A batter strikes a thrown ball (m=0.15kg) that was moving horizontally at 40 m/s. The ball leaves the bat moving at 40m/s in the opposite direction. The bat was in contact with the ball for 15ms. Find:
(A) the impulse exerted by the batter;
(B) the magnitude of the average force exerted by the bat on the ball.
(C) the baseball's change in momentum;
(C) Momentum, like velocity, is a vector. If the direction of the initial velocity of the ball is considered positive:
Therefore, the change in momentum is
Note: since direction matters, the answer is not zero.
(b) J=Δp=-12kg*m/s or -12N*s
(c) Since, by definition,
The magnitude of the force is therefore 800N (the answer was positive because the bat is hitting the ball in the negative direction)
(this multiple choice question has been scrambled)