more and more energy.

Assuming constant energy per jump gives a different conclusion.
 
  • #17
PeroK said:
What if each jumper was a coiled spring? That would give a constant energy increase to the system. Not a common impulse.
I think that is a very relevant idea. The characteristics of a 'jumper' haven't been defined. I was assuming that the parting speed would be the relevant quantity but force times time is not the same as force times distance (impulse vs Energy supplied). There are some apparent paradoxes here unless we are tighter in our description of the problem. My comments about space rocketry (post #11) would come in here.
 
  • #18
sophiecentaur said:
I think that is a very relevant idea. The characteristics of a 'jumper' haven't been defined. I was assuming that the parting speed would be the relevant quantity but force times time is not the same as force times distance (impulse vs Energy supplied). There are some apparent paradoxes here unless we are tighter in our description of the problem. My comments about space rocketry (post #11) would come in here.
Yes, the problem is that all the kinetic energy must ultimately come from some potential source. It's not clear what happens with a human jumper, in terms of how.much energy they could release. In general, however, it would be the energy from a spring or a fuel source that would be constant. Or, at least, a limiting factor.

In your rocket example, I doubt it would be more effective to burn all the fuel at once. I think the slow burn alternative would be better.
 
  • #19
@Winter_Dew as you can see, this is a deceptively tricky problem!
 
  • #20
PeroK said:
It's not explicitly stated one way or the other.
Post #1 says same force, and post #5 confirms it's over the same time, after I asked about it.
 
  • #21
sophiecentaur said:
That would be a different matter. The first jumper would give the second jumper v/2 velocity and the second number would impart almost the whole of his v. Two jumpers together would only impart v.
There must be a M:m ratio where the two strategies give the same resultant velocity. Go on go on go on - you know you want to.

After testing out different m and M’s i also found this to be the case where there was a point when the higher velocity situation changed to the otherSo is it to my understanding that the answer depends on the masses, and there is not on situation that causes higher velocity over the other?
 
  • #22
PeroK said:
@Winter_Dew as you can see, this is a deceptively tricky problem!

Yeah, it seems so. I have to find some time to ask my friend about this problem again.
 
  • #23
A.T. said:
Post #1 says same force, and post #5 confirms it's over the same time, after I asked about it.
I didn't see post #5. That may be the OP's interpretation of the problem, which may not be what the original question setter intended. Especially given that the quoted answer was the opposite conclusion (better one at a time)!
 
  • #24
Winter_Dew said:
Yeah, it seems so. I have to find some time to ask my friend about this problem again.
It also depends on the parameters you use for 'jumping off'. That can sound daft but you can specify the Impulse input (which is what we assumed) and that gives a fixed separation speed or you can specify a fixed amount of Energy injected and get a different answer. Considering the length of the legs is fixed, with two jumpers the acceleration would be greater so the time for leg stretching could be less (would be - actually) so the impulse delivered would be different but the Work done may not change.
You will have to give your 'friend(?)' a grilling and tell him his question is incomplete.
Great question to make us think. :biggrin:
 
  • #25
Winter_Dew said:
After testing out different m and M’s i also found this to be the case where there was a point when the higher velocity situation changed to the otherSo is it to my understanding that the answer depends on the masses, and there is not on situation that causes higher velocity over the other?
This can't be correct. The impulse analysis is independent of the relationship between the masses.

See post #10.
 
  • #26
PeroK said:
This can't be correct. The impulse analysis is independent of the relationship between the masses.

See post #10.
But the Impulse Analysis is not always applicable because of the physical constraints. The force is limited and so is the length over which it can be applied. A simple statement that a chosen speed of separation can be achieved for all situations. If the mass of the car is small enough, it is not possible to push it away with a chosen momentum because of the length and strength of the legs.
The OP contains a question that is answerable of we assume 'reasonable' parameters. The car needs to be relatively massive but a similar experiment with a skateboard would not give the same result. There's nothing wrong with the question as posed as long as we don't explore the boundaries too far.
 
  • #27
sophiecentaur said:
But the Impulse Analysis is not always applicable because of the physical constraints. The force is limited and so is the length over which it can be applied. A simple statement that a chosen speed of separation can be achieved for all situations. If the mass of the car is small enough, it is not possible to push it away with a chosen momentum because of the length and strength of the legs.
The OP contains a question that is answerable of we assume 'reasonable' parameters. The car needs to be relatively massive but a similar experiment with a skateboard would not give the same result. There's nothing wrong with the question as posed as long as we don't explore the boundaries too far.
I don't see this at all. If you assume a common impulse, then it's clear. It's always better to jump together, independent of the mass.

