**Problem Statement:**A 70 kg football player runs with a velocity of 3 m/s towards a 60 kg player running at 4 m/s. The two players stick to each other upon collision.**Questions:**a. What type of interaction is this? (explosion, elastic collision, or inelastic collision)b. What is the combined velocity of the players after the collision?### Solution:a. **Type of Interaction:**This interaction is an **inelastic collision**. In an inelastic collision, the colliding objects stick together after the collision.b. **Combined Velocity Calculation:**To calculate the combined velocity of the players after the collision, we can use the principle of conservation of momentum. The total momentum before the collision will be equal to the total momentum after the collision.Let:- Mass of the first player, \( m_1 = 70 \) kg,- Velocity of the first player, \( v_1 = 3 \) m/s,- Mass of the second player, \( m_2 = 60 \) kg,- Velocity of the second player, \( v_2 = 4 \) m/s.Since the players stick together, we can use the following formula for the final velocity (\( v_f \)):\[ v_f = \frac{m_1 \cdot v_1 + m_2 \cdot v_2}{m_1 + m_2} \]Given:- \( m_1 = 70 \) kg,- \( v_1 = 3 \) m/s,- \( m_2 = 60 \) kg,- \( v_2 = 4 \) m/s.Substitute the values into the formula:\[ v_f = \frac{70 \cdot 3 + 60 \cdot 4}{70 + 60} \]\[ v_f = \frac{210 + 240}{130}\]\[ v_f = \frac{450}{130}\]\[ v_f = 3.46 \text{ m/s}\]Thus, the combined velocity of the players after the collision is approximately \( 3.46 \) m/s.

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Description: **Problem Statement:**A 70 kg football player runs with a velocity of 3 m/s towards a 60 kg player running at 4 m/s. The two players stick to each other upon collision.**Questions:**a. What type of interaction is this? (explosion, elastic collision,

Given: The mass of the first player is 70 kg. The mass of the second player is 60 kg. The velocity…

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2. A 70 kg football player runs with a velocity of 3 m/s towards a 60 kg player running-4. The 2 players stick to each other upon the collision. a. What type of Interaction Is this? (explosion, elastic collision, or Inelastic collislon)? b. What Is the combined velocity of the players after the collision?

2. A 70 kg football player runs with a velocity of 3 m/s towards a 60 kg player running-4. The 2 players stick to each other upon the collision. a. What type of Interaction Is this? (explosion, elastic collision, or Inelastic collislon)? b. What Is the combined velocity of the players after the collision?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Topic Video

Question

**Problem Statement:**

A 70 kg football player runs with a velocity of 3 m/s towards a 60 kg player running at 4 m/s. The two players stick to each other upon collision.

**Questions:**

a. What type of interaction is this? (explosion, elastic collision, or inelastic collision)

b. What is the combined velocity of the players after the collision?

### Solution:

a. **Type of Interaction:**

This interaction is an **inelastic collision**. In an inelastic collision, the colliding objects stick together after the collision.

b. **Combined Velocity Calculation:**

To calculate the combined velocity of the players after the collision, we can use the principle of conservation of momentum. The total momentum before the collision will be equal to the total momentum after the collision.

Let:
- Mass of the first player, \( m_1 = 70 \) kg,
- Velocity of the first player, \( v_1 = 3 \) m/s,
- Mass of the second player, \( m_2 = 60 \) kg,
- Velocity of the second player, \( v_2 = 4 \) m/s.

Since the players stick together, we can use the following formula for the final velocity (\( v_f \)):

\[
v_f = \frac{m_1 \cdot v_1 + m_2 \cdot v_2}{m_1 + m_2}
\]

Given:
- \( m_1 = 70 \) kg,
- \( v_1 = 3 \) m/s,
- \( m_2 = 60 \) kg,
- \( v_2 = 4 \) m/s.

Substitute the values into the formula:

\[
v_f = \frac{70 \cdot 3 + 60 \cdot 4}{70 + 60}
\]
\[
v_f = \frac{210 + 240}{130}
\]
\[
v_f = \frac{450}{130}
\]
\[
v_f = 3.46 \text{ m/s}
\]

Thus, the combined velocity of the players after the collision is approximately \( 3.46 \) m/s.

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