In the simplest form of Atwood machine,two masses are 0.3 kg and 0.6 kg.We consider massless pulley,string and frictionless surfaces.we let the arrangement move at t=0.
You can calculate that the tension in the string is 3.9 N and the acceleration~3.26

The larger mass is stopped 2 seconds later the arrangement started moving.
We are to find the timee elapsed before the string is taut again.

I thought that the lighter mass would move upwards (at the instant bigger mass was stopped) with speed: v=at=(3.26x2) m/s=6.52 m/s

Then using total time of flight I got T=(2v/g)=1.33 s

Whereas the answer is 0.67 s

can anyone show me if I am wrong?

Uhm let me try (but I dont study physics, so don't trust me!)

I guess here is where you were going wrong:


Then using total time of flight I got T=(2v/g)=1.33 s

Whereas the answer is 0.67 s

can anyone show me if I am wrong?

This is a two way movement, so the total distance and therefore the
time should be halved (isn't it :confused: )

I don't know how to write sigma here, so I replace with this symbol: #

#F = ma = (m1 + m2)a

force affecting each mass:
#F = (m2g-T) + (T - m1g) = g (m2 - m1)

(m1 + m2) a = g (m2 - m1)

therefore:

a = g (m2 - m1) / (m1 + m2)
a = 10 (0.6 - 0.3) / (0.3 + 0.6) = 3.33 m/s2

final velocity: V = Vo + at = 0 + 3.33(2) = 6.66 m/s

time elapsed: t = 1/2 (2V/g) = V/g = 6.66/9.8 = 0.67 s

There is "two way" movement. The smaller weight has an upward velocity at the time the larger weight is stopped. The smaller weight will fly up a bit before it starts falling. The string will be taut again when the smaller weight once again is at the position it had at the moment the larger weight was stopped.

Neeklash, your answer is correct.

Neeklash, your answer is correct.

Yes...