Transfer of momentum

[noxid]

Lets say i am falling from a great height, and i am wearing a bag that is the same mass as me, and just before i hit the ground i throw the bag at the ground at a velocity (w.r.t. the ground) twice that which i am falling relative to the ground. Would i survive?

[DrKrettin]

Let's say your velocity downwards at that point is v, and your mass is m. You and the bag are a closed system so your total momentum downwards is 2mv. If you manage to do as you say, the momentum of the bag itself is 2mv, so your own momentum is zero. Your momentum at hitting the ground would then be very small, so you would survive that impact. I'm not sure, however, how your body would react to the g forces experienced when throwing the bag, apart from the physical impossibility of doing it.

[Strange]

Indeed. Stopping your motion suddenly by some other means would be just as destructive by hitting the ground.

[noxid]

Indeed. Stopping your motion suddenly by some other means would be just as destructive by hitting the ground.

lets say i am 62Kg and my velocity just before hitting the ground is 53m/s, about the terminal velocity of a person

then the total momentum of me and the bag is 62*2 *53=6572 Kgm/s, i throw the bag at 106 m/s at the ground. So its momentum is 106*53=6572 and my resulting momentum is zero, due to v=0 in p=mv

Since force, F=ma, and my acceleration here is zero, i should experience no force, shouldnt I?

[Strange]

lets say i am 62Kg and my velocity just before hitting the ground is 53m/s, about the terminal velocity of a person

then the total momentum of me and the bag is 62*2 *53=6572 Kgm/s, i throw the bag at 106 m/s at the ground. So its momentum is 106*53=6572 and my resulting momentum is zero, due to v=0 in p=mv

Since force, F=ma, and my acceleration here is zero, i should experience no force, shouldnt I?

 

Now calculate the force you experienced when you threw the bag.

[noxid]

F=ma, so, F=62*(53-0)/t, where 53 is the final velocity of the bag relative to me, but i am not sure what t should be

[swansont]

F=ma, so, F=62*(53-0)/t, where 53 is the final velocity of the bag relative to me, but i am not sure what t should be

t is however long it took to throw the bag. If it's longer than the deceleration time of impact, it will do less damage than hitting the ground.

[noxid]

t is however long it took to throw the bag. If it's longer than the deceleration time of impact, it will do less damage than hitting the ground.

So basically i take a long time to throw, the bag i i experience less force, explain...

give me a scenario where t is short vs where it is long

 

ie. how can i throw the bag to achieve different t

Edited August 8, 2016 by noxid

[Strange]

If you can do it slowly, like firing a jet engine rather than throwing a single object, then you can slow down gradually and survive.

[Delta1212]

You'd probably be better off with a beanbag cannon that could fire off the equivalent mass of beanbags over the final second or two of your descend rather than trying to throw a single bag all at once right at the end.

 

Hitting the ground doesn't hurt you because it's the ground. It hurts because it stops you very quickly. Anything that stops you that quickly is going to have the same effect.

 

It's the difference between gradually slowing down in a car and slamming on the breaks.

[swansont]

So basically i take a long time to throw, the bag i i experience less force, explain...

give me a scenario where t is short vs where it is long

 

ie. how can i throw the bag to achieve different t

There's not a lot of dynamic range here, but you can compare the long windup of a baseball pitcher with a quick flip where you don't draw you arm back. As long as the final speed is the same, the process that takes longer to get the ball up to speed results in a lower force.

 

(The issue here is that a longer arm motion usually results in a greater speed)

 

A better example is given by Strange. If instead of throwing a 62 kg rock you throw 62 1 kg rocks at the same speed but over a greater span of time.

[DrKrettin]

A better example is given by Strange. If instead of throwing a 62 kg rock you throw 62 1 kg rocks at the same speed but over a greater span of time.

 

 

I don't think that would not work very well, because by the time you had thrown all 62 rocks you will have gained terminal velocity again (apart from the effect of the last rock)

[imatfaal]

 

 

I don't think that would not work very well, because by the time you had thrown all 62 rocks you will have gained terminal velocity again (apart from the effect of the last rock)

 

I think the assumption is you "throw" with some sort of automated cannon - 30 per second and you have only two seconds firing that's only 20m/s gained.

 

I seem to remember mythbusters doing calcs for what sort of gun you would need to get off the floor like Yosemite Sam in the Bugs Bunny cartoons. I think it was a couple of miniguns one in each hand - not exactly safe to be around the ricochets.

 

The waterjets you can hire on the beach are obviously throwing back enough water to keep you stable - a little stronger one of those AND a very very long feed hose...

[sethoflagos]

Isn't opening a parachute essentially the same as throwing trillions of 'little rocks' (or air molecules) towards earth over a lengthy period?

 

No need for them to be 'thrown' any faster than you're falling just so long as you transfer momentum faster than you accumulate it until you're down to a safe descent velocity.

[swansont]

 

 

I don't think that would not work very well, because by the time you had thrown all 62 rocks you will have gained terminal velocity again (apart from the effect of the last rock)

 

 

Terminal speed wasn't a condition of the problem. In that case, momentum is momentum. You could even have cases where you stop, or even gain an upward velocity, from throwing some of the rocks in mid-flight. Then it's like you are jumping from a much lower height.

 

But you're right, if you include terminal velocity, there are additional constraints to the problem.

Isn't opening a parachute essentially the same as throwing trillions of 'little rocks' (or air molecules) towards earth over a lengthy period?

 

No need for them to be 'thrown' any faster than you're falling just so long as you transfer momentum faster than you accumulate it until you're down to a safe descent velocity.

 

A parachute increases your drag force, so while the result is similar, the analysis is not the same as with mass transfer.