1. The problem statement, all variables and given/known data A baseball with a mass of 0.125 kg is moving horizontally at 32.0 m/s [E], when it is stuck by a bat for 0.00200 seconds. The velocity of the ball just after the collision is 52.0 m/s [W 20 N]. 2. Relevant equations p = m Δv Fnet = ∆p/∆t there are also these equations Fnetx ∆t = m(v2x – v1x) Fnety ∆t = m(v2y – v1y) 3. The attempt at a solution i've gone over this a few times but nothing looks right to me... Given m for mass of baseball = 0.125 kg v1 for the velocity of the baseball before the collision = 32.0 m/s [E] v2 for the velocity of the baseball after the collision = 52.0 m/s [W 20 N] Δt for duration of time the baseball struck the bat = 0.00200 s a. Find the impulse experienced by the ball. (6 marks) Let [E] be positive To solve for the horizontal x-component of v2 v2x = 52.0 m/s * cos (20) v2x = 48.86 m/s To solve for the vertical y-component of v2 v2y = 52.0 m/s * sin (20) v2y = 17.79 m/s Since v2x is in the [W] direction, let v2x be negative To calculate the impulse, if p = m Δv Then, Δv = √((v2x – v1x)2 – (v2y – v1y)2) Δv = √((-48.86 m/s – 32.0 m/s)2 – (17.79 m/s – 0 m/s)2) Δv = √((-80.86 m/s)2 – (17.79 m/s)2) Δv = √((6538.34 m/s) – (316.48 m/s)) Δv = √(6221.87 m/s) Δv = 78.88 m/s So, p = m Δv p = (0.125 kg)(78.88 m/s) p = 8.86 kgm/s b. Find the average net force of the ball. (2 marks) Fnet = ∆p/∆t Fnet = (8.86 kgm/s)/(0.00200 s) Fnet = 4430 N <--- that's WAY too much force I also know there's an angle in there somewhere that i SHOULD be using force to calculate for the direction of the impluse tan θ = Fnety/Fnetx? i don't know honestly i'm so confused at this point...PLEASE help!