Question

Running at 2.0 m/s, Burce, the 45.0 kg quarterback, collides with Max, the 90.0 kg tackle, who is traveling at 7.0 m/s in the other direction. Upon collision, Max continues to travel forward at 1.0 m/s. What is the new velocity of Bruce?(list unknown variable and known variables, write an equation, plug in numbers, and get answer with unit.)

Answer 1

Using the principle of conservation of momentum to determine Bruce's new velocity after the collision with Max, we find that Bruce's velocity after the collision is approximately -8.0 m/s, which means he is pushed backward.

Step-by-step explanation:

The subject of this question fits under Physics as it involves concepts of conservation of momentum and kinematics. To solve for the new velocity of Bruce, the quarterback, we can use the principle of conservation of momentum which states that the total momentum of an isolated system remains constant if no external forces act on it. Here are the steps:

Known Variables: Bruce's mass (m1) = 45.0 kg, Bruce's velocity (v1) = 2.0 m/s, Max's mass (m2) = 90.0 kg, Max's velocity (v2) = -7.0 m/s, Max's velocity after collision (v2') = 1.0 m/s.

Unknown Variable: Bruce's velocity after collision (v1').

Equation: Using the conservation of momentum m1v1 + m2v2 = m1v1' + m2v2'.

Calculation: Plug in the numbers: (45.0 kg)(2.0 m/s) + (90.0 kg)(-7.0 m/s) = (45.0 kg)(v1') + (90.0 kg)(1.0 m/s).

After calculating, we find that Bruce's velocity after the collision, v1', is approximately −8.0 m/s, indicating that Bruce is pushed backward after the collision.