JEE Advanced · 2006 · 09 Apr 2006 · Laws of Motion

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Two blocks A and B of masses 2m and m, respectively, are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in the figure. The magnitudes of acceleration of A and B, immediately after the string is cut, are respectively

(A)g, g/2
(B) $g /2, g$ ✓ Correct
(C)g , g
(D) $g /2, g /2$

Answer: (B)

Solution

Before the string is cut, the spring supports the total weight

(2 m + m) g = 3 m g

, so the spring force is

F_{s} = 3 m g

.

Immediately after the string is cut:

Block B (mass m) is free and falls with acceleration $a_{B} = g$ (free fall).
Block A (mass 2m) still has the spring force $F_{s} = 3 m g$ acting upward and weight $2 m g$ downward.
Net upward force on A: $F_{n e t} = 3 m g - 2 m g = m g$
Acceleration of A: $a_{A} = \frac{m g}{2 m} = \frac{g}{2}$ (upward)

Therefore, the magnitudes are

g /2

and

g

respectively, which is option (B).

JEE Advanced · 2006 · 09 Apr 2006 · Single correct option

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JEE Advanced · 2006 · 09 Apr 2006 · Laws of Motion

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