Question

JEE-Physics
7.
Three identical balls 1,2,3 are suspended on springs one below the other as
shown in the figure. OA is a weightless thread.
(a) If the thread is cut, the system starts falling. Find the acceleration of all the
balls at the initial instant
(b) Find the initial accelerations of all the balls if we cut the spring BC which is
supporting ball 3 instead of cutting the thread.
20
3
TL 11.1​

Answer 1

Answer:

It is assumed that all balls are of same mass m and both springs have same force constant k.  

Let x1 be the elongation of spring between A and B. Let x2 be the elongation of spring between B and C.

 Let aA, aB and aC are acceleration of balls A, B and C respectively

 

(1) equilibrium  

Let T is the tension in the string between Ball-A and support

for Ball-C :-                  k×x2 = m×g  ..............(1)

for Ball-B :-                  m×g + k×x2 = k×x1 .................(2)

                  using eqn.(1),       2×m×g = k×x1 ...................(3)

 for Ball-A :-               T = m×g + k×x1  .........................(4)

                   using eqn.(3)         T = 3×m×g  ........................(5)

 (2) When string between ball-A and support is broken

for Ball-A :-          m×g + k×x1 = m×aA .............(4)

          using eqn.(3), it can rewrite eqn.(4) as ,   3×m×g = m×aA  , aA = 3×g

for Ball-B :-    m×g + k×x2 - k×x1  = m×aB ...................(5)

   using eqn.(2), we know that LHS of  eqn.(5) is zero, hence aB = 0

for Ball-C:-    m×g - k×x2 = m×aC .......................(6)

        using eqn.(1), we know that LHS of  eqn.(6) is zero, hence aC = 0  

(3) When spring between ball-A and ball-B is broken

 Let T1 is the tension in the string between Ball-A and support  

for Ball-A :-          m×g  =  T1  .............(7)

                      m×g - T1 = m×aA  .............(8)

                      LHS of eqn.(8) is zero because of eqn.(7), hence aA = 0

for Ball-B :-    m×g + k×x2   = m×aB ...................(9)

   using eqn.(1), we rewrite eqn.(9) as,   2×m×g    = m×aB

Hence aB = 2×g

 for Ball-C:-    m×g - k×x2 = m×aC .......................(10)

              using eqn.(1), we know that LHS of  eqn.(6) is zero, hence aC = 0

(3) When spring between ball-B and ball-C is broken

Let T2 is the tension in the string between Ball-A and support

 

for Ball-A :-          m×g + k×x1  =  T2  .............(11)

                      m×g + k×x1 - T2 = m×aA  .............(12)

                      LHS of eqn.(12) is zero because of eqn.(11), hence aA = 0

 

for Ball-B :-    k×x1 - m×g    = m×aB ...................(12)

   using eqn.(3), we rewrite eqn.(9) as,   m×g  = m×aB

Hence aB = g (upwards)

 

for Ball-C:-    m×g  = m×aC .......................(6)

                      hence aC = g  (downwards)