Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

67 of 130

Question 67 of 130

Two blocks of masses 22 kg and 11 kg respectively, are tied to the ends of a string which passes over a light frictionless pulley as shown in the figure below. The masses are held at rest at the same horizontal level and then released. The distance traversed by the centre of mass in 22 s is _______ m. (Take g=10g = 10 m/s2^2)

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Answer: (C) 2.222.22

The acceleration of the blocks in the Atwood machine is given by:

a=m1−m2m1+m2ga = \dfrac{m_1 - m_2}{m_1 + m_2} g

Substituting the given values (m1=2m_1 = 2 kg, m2=1m_2 = 1 kg):

a=2−12+1g=g3a = \dfrac{2 - 1}{2 + 1} g = \dfrac{g}{3}

Taking the downward direction as positive, the acceleration of the 22 kg block is a1=g3a_1 = \dfrac{g}{3} and the acceleration of the 11 kg block is a2=−g3a_2 = -\dfrac{g}{3}.

The acceleration of the centre of mass is:

acm=m1a1+m2a2m1+m2a_{cm} = \dfrac{m_1 a_1 + m_2 a_2}{m_1 + m_2}

acm=2(g3)+1(−g3)2+1=g33=g9a_{cm} = \dfrac{2 \left(\dfrac{g}{3}\right) + 1 \left(-\dfrac{g}{3}\right)}{2 + 1} = \dfrac{\dfrac{g}{3}}{3} = \dfrac{g}{9}

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=2t = 2 s is:

Scm=12acmt2S_{cm} = \dfrac{1}{2} a_{cm} t^2

Scm=12(109)(2)2=209≈2.22 mS_{cm} = \dfrac{1}{2} \left(\dfrac{10}{9}\right) (2)^2 = \dfrac{20}{9} \approx 2.22 \text{ m}

Answer: 2.222.22

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