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

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ramyani chakrabarty (3105)

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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

NIT silchar electrical engineering

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