Answer to Question #106026 in Mechanics | Relativity for Valentine Agun

Question

A pitcher throws a 0.14-kg baseball toward the batter so that it crosses home plate horizontally and has a speed of 42 m/s just before it makes contact with the bat. The batter then hits the ball straight back at the pitcher with a speed of 48 m/s. Assume the ball travels along the same line leaving the bat as it followed before contacting the bat. (a) What is the magnitude of the impulse delivered by the bat to the baseball? (b) If the ball is in contact with the bat for 0.005 0 s, what is the magnitude of the average force exerted by the bat on the ball? (c) How does your answer to part (b) compare to the weight of the ball?

Answer 1

The moment of motion of the ball before hitting the bat is $P_1=mV_1$ . After ball leaving the bat it moves in opposite direction with the momentum $P_2=-mV_2$. It means that during the strike, the interaction of the ball with the bat, its momentum changed by an amount

(1) $\Delta P=P_2-P_1=-m(V_2+V_1)=-0.14 kg\cdot (48 ms^{-1}+42ms^{-1})=-12.6 Ns$

The minus sign in this expression means that the change in momentum occurred in the direction of the pitcher, since we attributed the sign plus to the original speed. Since the change in momentum occurred during the time interval $\Delta T=0.005 s$ of collision, the average force with which the bat exert on the ball is

(2) $F=\frac {\Delta P}{\Delta T}=\frac {-12.6 Ns}{0.005s}=-2.52\cdot 10^3N=-2.52kN$

The minus sign in this expression means that the average force exerted by the bat on the ball occurred in the direction from the batter to the pitcher.

Answer: a) The magnitude of the impulse delivered by the bat to the baseball is 12.6Ns.

b) The magnitude of the average force exerted by the bat on the ball is 2.52kN.

c) The answer to part (b) is in direct ratio compared to the weight of the ball. As the ball weight increase the average force exerted by the bat on the ball must increased too.