You may have to assume a relatively small common impulse in the case of a light object, but there's no point at which the result reverses, as it were.
 
  • #28
sophiecentaur said:
But the Impulse Analysis is not always applicable ...
Whether the assumption of equal impulses is realistic is a completely different question from whether that assumption can lead to opposite results for different mass ratios.
 
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  • #29
A.T. said:
Whether the assumption of equal impulses is realistic is a completely different question from whether that assumption can lead to opposite results for different mass ratios.
If you assume a given parting velocity for all conditions then you get a result that's mass dependent. Which assumption is better? I think the fixed v is more realistic and no one specified which assumption to use. Do we go with Impulse, right or wrong, or inject some realism? Reality must be somewhere in between because the parting speed could be affected by contributions from multiple jumpers. I can't accept that a model that wouldn't work for a skateboard can be considered as suitable.
 
  • #30
sophiecentaur said:
If you assume a given parting velocity for all conditions then you get a result that's mass dependent.
Which jump tactic gives the most final car speed depends on the mass ratio? Show your math please.
 
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  • #31
A.T. said:
Which jump tactic gives the most car speed depends on the mass ratio? Show your math please.
well. i was looking at the issue and in my calculation if the jumping off person is moving in

reverse direction then the change in speed of the car will vary directly as the the number of persons jumping together increases.

if M is car's mass and m is the person's mass then
the change in speed(slowing down) occurs if the person is jumping in backward direction

the speed change comes to = (n-1) x (m/M) x v
where n is the number of persons jumping together and v is the velocity of jump.
The above have been calculated using conservation of momentum only.
so naturally if (m/M) is larger the change in velocity will be larger .
 
  • #32
A.T. said:
Which jump tactic gives the most final car speed depends on the mass ratio? Show your math please.
The claim was under the assumption of a fixed parting velocity. One assumes that this is the relative velocities of car and jumper, post jump. Let us take a nice extreme test case and see what happens with two different jump orders.

Jumper 1: 1 kg.
Jumper 2: 2 kg.
Jumper 3: 3 kg.
Car: 1 kg.
Parting velocity: 1 m/sec.​

Test A: 1 then 2 then 3.
  1. Mass ratio is 1 to 6. Jumper 1 moves left at 6/7 m/sec. Car+2+3 moves right at 1/7 m/sec.
  2. Mass ratio is 2 to 4. Jumper 2 moves left at 1/2 m/sec. Car+3 moves right at 1/2 m/sec. [Both in addition to the starting 1/7 right]
  3. Mass ratio is 3 to 1. Jumper 3 moves left at 3/4 m/sec. Car moves right at 3/4 m/sec. [Both in addition to starting 1/7 + 1/2 right]
Total delta v for car = 1/7 + 1/2 + 3/4 ~= 1.393 m/sec.
Test B: 3 then 2 then 1.
  1. Mass ratio is 3 to 4. Jumper 3 moves left at 4/7 m/sec. Car+1+2 moves right at 3/7 m/sec.
  2. Mass ratio is 2 to 2. Jumper 2 moves left at 1/2 m/sec. Car+1 moves right at 1/2 m/sec [Both in addition to the starting 4/7 right]
  3. Mass ratio is 1 to 1. Jumper 1 moves left at 1/2 m/sec. Car moves right at 1/2 m/sec [Both in addition to the starting 4/7 + 1/2 right]

Total delta v for car = 4/7 + 1/2 + 1/2 ~= 1.571 m/sec.
So yes, under the given assumption, the winning jump tactic appears to depend on the jumper's masses.
 
  • #33
drvrm said:
...if (m/M) is larger the change in velocity will be larger .
jbriggs444 said:
...different jump orders...

The question in the OP is about:

All at once
vs.
One after the other

Does which of these two options is better depend on the mass ratios, assuming common separation speed?
 
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  • #34
A.T. said:
Does which of these two options is better depend on the mass ratios, assuming common separation speed?
This is discussed in posts #10 and #13 and the answer depends on the mass ratio. Common separation speed is about the only one that can be realized with actual legs and bodies.
 
  • #35
A.T. said:
T
All at once
vs.
One after the other
OK. Now I'm picking up what you're putting down.
 

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