Solving for Tension & Acceleration in M1 & M2

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To solve for tension and acceleration in the system with m1 = 300g, m2 = 200g, and F = 0.4N, the equations T = m2*a2 and F - 2T = m1*a1 are established. The relationship a2 = 2a1 arises from the connection of the masses via a pulley, indicating that for every meter m1 moves, m2 moves half that distance. This acceleration constraint is crucial for correctly calculating the system's dynamics. Understanding this relationship can be aided by visualizing the motion with a string or paper strip. The correct application of these principles will yield the desired tension and acceleration values.
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Assume there is negligible friction between the blocks and the table. Compute the tension in the cord and the acceleration of m2 if m1 = 300g, m2 = 200g, and F=0.4N

I figured out that T=m2*a2, F-2T=m1a1.
According to the answer, 2a1=a2. But I don't undestand why.
 

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acceleration constraint

Your force equations are correct, but incomplete. You need to incorporate the constraint imposed by the fact that the masses are connected via a pulley and ropes--that's what will give you the additional equation a2 = 2 a1. To understand where that relationship comes from, answer this question: If m1 moves one meter to the right, how far does m2 move? (If you have trouble visualizing the motion, use a piece of string or a paper strip.)
 
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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