1.66 seconds ???
In a simple fixed pulley system, two unequal masses m1 and m2 are connected by a light string passing over a light fixed pulley.
the larger mass is stopped for a momment 2 secs after the system is set into motion. Find the time elapsed before the string is tight again...
m1=300g
m2=600g
now the words a momment or an instant are tricky.....by these how much time is meant other wise the problem is not a problem!!
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15 Answers
49what about this???
i just want to know what is the meaing of the phrase the larger mass is stopped for a momment.........how much time is that supposed to mean???help please!![2][2]
49
now if this describes the situation described in the sum then, what i ask is what is the value of dt???
49yea red lines signify the direction of velocity and black lines signify the direction of acceleration!! and i cannot edit posts!![2][2]
1SIMPLY SUPPOSE AT t=2 sec the large block starts moving from rest and small block has velocity which it gained in first 2 seconds
after 2 seconds they become free from each other for some time
for finding time assume length of string is constant and equate distance travelled by 300 kg in upward = distace travelled by 600 kg in downward from rest.
11How do u suppose that at t=2 sec, the larger block starts to move?
I'm getting the ans as \frac{3Mv}{2g(m+M)} where v=\frac{4g}{3} & M is the larger block.
1at t=2sec the large block is STOPPED FOR A MOMENT, NOT PERMANENTLY STOPPED..
u assumed that the block is permanently stopped so u are getting twice the answer from my answer....
11for a moment means the velocity of that mass will be zero.
Then it gain starts accelerating
11I have assumed it to be at rest only for a moment harsh - the assumption that u are talking would give an entirely different ans.
I did the way virang is thinking on.
1I THINK YOUR VELOCITY IS WRONG
CHECK UR VELOCITY
I GET IT AS 20/3 BUT U GET AS 40/3