Question

A 0.15kg baseball pitched at 30 m/s is hit on a horizontal line drive
straight back at the pitcher at 46.0 m/s. If the contact time between bat and ball is 5x10^-3 s, calculate the force (assumed to be constant) between the ball and bat. [Hint: impulse!]


I did FΔt = Δp = mΔv
for Δv I'm not sure if it is (30)-(46)=(-16) or (30)-(-46)=(76).

I think it is 76, so then I did:
(0.15)(76) = 11.4 kg *m/s
then 11.4/(5x10^-3) = 2280 N
a = (V-Vo)/t = (-46-30)/5*10^-3 =
-15.2*10^3 m/s^2.

F = m*a = 0.15 * (-15,200) = -2280 N.

Related Questions

A 0.145 kg baseball pitched at 38.0 m/s is hit on a horizontal line drive straight back toward the p... A .145kg baseball pitched at 39m/s is hit on a horizontal line drive straight back toward the pitche... A 0.130kg baseball pitched at 35.5 m/s is hit on a horizontal line drive straight back at the pitch...