Question

Help! 50 points! Class 11 Physics. Q. 43

Answer 1

After burning thread PA, Block B moves downward and spring moves leftward .(see the attachment)
Let ∅ is the angle formed by spring with vertical
Then, final length of string ( I mean after extension of string ) is ( lo +x ) where, x is extension of string .
Btw here, (lo+x ) = lo.sec∅ ,as you can see that .but I took ( lo+x) for simplification.
Now, break the components of forces according to figure .
Upward force balanced with downward force
Vertical component of Spring force balance with weight .( Because when we burn the thread, after sometimes block A lift up and loose normal reaction. At instant , spring force = weight.)
e.g. N = Fs.cos∅ = K.xcos∅ = mg
Where x is elongation of spring .
Here, cos∅ = lo/(lo+x)
k.xlo/(lo+x) =mg
from equation ,
5mg/lo = k,
5mg.x/(lo+x) = mg
5x = lo + x
x = lo/4 -----------(1)
again, for y ,
use Pythagoras theorem,
y= √{(lo+x)² - lo²}
= √{lo² + x² + 2lo.x - lo²}
= √{lo²/16 + 2lo.lo/4}
= √{lo²/16 + lo²/2}
= 3lo/4 -----------------(2)
Now, use the conversation of energy ,
∆K.E + ∆P.E = 0
1/2mv² ( for block A) + 1/2mv² ( for block B) + 1/2kx² - m.gy ( for block B) - mg.0(for block A) = 0
mv² + 1/2× 5mg/lo.lo²/16 = mg.3lo/4 { from eqns (1) and (2) }
mv² + 5mg.lo/32 = 3m.glo/4
mv² = 19mg.lo/32
v² = 19g.lo/32
v = (19glo/32)½
now put the value of lo and g and get answer .
e.g v = ( 19× 9.8 × 0.5/32)½
v =(2.9)½ = 1.7 m/s

Answer 2

Answer is in attachment

thank you