a football player runs at 8m/s and plows into a 80 kg referee standing on the field causing the referee to fly forward at 5.0 m/s. if this were a perfectly elastic collision, what would the mass of the football player be?

We can solve this problem using the principles of **conservation of momentum** and **conservation of kinetic energy**, which both apply in perfectly elastic collisions.

### Step 1: Conservation of momentum
In an elastic collision, the total momentum before and after the collision is conserved. The formula for momentum is:

\[
p = mv
\]

where:
- \( m \) is the mass of the object
- \( v \) is the velocity of the object

Before the collision:
- The football player has a mass \( m_1 \) and velocity \( v_1 = 8 \, \text{m/s} \).
- The referee has a mass \( m_2 = 80 \, \text{kg} \) and is initially stationary, so \( v_2 = 0 \, \text{m/s} \).

After the collision:
- The football player's velocity becomes \( v_1' \) (which we don't know yet).
- The referee's velocity becomes \( v_2' = 5.0 \, \text{m/s} \).

Using conservation of momentum, the total momentum before and after the collision must be equal:

\[
m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'
\]

Substituting the known values:

\[
m_1 \cdot 8 + 80 \cdot 0 = m_1 v_1' + 80 \cdot 5
\]

This simplifies to:

\[
8m_1 = m_1 v_1' + 400
\]

### Step 2: Conservation of kinetic energy
In a perfectly elastic collision, the total kinetic energy before and after the collision is also conserved. The formula for kinetic energy is:

\[
KE = \frac{1}{2} mv^2
\]

So, using the conservation of kinetic energy:

\[
\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} m_1 v_1'^2 + \frac{1}{2} m_2 v_2'^2
\]

Substituting the known values:

\[
\frac{1}{2} m_1 (8^2) + \frac{1}{2} (80)(0^2) = \frac{1}{2} m_1 v_1'^2 + \frac{1}{2} (80)(5^2)
\]

Simplifying:

\[
32m_1 = \frac{1}{2} m_1 v_1'^2 + 1000
\]

Now we have two equations:

1. \( 8m_1 = m_1 v_1' + 400 \)
2. \( 32m_1 = \frac{1}{2} m_1 v_1'^2 + 1000 \)

Let me solve this system of equations for \( m_1 \).The mass of the football player is approximately \( 36.36 \, \text{kg} \), and after the collision, his velocity becomes \( -3.0 \, \text{m/s} \). The negative sign indicates that the football player is moving in the opposite direction after the collision.

Would you like further clarification or details on this problem?

Here are some related questions:
1. What happens to kinetic energy in an inelastic collision?
2. How does the velocity of both bodies change in an elastic collision?
3. How would the results differ if the collision were perfectly inelastic?
4. What is the role of impulse in momentum change during collisions?
5. How would angular momentum be conserved in a rotational system?

**Tip:** When solving physics problems involving collisions, always check if momentum and kinetic energy conservation apply, as they depend on whether the collision is elastic or inelastic.

A football player runs at 8m/s and plows into a 80 kg referee standing on the field causing the referee to fly forward at 5.0 m/s. If this were a perfectly elastic collision, what would the mass of the football player be?

Solve a problem involving a football player running at 8m/s colliding with an 80 kg referee in a perfectly elastic collision. Learn about the conservation of momentum and kinetic energy, and calculate the football player's mass. Ideal for Grades 10-12 physics students.

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