How Do Newton's Laws Explain the Motion of Two Men Pushing Each Other?

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In the scenario where an 80 kg man (A) pushes another 80 kg man (B) with a force of 500N, the application of Newton's Second and Third Laws is crucial for understanding their motion. The discussion raises questions about the conditions under which the push occurs, such as whether they are on a frictionless surface or how their stance affects the outcome. It is suggested that if the force is applied continuously, both men would accelerate together rather than fly apart. The ambiguity in the scenario highlights the need for additional information to accurately determine the resulting motion. Overall, the interaction exemplifies the complexities of applying Newton's laws in real-world situations.
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Homework Statement



An 80 kg man (A) pushes another 80kg man (B) with a force of 500N, describes what occurs.

Homework Equations


Newtons 2nd and 3rd Law ?



The Attempt at a Solution


I suspect the answer is both men fly apart with an acceleration of (500/80) m/s2 ? But in real life I would imagine man A would stay put ?
 
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Hi pkc111! :smile:
pkc111 said:
An 80 kg man (A) pushes another 80kg man (B) with a force of 500N, describes what occurs.

Answer 1:

Are they on ice? or in space?

Is each man standing with feet side-by-side, or in-line?

Are their weights being supported on their heels or their toes?

Is the force applied to B at this centre of mass?

If this is a physics question, there's nowhere near enough information. :rolleyes:

If it's a sociology question, I'd say they're equally matched, so there'll be a fight! :wink:​

Answer 2:

The force is given in Newtons, so it's not an impulse (which would be in Newton.seconds).

So the question must mean that the force is being applied continuously, which means that A and B must be staying together, with acceleration of … ? :smile:
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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