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tarun_bits

hcv ques--50

tarun_bits

5 pointer 2 one who will solve...

svj29

m=320g=3.2kg
k=40
h=40cm
plz form d FBD or interpret frm d soln:
from FBD
kx cosz=mg (theta=z)
(when the block breaks of, R=0)

cosz=mg/kx
0.4/0.4+x=3.2/40x

x=0.1m

s=underroot of (h+x)square - h square
=underroot of 0.5square - 0.4square=0.3m

let the velocity of the body at B be v
change in KE=wrk done (4 d systm)

(1m into u square/2 + 1m into v square/2) = -1k into x square/2
+mgs

=> 0.32 into v square = u knw all d values jz put n u'll get
v=1.5m/s

frm d figure.
delta l =h(secz - 1) 1
frm energy consrvatn:
mgs=2[half m into v square] + half into K into (delta l) square

mgh tanz=m into v square + half k into h square into (secz-1)square

dis is eqn 2

frm the law of motion at break of
mg=kh (secz-1)cosz
1-cosz=mg/kh
cosz=kh-mg/kh
=0.8
putting value of z in eqn 2
0.32 into 10 into 0.4 into 0.75
=0.32 into v square + half into 40 into (0.4)square(1.25-1)square

solve for v and u'll get 1.5m/s....

phew!!!!

BALGANESH

waterdemon posted this soln

The Energy is stored in the following three forms:

(a)Elastic Potential Energy of the Spring:

(Stretched by a length of "dL")

= (1/2)*K*(dL)²

(b)K.E of Block A = (1/2)mv²

(c)K.E of Block B = (1/2)mv²

Now Applying the law of conservation of energy,

mgx = (1/2)k(dL)² + (1/2)mv²+ (1/2)mv²

v² = gx - (k/2m)(dL)² ..........(1)

Considering the Equilibrium, of Block A at position A :

R + FCos@ = mg ...........(2)

For a Spring,

F = K(dL) and for Breaking off R = 0.

Substituting the value of R and F in Eq.(2)

K(dL)Cos@ = mg ..........(3)

From the Geometry of the problem ,

dL = (L/Cos@) - L = L{ (1/Cos@) - 1 }

Substituting in Eq.(3)

K{ L [ (1/Cos@) - 1 ]Cos@ = mg

On solving we get,

1 - Cos@ = mg/KL

Cos@ = [1-(mg/KL)]

Cos@ = 1 - (0.32*10/40*0.40) = 4/5

and dL = 0.4(5/4-1)

dL = 0.1 m from above.

Now x=L(tan@) = 0.4*(0.3/0.4) = 0.3m.

Subsitituting the values in eq.(1)

v² = 10*0.3 (40*(0.1)²/2*0.32)

v² = 3 - 0.64

v² = 2.36

v = 1.5 m/s.

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