People Jumping Off a Car (Momentum Problem)

[drvrm]

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 .

 

[jbriggs444]

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.

 

[A.T.]

...if (m/M) is larger the change in velocity will be larger .

...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?

 

[sophiecentaur]

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.

 

[jbriggs444]

T
All at once
vs.
One after the other

OK. Now I'm picking up what you're putting down.

 

[A.T.]

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.

Post #10 assumes common momentum transfer, not common separation speed, and shows that jumping separately is never better, regardless of mass ratio. So the answer to the OPs question (which of the two tactics is better?) does NOT depend on the mass ratio.

Post #13 contains no complete derivation, just an assertion:

There must be a M:m ratio where the two strategies give the same resultant velocity.

Are you talking about a non-zero, non-infinite ratio here? If yes then what is that ratio?

 

[PeroK]

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.

Common separation speed doesn't make sense for the jumping together option. In that case there is one large separation velocity to compare with two or more smaller ones.

 

[PeroK]

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.

I think it's realistic, but the problem is that the two jumping together expend a lot more energy than the two jumping separately.

If the two jumping together put maximum effort in, then the two jumping individually would have to take it easy to get the same impulse.

It's not, therefore, so much a better tactic, but simply putting more effort into the jump.

I would go back to the common energy input as the most logical assumption.

 

[sophiecentaur]

Common separation speed doesn't make sense for the jumping together option. In that case there is one large separation velocity to compare with two or more smaller ones.

How doesn't it make sense any more than the other options don't make sense? We are not able to meter our muscle output in terms of v, impulse or Work so we just have to pick one and work with that. No one is more right or wrong. However, I have already pointed out that there are real limits to the Impulse that can be transferred with human legs, independent to the mass involved. So perhaps Impulse is not suitable.

 

[PeroK]

How doesn't it make sense any more than the other options don't make sense? We are not able to meter our muscle output in terms of v, impulse or Work so we just have to pick one and work with that. No one is more right or wrong. However, I have already pointed out that there are real limits to the Impulse that can be transferred with human legs, independent to the mass involved. So perhaps Impulse is not suitable.

Yes, you're right, it is another option.

 

[sophiecentaur]

I would go back to the common energy input as the most logical assumption.

I think that's probably the most realistic, bearing in mind the distance moved by the leg is the same. But there is still the issue of whether the Force could be the same when the leg is extending at high speed (when pushing on a skateboard).
Someone earlier suggested a spring as a good model. That would make the analysis easier and wouldn't have a change of sign of the advantage of multiple jumpers.

I would say we've had quite a good value forty posts this time!

 

[Ben2]

Using A.T.'s 3rd post, let (delta v)d indicate change in velocity with delayed jumps and (delta v)t indicate change
in velocity with a jump together. For two jumpers, A.T.'s formulation gives me
(delta v)d > (delta v)t if and only if M^2 > 2m^2 or M > sqrt(2) * m.
Thanks for a great problem!

 

[DonDiablo]

I think the impulse solution is the more realistic one. My assumptions here are that all jumpers are equal. If you took an energy (work on car model) for the solution it would make no sense. If all jumpers are equal and they apply a force F on the car for the same duration of time (which makes sense no matter how how fast the car goes or what its mass was) because the jumpers are standing on the car anyways.. then they all apply the same amount of impulse on the car leading to the solution mentioned in post #10! But I still have problems applying the laws to this case. Let's say they all put in the same amount of kinetic energy onto the car which would make sense since they all have the same power etc. if they jumped off one by one it meant that for the second jumper the car would already move faster than for the first jumper. Now if every jumper applied the same amount of force onto the car during the same amount of time (which we sort of agreed on stating that all jumpers are equal) - Like if the energy put on the car was given by the force of the jumper during a way which way would that be- the way the jumpers legs bend or the distance the car would have taken during the process of jumping and applying the force? because the distance the car would have taken would differ from jumper to jumper if they jumped off the car one after another... which would in return mean they all applied a different amount of kinetic energy to the car... which we agreed upon on they wouldn't because why would they given that they were equal. Also the outcome of a given experiment in thought must not differ in dependence on which method of calculation is being used right? Since this would undermine the integrity of the scientific value of any given prediction.

@Ben2 I don't know how you came to that solution but given @A.T. 3d post post #10 in this thread the only time it doesn't make a difference if they jumped off together or not is when the mass of the car was 0 and the equations are no longer properly defined!

