# Dynamics of Particles : Problems involving pulleys.

## Homework Statement

Two particles P and Q of masses 1 kg and 2kg respectively are hanging vertically from the ends of a light inextensible string which passes over a smooth fixed pulley. The system is released from rest with both particles a distance of 1.5 m above a floor. When the masses have been moving for 0.5s the string breaks. Find the further time that elapses before P hits the floor.

2g - T = 2a
T- g = a
s=ut+1/2at^2

## The Attempt at a Solution

First, I found the acceleration of the two the particles, and then found the distance particle Q moves in the 0.5seconds. Then, added this distance to 1.5m as it summed up to be the distance that P had to cover in order to hit the floor. But I couldn't proceed to solve the rest of the sum as I don't understand the string breakage theory. Help, anyone?

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tbh i think you've done the hard part. so now you know how high P is above the floor when the string breaks. now P is in freefall. so its acceleration is 9.8 m/s^2, you've already calculated the height above the floor it is at and you only need one more thing to be able to find the time and that would be its initial velocity. you can find using v^2=u^2+2as and the values from the 1st part.

Tried that way, but the answer doesn't match. The answer is 0.813 seconds...

ideasrule
Homework Helper
Can you show all of your work? Did you use the height of P above the floor for the "s" in v^2=u^2+2as, or did you use the distance for which P accelerated (which is height-1.5 m)?