A 2, kg block moving with 10, m/s strikes a s | vedclass

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(C) Let the mass of both blocks be $m = 2\, kg$. The initial velocity of the first block is $u_1 = 10\, m/s$ and the second block is $u_2 = 0\, m/s$.S

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$10$

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(C) Let the mass of both blocks be $m = 2\, kg$. The initial velocity of the first block is $u_1 = 10\, m/s$ and the second block is $u_2 = 0\, m/s$.Since the floor is smooth,there is no external horizontal force on the system. Thus,the linear momentum of the system is conserved.Initial momentum $P_i = m u_1 + m u_2 = 2(10) + 2(0) = 20\, kg\cdot m/s$.When the spring is maximally compressed,both blocks move with the same velocity $v_{cm} = \frac{m u_1 + m u_2}{m + m} = \frac{20}{4} = 5\, m/s$.When the spring returns to its natural length,the blocks separate. Let the final velocities be $v_1$ and $v_2$.By conservation of momentum: $m v_1 + m v_2 = 20 \implies v_1 + v_2 = 10$.By conservation of energy: $\frac{1}{2} m u_1^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 \implies v_1^2 + v_2^2 = 100$.Substituting $v_2 = 10 - v_1$ into the energy equation: $v_1^2 + (10 - v_1)^2 = 100 \implies v_1^2 + 100 - 20 v_1 + v_1^2 = 100 \implies 2 v_1^2 - 20 v_1 = 0$.This gives $v_1 = 0$ or $v_1 = 10$.Since the first block cannot pass through the second,$v_1 = 0$ and $v_2 = 10\, m/s$.

$A$ $2\, kg$ block moving with $10\, m/s$ strikes a spring of constant $\pi^2\, N/m$ attached to a $2\, kg$ block at rest kept on a smooth floor. The velocity of the rear $2\, kg$ block after it separates from the spring will be ..... $m/s$.

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