Physics Q28 – JEE-MAIN 2026

Home Physics Laws of MotionQuestion #211

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}

)

Clue 1 of 3: System Dynamics

First, determine the acceleration of the individual blocks in the Atwood machine setup.

🧠 Full Solution Path

Step 1: Calculate the acceleration of the individual blocks✦ Active

For an Atwood machine, the acceleration of the blocks is given by $a = \frac{(m_{1} - m_{2}) g}{m_{1} + m_{2}}$ . Given $m_{1} = 2 kg$ , $m_{2} = 1 kg$ , and $g = 10 {m/s}^{2}$ .

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

Taking the upward direction as positive, $m_{1}$ accelerates downwards ( $a_{1} = - \frac{10}{3} {m/s}^{2}$ ) and $m_{2}$ accelerates upwards ( $a_{2} = + \frac{10}{3} {m/s}^{2}$ ).

Step 2: Determine the acceleration of the center of mass○ Expand

The acceleration of the center of mass is given by $a_{C M} = \frac{m_{1} a_{1} + m_{2} a_{2}}{m_{1} + m_{2}}$ .

a_{C M} = \frac{(2 kg) (- \frac{10}{3} {m/s}^{2}) + (1 kg) (+ \frac{10}{3} {m/s}^{2})}{2 kg + 1 kg}

a_{C M} = \frac{- \frac{20}{3} + \frac{10}{3}}{3} = \frac{- \frac{10}{3}}{3} = - \frac{10}{9} {m/s}^{2}

The negative sign indicates that the center of mass moves downwards.

💡 Teacher's Secret Hint

Remember to account for the direction of acceleration for each mass when calculating the center of mass acceleration.

Step 3: Calculate the distance traversed by the center of mass○ Expand

Since the system starts from rest, the initial velocity of the center of mass is $u_{C M} = 0$ . The distance traversed in time $t = 2 s$ is given by $S_{C M} = | u_{C M} t + \frac{1}{2} a_{C M} t^{2} |$ .

S_{C M} = | 0 \times 2 + \frac{1}{2} (- \frac{10}{9}) (2)^{2} |

S_{C M} = | \frac{1}{2} \times (- \frac{10}{9}) \times 4 | = | - \frac{20}{9} | = \frac{20}{9} m

Converting to decimal, $S_{C M} \approx 2.22 m$ . This matches option 3.

💡 Teacher's Secret Hint

The question asks for the distance traversed, which is the magnitude of the displacement.

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Physics Question 28 – JEE-MAIN 2026

Clue 1 of 3: System Dynamics

🧠 Full Solution Path