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Ask iit jee aieee pet cbse icse state board experts Expert Question: simple harmonic motion(blocks spring problem).
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rini (221)

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Two blocks A(5kg) and B(2kg) attached to the ends of a spring constant 1120N/m are placed on smooth horizontal plane with the spring undeformed. simultaniously velocities of 3m/s and 10m/s along the line of the spring in the same direction are imparted to A& B then:
 
(a)what is the maximum extention of the spring?
(b)when does the first maximum compression occurs after start?

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sboakarthik (2)

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maximum extension of the spring will be 2.23cm

if it its wrong pls notify me
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rini (221)

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the correct ans is 25cm.. but ur proceedure may be correct. plz can u help me out by some usuful hints abt this ques which will help me to get the ans(not neccessorily the whole solution but imp hint)..
thanks..

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rini (221)

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some help plzzz

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catch_arnnie (521)

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solution::
 
for part a)

see, in my method, i hav considered that if we find the amount of maximum compression then the same amount will be the maximum extension.

so, to find maximum compression, we use two equations

1)conservation of momentum

m1v1 + m2v2 = m1v + m2v
at the time of maximum compression, the velocities of the blocks will be same.

5 * 3 + 2 * 10 = 5v + 2v

=> v=5m/s

2)conservation of energy

1/2 m1 v1^2 + 1/2 m2 v2^2 = 1/2 m1 v^2 + 1/2 m2 v^2 + 1/2 k x^2 where x is the amount compression

putting the values & solving,

=> 45 + 200 = 175 + 1120 x^2
=>x^2 = 70/1120

=>x = 1/4 m or 25cm

i hope you got the solution..
if not then plz ask...

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rini (221)

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thanks a lot........
i got ur solution..
thanks once again 4 ur help..

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catch_arnnie (521)

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your welcome....

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