Question

Let A represent the ball moving at 2.00 m/s, and call that direction the positive direction. Let B represent the ball moving at 3.60 m/s in the opposite direction. Thus, VA =2.00 m/s and VB=-3.60 m/s. Use Eq. 7-7 to obtain a relationship between the velocities.

VA-VB=-(V’A-V’B) → V’B = 5.60 m/s+ V’A

Substitute this relationship into the momentum conservation equation for the collision, noting that MA=MB

Answer 1

So the velocity of ball A after the collision is 1.80 m/s.

Step-by-step explanation:

The momentum conservation equation for a collision is given by the equation:

MAVA + MBVB = MAV'A + MBV'B

Substituting in the given values, we have:

MA2.00 m/s + MB(-3.60 m/s) = MAV'A + (MB(5.60 m/s + V'A))

Since MA=MB, we can simplify the equation further:

2.00 m/s + (-3.60 m/s) = V'A + (5.60 m/s + V'A)

Solving for V'A, we get:

V'A = 1.80 m/s

So the velocity of ball A after the collision is 1.80 m/s.