Bat Hits Baseball - Force & Time Calculation

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A pitcher throws a baseball with a mass of 1.5 kg at a speed of 25 m/s, which then hits a bat and returns at the same speed. The force exerted by the bat on the ball is 50 N. The calculation for the time of contact between the bat and the ball is based on the change in momentum and the force applied. Despite the initial confusion over the baseball's mass, the calculations confirm that the time of contact is indeed 1.5 seconds. The discussion highlights the importance of accurate data in physics problems while affirming the correctness of the original calculation.
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<< Moderator Note -- thread moved to the HH forums >> [/color]

a pitcher throws a 1.5kg baseball with speed 25m/s. it hits the bat and return in opposite direction with same speed. the force that bat exerted on ball is 50N. For how long was the bat touching(in contact) the ball? I used: delat t * F = m*delta velocity delta t 50N = 1.5(25-(-25)) and i got t=1.5 s. Is it correct? It does not seem like.
 
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gnome222 said:
a pitcher throws a 1.5kg baseball with speed 25m/s. it hits the bat and return in opposite direction with same speed. the force that bat exerted on ball is 50N. For how long was the bat touching(in contact) the ball? I used: delat t * F = m*delta velocity delta t 50N = 1.5(25-(-25)) and i got t=1.5 s. Is it correct? It does not seem like.
That's a pretty happy baseball. Are you sure of that 1.5 kg?

Chet
 
Chestermiller said:
That's a pretty happy baseball. Are you sure of that 1.5 kg?

Chet
yes
 
gnome222 said:
yes
A major league baseball has a mass of 5.25 oz (0.15 kg). That 1.5 kg is an obvious typo.

And, please, don't try to play baseball with a 1.5 kg baseball. If you got hit by a fastball, it would be "lights out."

Chet
 
o
Chestermiller said:
A major league baseball has a mass of 5.25 oz (0.15 kg). That 1.5 kg is an obvious typo.

And, please, don't try to play baseball with a 1.5 kg baseball. If you got hit by a fastball, it would be "lights out."

Chet
ok, but under the conditions described what would be time?
 
gnome222 said:
o

ok, but under the conditions described what would be time?
If you are asking whether you solved the problem correctly, given the obviously faulty data, then the answer is yes.

Chet
 
force is the rate of change of momentum.
Thus, you have been given the change in momentum, the force excerted.
So,
t=Change in momentum/force
=2*1.5*25/50=75/50=1.5

Yup, your answer is correct
 
This is the sort of question you have to answer on impulse. Boom boom.
 
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