A block is attached to a horizontal spring and is initially at rest at the equilibrium position $ x = 0 $, as shown in Figure $ 1 $. The block is then moved to position $ x = -A $, as shown in Figure $ 2 $, and released from rest, undergoing simple harmonic motion. At the instant the block reaches position $ x = +A $, another identical block is dropped onto and sticks to the block, as shown in Figure $ 3 $. The two-block-spring system then continues to undergo simple harmonic motion. Which of the following correctly compares the total mechanical energy $ E_{\text{tot},2} $ of the two-block-spring system after the collision to the total mechanical energy $ E_{\text{tot},1} $ of the one-block-spring system before the collision?

Question

A block is attached to a horizontal spring and is initially at rest at the equilibrium position $ x = 0 $, as shown in Figure $ 1 $. The block is then moved to position $ x = -A $, as shown in Figure $ 2 $, and released from rest, undergoing simple harmonic motion. At the instant the block reaches position $ x = +A $, another identical block is dropped onto and sticks to the block, as shown in Figure $ 3 $. The two-block-spring system then continues to undergo simple harmonic motion. Which of the following correctly compares the total mechanical energy $ E_{\text{tot},2} $ of the two-block-spring system after the collision to the total mechanical energy $ E_{\text{tot},1} $ of the one-block-spring system before the collision?(a) $ E_{\text{tot},2} = E_{\text{tot},1} $
, (b) $ E_{\text{tot},2} = \dfrac{1}{2} E_{\text{tot},1} $
, (c) $ E_{\text{tot},2} = 2 E_{\text{tot},1} $
, (d) $ E_{\text{tot},2} = \dfrac{1}{4} E_{\text{tot},1} $
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