Find the average force exerted by the bat on the ball if the two are in contact for 0.0028 s. Answer in units of N.
A pitcher throws a 0.15kg baseball so that it crosses home plate horizontally with a speed of 14m/s. It is hit straight back at the pitcher with a final speed of 32m/s.
Given: Choose the positive direction to be from the pitcher to the plate (some answers may be negative).
The mass of the ball the pitcher throws is 0.15 kg. The velocity of the baseball is 14m/s. The velocity of the ball after hit = -32m/s after hit(negative as,the direction of the velocity is opposite and towards the pitcher).
Therefore , the change of momentum = mass of the ball*(finalal velocity)-mass of the ball*(initial velocity of the ball after hit))= 0.15kg(-32m/s)-0.15kg(14m/s) = -6.9kg m/s. As the change of momentum took in 0.0028 seconds, the force on the ball is the rate of change in the momentum =[-6.9kg m/s]/(0.0028s) = -2464.2857 N, the negative indicates the direction is from plate towards the pitcher.
Initial velocity = u = 14 m/s
Final velocity = v = -32 m/s
Time taken for change in velocity = t = 0.0028 s
Mass of ball = m = 0.15 kg
Force exerted = acceleration*mass = a*m, and
Acceleration = a = (v - u)/t
Substituting given value of u,v and t we get
a = (-32 - 14)*0.0028 = -46*0.0028 = 0.1288 m/s^2
Substituting this value of acceleration in equation for force
Force exerted = 0.1288*0.15 = 0.01932 N
Answer: Average force exerted by the bat on ball is 0.01932 N