Final answer:
The x-component of the baseball's velocity just after it leaves the bat is 33.1 m/s in the +x direction, calculated using the impulse-momentum theorem.
Step-by-step explanation:
The question asks us to calculate the x-component of the baseball's velocity just after leaving the bat when an impulse of 6.9 N⋅s is applied to it. Since this is an intergalactic baseball game, there are no gravitational forces acting on the ball, and the problem simplifies to a one-dimensional impulse-momentum problem.
The impulse applied to the ball can be calculated using the impulse-momentum theorem, which states that the change in momentum (Δp) is equal to the impulse (J) applied to an object. Mathematically, this can be expressed as J = Δp. Since momentum is the product of mass (m) and velocity (v), we can rewrite this as J = m(vf - vi), where vf is the final velocity and vi is the initial velocity.
The initial momentum of the baseball is its mass times its initial velocity, which is 0.135 kg × (-18.0 m/s) = -2.43 kg⋅m/s. The impulse given to the ball is 6.9 N⋅s, which will change the baseball's momentum. Using the impulse-momentum theorem, we add the impulse to the initial momentum to find the final momentum:
Δp = J = 6.9 N⋅s
pf = pi + J = -2.43 kg⋅m/s + 6.9 kg⋅m/s = 4.47 kg⋅m/s
Finally, to find the final velocity, vf, of the baseball, we divide the final momentum by the mass of the baseball:
vf = pf / m = 4.47 kg⋅m/s / 0.135 kg = 33.1 m/s.
Hence, the x-component of velocity of the baseball just after leaving the bat is 33.1 m/s in the +x direction, i.e., back toward the pitcher.