Suppose we drop a 0.145-kg baseball and a 7.0-kg et bowling ball simultaneously

Question

Suppose we drop a 0.145-kg baseball and a 7.0-kg et bowling ball simultaneously from rest. They both fall for 0.50 s and both reach the same speed of 4.9 m/s just before hitting the ground. (a) What impulse (magnitude and direction) did each ball receive from the earth during the interval between being released and hitting the ground? (b) What is the rate at which each ball received momentum as they dropped (assuming the rate is constant)? (c) is this rate consistent with equation C2.6 in each case?

Explanation / Answer

mass of baseball m = 0.145 kg
mass of bwling ball M = 7 kg
final velocity = 4.9 m/s
time taken = 0.5 s = t

a. impulse = change in momentum = mv
   so for baseball, I = 0.145*4.9 = 0.7105 kg m/s
   for bowling ball, I = 7*4.9 = 34.3 kg m/s

b. tiem rate of momentum transfer = momentum change/time
   for baseball, 0.7105/0.5 = 1.421 N
   for bowling ball, 34.5/0.5 = 69 N

c. rate = mg
   baseball, mg = 1.421 N
   bowling ball, Mg = 68.67 N
   hence the rates arfe consistent

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