0
$\begingroup$

I know Newton’s Third Law states that every action has an equal and opposite reaction. I also understand that the less mass an object has, the faster it will accelerate relative to a more massive object if the same force is applied.

What if you are in the vacuum of deep space where gravitational forces from planets and stars are negligible and there is no air resistance/friction. If I am accelerating through space and collide with a lighter stationary object, exerting a force on it equal to my mass multiplied by my acceleration, shouldn’t that object exert an equal force on me and cause me to decelerate to zero and result in both me and the object traveling at a constant velocity? If not, why do me and the object I collided with continue to accelerate?

$\endgroup$

4 Answers 4

3
$\begingroup$

If I am accelerating through space and collide with a lighter stationary object, exerting a force on it equal to my mass multiplied by my acceleration, shouldn’t that object exert an equal force on me and cause me to decelerate to zero and result in both me and the object traveling at a constant velocity? If not, why do me and the object I collided with continue to accelerate?

The forces will be equal and opposite, but (because the stationary object is lighter than you are) it will accelerate more than you decelerate. Both you and the object will accelerate as long as you are in contact with the object and exerting forces on each other throughout the duration of the collision. Whether you come to rest or not depends on the relative mass of the other object and the coefficient of restitution for the collision.

$\endgroup$
2
  • $\begingroup$ Assuming both me and the object accelerate and remain in contact, why wouldn’t the force acting against me from the lighter object eventually cause me to decelerate if the force is equal and opposite to the force I’m exerting on it? $\endgroup$
    – Newbie223
    Jun 1, 2022 at 2:30
  • $\begingroup$ @Newbie223 you said to assume that you continue to accelerate. So the force from the lighter object cannot cause you to decelerate, by your assumption. $\endgroup$
    – Dale
    Jun 1, 2022 at 3:42
0
$\begingroup$

That's only half true.

Mass exist's - there is no way around. And since we have no other way - in space mass is not important - if collided in space. It is only the acceleration that count's.

So in your case let's say your weight is 100kg and you collide with 10kg. The force will be the same on both: you and the object. In Space it is only the acceleration that comes down to the impactvalue and destructionpower.

Or: Every Needle you punch - you hurt the needle the same as it hurts you. Einstein's "special r.t." included this and it was proven by Apollo11 as they hit the american flag "easy" into the moon. The Moon is a giant "rock", should be hard to put a flag in, but it was like putting a knife into butter.

That confirms and complete's your question. It is the acceleration - NOT - the mass. Mass is exising but has no value due to the destruction or colliding power.

$\endgroup$
1
  • $\begingroup$ What is only half true? And your analogy with the needle is not valid. Equal forces can have significantly different effects. $\endgroup$
    – nasu
    Dec 8, 2023 at 23:45
0
$\begingroup$

Your acceleration $a$ though space does not dictate the force of contact $F_c$. The acceleration through space depends on an external force $F_x$ applied according to Newton's 2nd law

$$a = \frac{F_x}{M}$$

At the moment of impact (shown below with the masses separated for clarity) there is an additional force $F_c$ applied in equal and opposite measure to both objects per Newton's 3rd law. Also, the speed of impact determines the magnitude of this force.

fig1

This force is not instantaneous but must act over a period of time $\Delta t$ such that it results in a change in motion after the impact. We call such forces, impulses and have the units of momentum.

And if such an impulse is acting on the smaller body $m$ an equal and opposite impulse must act on the bigger body $M$ (you).

Since the bodies cannot interpenetrate, the minimum contact impulse $J_c = F_c \Delta t$ needed to merge the two bodies is calculated below:

  • Just before the impact the bigger body has speed $V$ and thus carries momentum $P = M V$.

  • During the impact, some momentum $J_c$ is removed from the big body and transferred to the smaller body keeping the total momentum of the system the same. Well, this assumes that the impact happens in finite time, but this time is so small that we can ignore the external force $F_x$ and its effects for now.

  • So the momentum after the impact of the big body is $P' = M V - J_c$

  • And for the smaller body $p' = J_c$

  • If after the impact the speeds of the two bodies match to some value $V'$.

  • For the big body $M V - J_c = M V'$ is the momentum after

  • and for the smaller body $J_c = m V'$ is the momentum after

  • Add the two together to make $M V = M V' + m V'$ which is solved for $$V' = \frac{M V}{M+m}$$

  • Use in the smaller body equation to get $$J_c = \frac{m M}{M+m} V$$

  • Now consider the case of purely elastic impact. The impulse is twice as much as the first case $J_c = 2 \frac{m M}{M+m} V$

  • From the momentum of the big body after $P' = M V - J_c = M V'$ you get the final velocity $$V' = \frac{M-m}{M+m} V$$

    As you can see when $M>m$ the big body will not stop and continue it its path but with a reduced speed.

  • Front the momentum of the small body after $p' = J_c = m v'$ (I introduced $v'$ is the final velocity here because it is different from $V'$ used before) you get the final velocity $$v' = \frac{2 M }{M+m} V$$

  • The difference in speeds equals to the impact speed $$v' - V' = V$$ and this is a result of the perfectly elastic collision. It will never exceed this value, because you cannot add energy to the system with a collision.

  • To get the actual force applied, you can estimate the average force $F_c$ from the impulse $J_c$ given a contact time of $\Delta t$ using the simple formula $$F_c = \frac{J_c}{\Delta t}$$

$\endgroup$
0
$\begingroup$

You are misunderstanding a few factors, which I will try to sort out for you. Firstly, if you are accelerating at a rate 'a' then there must be a force F=ma which is acting on you to cause your acceleration. If you collide with an object, the force that is accelerating you is not exerted in full on the object, so your fourth sentence is simply incorrect. If the object becomes lodged in the folds of your spacesuit, then after the collision you and the object will continue to accelerate together at a rate slightly less than a, since the force causing your acceleration is now having to accelerate not just your mass but the mass of the particle too. On the other hand, if the object was an elastic pellet that bounced off your visor, then after the jolt of the collision you would no longer be in contact with the pellet. Given that, your mass is the same as it was before, so you will continue to accelerate at a rate a as before. The pellet will continue through space at a fixed velocity independently of you.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.