So i try to make different kind of solutions each leading to the same outcome of course starting with the forces- each jumper putting the same amount of force for the same amount of time onto the remaining system.
Assumptions: mass car=10kg, mass jumper=1kg, mass jumper 2=1 kg mass jumper3=1 kg, F(Jumper)=10N (1kg*10(m/s²) jumping duration (Duration during which force is implied on remaining system) = 1s; velocity of car at tha start of the experiment=0

Case 1 - jump one after another:F(jumper)=F(Remain)
F(jumper)*t=F(Remain)*t
m(jumper)*a(j)*t=m(Remain)+a(R)*t
1kg*10m/s²*1s=12kg*a*1s
a=10/12 m/s² - ergo - speed after jumper 1 is 10/12m/s F(jumper)/(mass of Ra

F(jumper2)=F(Remain)
(...)
1kg*10m/s²*1s=11kg*a*1sok kinda lost interest into writing this down... however if they all jumped together at the end of the day their relative speed to each other would be 0 (the relative spped of one jumper to another) - if they jumped one after another then jumper 2 would move relative to jumper 1 and jumper 3 would move relative to jumper 2 (and to a greater extent to jumper 1)... its a matter of distribution really -if they jumped off at different times with the same force applied for the same time theyd reach the same impulse relative to the car but not relative to another reference system. Some of the impulse of jumper1 on would be used to move jumper 2 and 3 thus lacking in the spped of the car in the final result...

if the all jumped together the total force applied on the car would be (F(j1)+F(j2)+F(j3)=30N - but if they jumped one after each other the total force applied on the car would be F(j1)*m(car)/(m(car)+m(j2)+m(j3))+F(j2)*(m(car)/(m(car)+m(j3))+F(j3)*(m(car)/m(car) which is obv. less and becomes lesser and lesser the bigger the masses of the jumpers are.

If you put it down with energies.. the result would be exactly the same... note that this is just the equivalent of A.T.'s post on page 1... you could show that if one used energies instead of forces... because the kinetic energy "applied" on the car would be the same for each jumper if their weight, force (used during jump) and jump time would be the same... the energy applied or the work done during jump must equal the kinetic energy of the car after the jump. the way along which the force is applied would be the direction of the jump and its length would be the dependent on the length of the legs of the jumpers for example. It wouldn't even matter if they were of different height because given they apply the same force for the jump and take the same amount of time to do so one could argue that if they were of different size one would bent their knee less than the other on so forth but I don't think one should engage in such figurative ways when approaching a rather simple physical problem!
Lg Don

Summary: A.T. s post on page one solved the problem fairly easy.

 

[A.T.]

...given @A.T. 3d post post #10 in this thread the only time it doesn't make a difference if they jumped off together or not is when the mass of the car was 0 ...

Note that while the mass ratio affects the delta v in post #10, it does not make a difference for the question, which of the two tactics is better. Given the assumptions of post #10, jumping together gives greater delta v for any finite / non-zero mass ratio.

 

[sophiecentaur]

Note that while the mass ratio affects the delta v in post #10, it does not make a difference for the question, which of the two tactics is better. Given the assumptions of post #10, jumping together gives greater delta v for any finite / non-zero mass ratio.

Only for the delta v assumption. Two people jumping would not both achieve the same delta v on a skateboard. We just aren't built for it. With that caveat, I accept the result.
Edit: Actually, I have thought again and it would be possible to 'push less hard' and produce the same delta v on a skateboard as on a car. What would be impossible for a human leg is transferring a given amount of momentum or energy, independent of mass. A skateboard would require an impossible leg velocity.

 

[Shadow89]

Disclaimer: I have not read all the posts, so I don't know the history of this topic. I wish only to reply to the OP. Also I will not cite sources other than myself.

I am not sure which is correct. you can use classical dynamics to advocate both s. But my intuition tells me it is possible that jumping off one by one will allow greater speed. My explanation follows:

There are three (dumb) guys on the car. When the first guy jumps off, he is jumping off a car with two other guys on it. (Note: The mass of a car with two guys on it is greater than the mass of just a car).
Now let's take that to the utter extreme. There are two scenarios. One: You jump off a skateboard as far as you can. Two: You jump off a nimitz-class aircraft carrier as far as you can. Obviously, you will get further if you jump off of a more massive object. The skateboard will simply fly backwards and you will not get very far.
Now our intuition tells us that therefore the force between you and the skateboard was not as great as the force between you and the aircraft carrier.
We can then assume that the force between the 1st guy and the car with two guys, is greater than 1/3 of the force between three guys and the car.

Those where my $0.02. I hope it made some sense, and I reserve the right to be completely wrong.