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14 Jul 2009 00:40:33 IST

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pls solve hcv ques.23 exercise newton,s laws of motions.................................

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pls solve hcv ques.23 exercise newton,s laws of motions.................................

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adi

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14 Jul 2009 00:42:52 IST

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plz rit the question

Killer Kiran

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14 Jul 2009 00:44:58 IST

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@ namitha.. turn two pages.. and look at qn 23.. u ll get the answer.. pls post the qn..

namita lohani

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14 Jul 2009 01:09:38 IST

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in a simple atwood machine ,two unequal masses m1=300 g nd m2=600 g are connected by a string going over a clamped light smooth pulley .the larger mass ig stopped for a moment 2 s after the system is set into motion .find the time elapsed before the string is tight again.

ramyani chakrabarty

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14 Jul 2009 01:30:16 IST

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Q. in a simple atwood machine ,two unequal masses m1=300 g nd m2=600 g are connected by a string going over a clamped light smooth pulley .the larger mass ig stopped for a moment 2 s after the system is set into motion .find the time elapsed before the string is tight again.

Use the diagram of Q 22 of HCV u referred.

T-m1g = m1 a ......................(1)
m2g - T = m2 a ......................(2)

a= (m2 -m1)g / (m1 + m2 ) = (0.6 - 0.3 ) * 10 / (0.6 + 0.3 ) = 3.33 m /s sq
T = m1 (g + a ) = 4 N

[ g is taken 10]
After 2 sec the m1 has velocity v = u + at = 6.66 m /s [ u = 0 ]
This is upward.
At time 2 sec, the mass m2 stops for a moment. But it was also moving with 6.66 m /s velocity in downward dirn. This is stopped momentarily.

But m1 continues to move till its vel is 0. How long m1 moves ?
Here v = 0 , u = 6.66 m /s

But a is not 3.33 m /sec sq.
a = -- 10 m /sec sq.

The vel of m1 is zero. It means is being retarded. The retarding force is gravity. The retarding accln is 10 m /sec sq.

The rest is simple.

v = u + gt

0= 6.66 - 10 t

t = 0.66 = 2/3 sec

During this 2 /3 sec m2 is also moving downward. So the string will be tight after 2/3 sec.

http://www.goiit.com/posts/list/mechanics-q-22-28286.htm

adi

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14 Jul 2009 01:35:29 IST

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@ ramyani chakrabarty,i think ur magnetic field question is rightly answered by me ........wats d mistake in da answer

Akshay

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14 Jul 2009 01:54:46 IST

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When one of the blocks is held, the string goes slack. So the tension in the string becomes zero. So the other block will movupward under the influence of gravity. It's basically uniformly retarded motion, while the

So, m1 , m2 be the masses. Then, acceleration of each block is thus {(m2-m1)/(m1+m2)}g=a

Now, the upper block has a velocity of 2a after 2 secs, hence, the the time taken to reach the maximum height in its motion under gravity

is given by 2a-gt=0

t=2a/g.

It is at this point that the system regains its tension

Solving for t it comes out to be 2/3 seconds.

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