Question

A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 49.0 m/s ; when it leaves the bat, the ball is traveling to the left at an angle of 31.0 ∘ above horizontal with a speed of 54.0 m/s . The ball and bat are in contact for 1.61 ms . Find the horizontal and vertical componenets of the average force on the ball

Answer 1

average force = ΔP/Δt = change in momentum/ time

i took left as -x and right ar +x and up as +y  

Fx = ΔPx/Δt

Δt = 1.61*10^-3

Pix = m*Vix = 0.145*49 = 7.105kg*m/s

Pfx = m*Vfx = 0.145*( -54*cos31) = -6.711  kg*m/s so

Fx = (Pfx-Pix)/ t = ( -6.711- 7.105) /(1.61*10^-3) =-8581.36646 N

Fx= 8581.36646 N

Fy = ΔPy/Δt

Pyi = m*Vyi =0.145*0 = 0  kg*m/s

Pyf = m*Vyf = 0.145*54*sin(31) = 4.03 kg*m/s

Fy = (Pfy-Piy)/ t = 4.03/ ((1.61*10^-3)) =2503.10559 N

Fy =2503.10559 N