A small paint bucket of mas 5.0 kg is attached to a larger bucket of mas 15.0 kg by a light rope of 25 m long. The small bucket is at rest on the ground, and the rope goes up and over a small frictionless pulley and is attached to the larger bucket, which is on a window sill, 11 meters above the ground. There is no slack in the rope. If the heavy bucket slips off the window sill, causing the smaller bucket to rise and the two to collide in mid air, after what time will the big spash occur?
I got the first step of the question; i wrote the two Fnet equations and solved for acceleration, but i don't know where to go after that.
d = 11 meters
They both go the same acceleration, speed, and distance.
Therefore each goes half the distance.
d/2 = (1/2) a t^2
After obtaining the acceleration, you can use it to find the time it takes for the big splash to occur. To do this, you need to find the time it takes for the smaller bucket to reach the window sill.
Let's denote the acceleration as "a" (which you have already calculated). Now, the acceleration of the smaller bucket is the same as the acceleration of the larger bucket since they are connected by the rope.
Since the smaller bucket is initially at rest, you can use the second equation of motion (assuming no initial velocity) to find the time it takes for it to reach the window sill:
S = ut + (1/2)at^2
Here, S is the vertical displacement (11 m), u is the initial velocity (0 m/s), t is the time, and a is the acceleration.
Substituting the values into the equation:
11 = 0t + (1/2)at^2
22 = at^2
Now, you can solve for t:
t^2 = 22/a
Taking the square root of both sides:
t = √(22/a)
Finally, substitute the value of "a" that you calculated earlier, and calculate the square root to find the time it takes for the smaller bucket to reach the window sill.