As shown in the figure,two blocks are connect | vedclass

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(C) The velocity of the center of mass $(v_{cm})$ is given by: $v_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{20 	imes 6 + 10 	imes (-3)}{20

Vedclass · Accepted Answer

When the spring is at maximum elongation,the velocity of the $10 \, kg$ block is $3 \, m/s$ (towards left).

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(C) The velocity of the center of mass $(v_{cm})$ is given by: $v_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{20 	imes 6 + 10 	imes (-3)}{20 + 10} = \frac{120 - 30}{30} = \frac{90}{30} = 3 \, m/s$ (towards right). Thus,option $(A)$ is correct. At the time of maximum elongation or maximum compression,the relative velocity of the blocks is zero,meaning both blocks move with the same velocity equal to the velocity of the center of mass ($v_{cm} = 3 \, m/s$ towards right). Therefore,the velocity of both the $10 \, kg$ block and the $20 \, kg$ block at these instances is $3 \, m/s$ (towards right). Comparing this with the given options: Option $(B)$ states the velocity of the $20 \, kg$ block is $3 \, m/s$ (towards right),which is correct. Option $(C)$ states the velocity of the $10 \, kg$ block is $3 \, m/s$ (towards left),which is incorrect. Since the question asks to choose the wrong alternative,$(C)$ is the wrong statement.

As shown in the figure,two blocks are connected with a light spring. When the spring is at its natural length,velocities are given to them as shown in the figure. Choose the wrong alternative.

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