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).
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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
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.
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