A U shaped tube of mass 2 m is placed on a horizontal surface. Two identical spheres each of diameter d (just less than the inner diameter of tube) and mass $m$ enter into the tube with velocity $v$ as shown in the figure taking all collisions to be elastic and all surfaces smooth.

Question

TestHub · Accepted Answer

(A) Apply momentum conservation in the direction of initial velocity of balls. $mv + mv = mv^{'} + mv^{'} + 2 mv^{'}$

$v^{'} = \frac{2 v}{4} \Rightarrow \frac{v}{2}$

(B) For speed apply energy conservation

$\frac{1}{2} m v^{2} + \frac{1}{2} m v^{2} = \frac{1}{2} 2 m v^{'2} + 2 (\frac{1}{2} m v_{b}^{2})$

$mv^{2} = mv^{'2} + mv_{b}^{2}$

$v^{2} - \frac{v ^{2}}{4} = v_{b}^{2} \Rightarrow \frac{3 v ^{2}}{4} = v_{b}^{2}$

$\Rightarrow v_{b} = \frac{3}{2} V$

Physics - Center of Mass Question with Solution | TestHub

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JEE Advanced 2026