2. A 0.50 kg baseball approaches a bat at 25 m/s. If it is in contact
with the bat for 0.005s, then rebounds at 25 m/s, what is the average
contact force between the ball and the bat?
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Answered by
0
Answer:
impulse = change in momentum
F × t = change in momentum
F × t = m ∆V
F × t = m { 25 - ( - 25 ) }
F × 0.005 = 50 m
F × 5 / 1000 = 50 × 0.5
F = 5000 N
YOOOODA:
my answer key says its -500 N ?
don't worry about the negative sign , it just tells u the direction,.. the direction of force is same as the direction in which ball rebounds , I have taken that direction to be positive so I have got a negative answer.. but instead of 500 I am getting 5000 , use the formula and check the calculation once..
how to get the negative answer is it just the same formula?
ha. just take { -25 - 25} .. in place of { 25 - (-25)} in the solution I have provided..
u get an answer with a negative sign..
hi sir do you know how to solve my second question?
which one?
oh no i answered it myself thanks for all the help
ok.. good luck
Answered by
0
Answer:
5000 N
Explanation:
mΔv = Ft
m(final velocity - initial velocity) = Ft
0.50 [25-(-25)] = F(0.005)
25 = F(0.005)
5000 = F
the initial velocity is negative because it is going in the opposite direction was the final, it can be though of as going south or down
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