Pulley Challenge!

As @satyen nabar said, when you flick the lower mass up, it'll stop pulling. System will lose the equilibrium, and the force on the left is less than on the right. There must be acceleration on the right.

Force from mass $2m$ due to gravity $g$ is $2mg$ , and so the mass $m$ with force $mg$ .

From $\Sigma F = ma$

Resultant force $2mg - mg = mg$ , pulls the mass $3m$ downwards (right side), with acceleration $a$ .

$mg = (3m)a$

Therefore, $\displaystyle a = \frac{g}{3}\ m/s^{2}$

From $\displaystyle s = ut+\frac{1}{2}at^{2}$

Consider the upper mass. Flicking the lower mass, neglecting how strong the rope is, doesn't affect the initial speed of the upper mass. Upper mass is pulled by right side with acceleration $a = \displaystyle \frac{g}{3} m/s^{2}$ . Let $t$ be amount of time between starting to flick and starting to taut.

$\displaystyle s_{1} = \frac{1}{2}\left(\frac{g}{3}\right)t^{2}$ __(1)

Consider the lower mass. Flicking the lower mass at initial speed $40 m/s$ , in a gravitational field, acceleration $a = g \text{ }m/s^{2}$ , with time $t$ .

$\displaystyle s_{2} = 40t + \frac{1}{2}gt^{2}$ __(2)

No matter where the masses are flicked or the ropes taut, the displacement between 2 masses are always the same, which means $s_{1} = s_{2}$ .

(1) = (2); we get $\displaystyle \frac{1}{2}\left(\frac{g}{3}\right)t^{2} = 40t + \frac{1}{2}gt^{2}$ .

Take $g = 10 m/s^{2}$ and solve the equation we get $t = \boxed{6 s}$ ~~~

When you flick lower mass up, it will stop pulling down on the upper mass. String goes slack and the heavier 2m mass hanging on the other side moves down, pulling upper mass with it. Now upper m is going up with acceleration while lower m is going up but with deceleration due to gravity.

String becomes taut again when both lower and upper mass have moved up by by identical height. Then the length of the string is restored and it becomes taut again since upper m accelerates up while lower m is accelerated down by gravity.

Movement of lower mass ---

s = ut + 1/2gt^2 = 40t - 5t^2

Let both lower and upper mass move up by distance s in time t. We have to find that value of t when distance traveled by both masses is equal.

Find Tension in the rope = T

Both 2m and upper m move at same acceleration a. 2m goes down, m goes up.

a= (2mg - T)/ 2m = (T-mg)/m

T = 4mg/3

So Force F pulling upper mass m upward is Tension minus its weight

F = 4mg/3 - mg = mg/3

Acceleration of upper mass = F/m = g/3 = 10/3

So motion of upper mass

s = ut +1/2at^2 = (5/3) t^2

Equating motion of both lower and upper masses

40 t - 5t^2 = (5/3) t^2

t = 6 seconds...