This is a problem involving conservation of momentum, where the momentum before the collision equals the momentum after the collision since no external forces are acting on the system.

Step 1: Define the variables

• Tackler:

  • Mass ($m_1$) = 130 kg
  • Velocity ($v_1$) = 2.7 m/s

• Halfback:

  • Mass ($m_2$) = 100 kg
  • Velocity ($v_2$) = -5.8 m/s (negative because it’s in the opposite direction)

• Final velocity ($v_f$) is the same for both since they stick together after the collision.

Step 2: Conservation of momentum equation

$m_1 v_1 + m_2 v_2 = (m_1 + m_2)v_f$

Step 3: Plug in the known values

$(130 \, \text{kg})(2.7 \, \text{m/s}) + (100 \, \text{kg})(-5.8 \, \text{m/s}) = (130 \, \text{kg} + 100 \, \text{kg})v_f$

$351 - 580 = 230v_f$

$-229 = 230v_f$

Step 4: Solve for $v_f$

$v_f = \frac{-229}{230} \approx -1.0 \, \text{m/s}$

Final Answer:

The mutual speed immediately after the collision is approximately: $\boxed{-1.0 \, \text{m/s}}$ (Negative indicates the direction is toward the halfback's original direction.)

Would you like further clarification or additional examples?

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Relative Questions:

1. What if the halfback had a greater mass than the tackler? How would the final velocity change?
2. How does the conservation of momentum apply if one of them was stationary?
3. Why is the velocity negative in this case, and what does it signify in terms of direction?
4. If the collision was elastic, how would the situation differ?
5. How does momentum conservation relate to energy conservation in such collisions?

Tip: Always pay attention to the direction of velocities in momentum problems—they're critical for accuracy!