JEE Main 2026 — Apr 2, Shift 2 — Physics solution

JEE Main2026Apr 2, Shift 2Physics

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
$2.22$
D
$1.42$

Solution

For an Atwood machine, the acceleration of the masses is $a = \frac{( m _{2} - m _{1} ) g}{m _{1} + m _{2}} = \frac{( 2 - 1 ) ( 10 )}{3} = \frac{10}{3}$ m/s $^{2}$ .

The heavier block ( $2$ kg) accelerates downward at $\frac{10}{3}$ ; the lighter block ( $1$ kg) accelerates upward at $\frac{10}{3}$ .

Centre-of-mass acceleration (taking downward positive): $a_{c m} = \frac{2 \cdot ( 10/3 ) + 1 \cdot ( - 10/3 )}{3} = \frac{10/3}{3} = \frac{10}{9}$ m/s $^{2}$ downward.

Starting from rest, in $t = 2$ s: $S_{c m} = \frac{1}{2} a_{c m} t^{2} = \frac{1}{2} \cdot \frac{10}{9} \cdot 4 = \frac{20}{9} \approx 2.22$ m.

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