The acceleration of the blocks in the Atwood machine is given by:
a=m1+m2m1−m2g
Substituting the given values (m1=2 kg, m2=1 kg):
a=2+12−1g=3g
Taking the downward direction as positive, the acceleration of the 2 kg block is a1=3g and the acceleration of the 1 kg block is a2=−3g.
The acceleration of the centre of mass is:
acm=m1+m2m1a1+m2a2
acm=2+12(3g)+1(−3g)=33g=9g
Since the system is released from rest, the initial velocity of the centre of mass is zero.
The distance traversed by the centre of mass in t=2 s is:
Scm=21acmt2
Scm=21(910)(2)2=920≈2.22 m
Answer: 2.22
