Average force on a baseball when struck by a bat

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SUMMARY

The discussion focuses on calculating the average force exerted on a baseball (0.145 kg) when struck by a bat, with the ball initially pitched at 32 m/s and reaching a height of 36.5 m after contact. The average force is derived using the equation Favg = Δp/t, where Δp is the change in momentum and t is the contact time of 2.5 ms. The correct approach involves separating the velocity components into horizontal and vertical vectors and applying the Pythagorean theorem to find the resultant velocity after impact. The confusion regarding the use of kinetic energy equations is clarified, emphasizing the importance of vector analysis in force calculations.

PREREQUISITES
  • Understanding of momentum and the equation Favg = Δp/t
  • Knowledge of vector components and Pythagorean theorem
  • Familiarity with basic physics concepts such as kinetic energy (KE) and gravitational potential energy (PE)
  • Ability to perform calculations involving time, mass, and velocity
NEXT STEPS
  • Study the application of vector decomposition in physics problems
  • Learn about momentum conservation in collisions
  • Explore the relationship between kinetic energy and potential energy in projectile motion
  • Practice calculating forces in various physics scenarios using Favg = Δp/t
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators looking for examples of force calculations in real-world applications.

brycenrg
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Homework Statement


Baseball(0.145kg) pitch is pitched horizontally at 32m/s, ball pops straight up to a height of 36.5m. If the contact time between bat and ball is 2.5ms, calculate the average force between the ball and the batt during contact.

Homework Equations


Favg = Δp/t
KE=KE

The Attempt at a Solution


1/2mv^2 = mgh
v= squar(2gh)
Find Vo when ball leaves bat
((0.145kg*27m/s)-(0.145kg*32m/s))/Tcontact = -290 N

My answer is wrong, in my solution manual it breaks the variables up into x and y. Then it uses Pythagoras equation for the hypotenus. Is that okay to do for forces? i thought that was only used for velocity vectors or.. Ima little confused.
 
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Force is a vector, momentum is a vector, so ##\displaystyle \vec F = {\Delta \vec p \over \Delta t}## is the equation to work out. It's also the equation you mention under 2. (KE=KE is not all that interesting; why do you list it there ?).

Under 3) attempt at solution you don't continue with the relevant equation, but drag in new stuff that may or may not be applicable. (in fact it is: you need it to find the vertical speed after the hit). But the 27m/s points straight up, whereas the 32 m/s is pointing in the horizontal direction. To go from (32, 0) to (0, 27) you need to change by (-32,+27). The magnitude is found with Pythagoras.
 
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Oh okay thank you BvU. Whoops i Meant to write P=P or ke+u=ke+u to find v in the direction up. thank you.
 

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