---
title: "A block of mass \\(M\\) slides along a horizontal frictionless surface at a constant speed \\(v_0\\). The block is moving toward an ideal spring with spring constant \\(k\\) that is attached to a rigid wall. Two cases are considered:  Case 1: The block of mass \\(M\\) hits the spring and compresses it to a maximum potential energy \\(U_1\\).  Case 2: The block of mass \\(M\\) first collides with and sticks to a second stationary block of mass \\(M\\). The two-block system then hits the spring and compresses it to a maximum potential energy \\(U_2\\).  Which of the following correctly compares \\(U_2\\) to \\(U_1\\) and provides a correct justification?"
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url: "https://nerd-notes.com/ubq/109143/"
date_modified: "2026-03-19T09:11:38+00:00"
---

# A block of mass \(M\) slides along a horizontal frictionless surface at a constant speed \(v_0\). The block is moving toward an ideal spring with spring constant \(k\) that is attached to a rigid wall. Two cases are considered:

Case 1: The block of mass \(M\) hits the spring and compresses it to a maximum potential energy \(U_1\).

Case 2: The block of mass \(M\) first collides with and sticks to a second stationary block of mass \(M\). The two-block system then hits the spring and compresses it to a maximum potential energy \(U_2\).

Which of the following correctly compares \(U_2\) to \(U_1\) and provides a correct justification?

A block of mass \(M\) slides along a horizontal frictionless surface at a constant speed \(v_0\). The block is moving toward an ideal spring with spring constant \(k\) that is attached to a rigid wall. Two cases are considered:

Case 1: The block of mass \(M\) hits the spring and compresses it to a maximum potential energy \(U_1\).

Case 2: The block of mass \(M\) first collides with and sticks to a second stationary block of mass \(M\). The two-block system then hits the spring and compresses it to a maximum potential energy \(U_2\).

Which of the following correctly compares \(U_2\) to \(U_1\) and provides a correct justification?

![The diagram shows two horizontal setups on a frictionless floor. In Case 1, a single block of mass M is moving to the right with speed v_0 toward a horizontal spring attached to a wall on the right. In Case 2, the same block of mass M is moving to the right with speed v_0 toward an identical stationary block of mass M. Further to the right in Case 2 is the same spring-wall assembly.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1773914269-PxNDYL.jpg)

- **A.** \(U_2 = U_1\) because the total momentum of the system is the same in both cases, and all momentum is eventually transferred to the spring.
- **B.** \(U_2 = \dfrac{1}{2} U_1\) because kinetic energy is dissipated during the inelastic collision between the two blocks.
- **C.** \(U_2 = \dfrac{1}{4} U_1\) because the final speed of the two-block system is half of the initial speed of the single block.
- **D.** \(U_2 > U_1\) because the two-block system has more mass and therefore more inertia to compress the spring a greater distance.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/109143/*