JEE ADVANCED LAWS OF MOTION QUESTION

Question

In the figure shown, C is a fixed wedge on horizontal surface. Blocks A and B are of
masses m and 2m respectively are kept as shown in figure. They can slide along the
incline plane smoothly. The pulley and string are massless. The inclined planes are
very long. A and B are released from rest. 2 seconds after the release, B is caught for a
moment and released again. Then (Take theta= 30 and g =10m/ s2 ) : Choose the correct
option(s).

A) the speed of ‘A’ just before the instant when the string becomes tight again will be
10/3 m/s
B) the speed of ‘A’ just before the instant when the string becomes tight again will be
zero.
C) the speed of ‘A’ just after the instant when the string becomes tight again will be
20/9 m/s
D) the speed of ‘A’ just after the instant when the string becomes tight again will be
10/3 m/s

Answer 1

Initially acceleration of B and A is same along the string which is given by
a=2mgsin30∘−mgsin30∘3m=g6

After 2 seconds, speed of both A and B is
VB=VA=g6×2=103 m/s

When B is caught for a moment and released again, speed of B becomes zero,while A is still having a speed 103m/s up the inclined due to which string will become slack. But A will decelerate and B will accelerate. Because of this the string will become tight again after A and B travel the same distance.

Let this time interval be t,
From, s=ut+12gt2
103×t−12gsin30∘×t2=12gsin30∘×t2
as t≠0, t=23sec

At this time speed of A is given by
v=103−g2×t=103−102×23=0