JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Conservation of Momentum

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Two blocks of masses 2 kg and 1 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 2 s is ____ m.
(Take

g = 10 m / s^{2}

)

(A)3.33
(B)3.12
(C)2.22✓ Correct
(D)1.42

Answer: (C)

Solution

The pulley system is an Atwood machine, so each block has acceleration

a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = \frac{( 2 - 1 ) \times 10}{3} = \frac{10}{3} m/s^{2}

, the 2 kg block moving down and the 1 kg block up. The centre of mass accelerates at

a_{c m} = \frac{m _{1} a - m _{2} a}{m _{1} + m _{2}} = \frac{2 - 1}{3} \times \frac{10}{3} = \frac{10}{9} m/s^{2}

directed downward. Starting from rest, the distance covered by the centre of mass in

t = 2

s is

S = \frac{1}{2} a_{c m} t^{2} = \frac{1}{2} \times \frac{10}{9} \times 4 = \frac{20}{9} m

. The centre of mass moves

\frac{20}{9} \approx 2.22

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Conservation of Momentum

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