Question
A 0.145 kg baseball is pitched at 42 m/s. The batter hits it horizontally to the pitcher at 58 m/s.
GIVEN:
m = 0.145 kg
v = 42 m/s
batter v = 58 m/s
a. Find the change in momentum of the ball.
p = mv
p = (0.145)(42)
p = 6.09
--> What velocity would I use? Would i use the velocity of the baseball
b. If the ball and bat are in contact for 4.6 x 10^-4 s, what is the average force during contact?
GIVEN:
m = 0.145 kg
v = 42 m/s
batter v = 58 m/s
a. Find the change in momentum of the ball.
p = mv
p = (0.145)(42)
p = 6.09
--> What velocity would I use? Would i use the velocity of the baseball
b. If the ball and bat are in contact for 4.6 x 10^-4 s, what is the average force during contact?
Answers
a) The sign of the velocity changes after the ball is hit. That makes the CHANGE in momentum equal to
M[42 -(-58)] = 0.145 kg * 100 m/s
b) Impulse
= (average force) x (time of contact)
= (change in momentum) = 14.5 kg m/s
Solve for the average force
M[42 -(-58)] = 0.145 kg * 100 m/s
b) Impulse
= (average force) x (time of contact)
= (change in momentum) = 14.5 kg m/s
Solve for the average force
Related Questions
I got all but this problem, I
A 0.145-kg baseball is pitched at 42.0 m/s. The batter hits it hori...
I got all but this problem.
A 0.145-kg baseball is pitched at 42.0 m/s. The batter hits it horiz...
a .415 kg baseball is pitched at 42 m/s. the batter hits it horizontally to the pitcher at 58 m/s....
A 0.149-kg baseball is pitched a 40 m/s. The batter hits it horizontally to the pitcher at 56 m/s....