A 10.0-kg block of ice is sliding due east at 9.00 m/s when it collides elastically with a 6.00-kg block of ice that is sliding in the same direction at 5.00 m/s. Define the positive direction as pointing due east, the 10.0-kg block as block A, and the 6.00-kg block as block B. Determine the velocity vaf of block A after the collision. Vaf = m/s due east Determine the velocity ÜBf Of block B after the collision. UBf = m/s due east

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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**Collision of Ice Blocks**

**Problem Statement:**

A 10.0-kg block of ice is sliding due east at 9.00 m/s when it collides elastically with a 6.00-kg block of ice that is sliding in the same direction at 5.00 m/s.

Define the positive direction as pointing due east, the 10.0-kg block as block A, and the 6.00-kg block as block B.

**Tasks:**

1. Determine the velocity \( \vec{v}_{Af} \) of block A after the collision.
   - \( \vec{v}_{Af} = \_\_\_\_ \) m/s due east

2. Determine the velocity \( \vec{v}_{Bf} \) of block B after the collision.
   - \( \vec{v}_{Bf} = \_\_\_\_ \) m/s due east

**Diagrams:**

If there were diagrams, they would illustrate the initial and final velocities of the blocks. The blocks are shown moving in the same direction (east) before and after the collision, with their velocities adjusting according to the principles of elastic collisions.

### Explanation

In an elastic collision, both momentum and kinetic energy are conserved. You will need these principles to solve for the velocities of the blocks after the collision.
Transcribed Image Text:**Collision of Ice Blocks** **Problem Statement:** A 10.0-kg block of ice is sliding due east at 9.00 m/s when it collides elastically with a 6.00-kg block of ice that is sliding in the same direction at 5.00 m/s. Define the positive direction as pointing due east, the 10.0-kg block as block A, and the 6.00-kg block as block B. **Tasks:** 1. Determine the velocity \( \vec{v}_{Af} \) of block A after the collision. - \( \vec{v}_{Af} = \_\_\_\_ \) m/s due east 2. Determine the velocity \( \vec{v}_{Bf} \) of block B after the collision. - \( \vec{v}_{Bf} = \_\_\_\_ \) m/s due east **Diagrams:** If there were diagrams, they would illustrate the initial and final velocities of the blocks. The blocks are shown moving in the same direction (east) before and after the collision, with their velocities adjusting according to the principles of elastic collisions. ### Explanation In an elastic collision, both momentum and kinetic energy are conserved. You will need these principles to solve for the velocities of the blocks after the collision.
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