Change in momentum of baseball being struck

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SUMMARY

The discussion focuses on calculating the change in momentum of a 0.33 kg softball struck by a bat. The initial velocity of the ball is 14 m/s at a 50° angle below the horizontal. Two scenarios are analyzed: (a) the ball exits the bat with a velocity of 15 m/s vertically downward, and (b) it exits horizontally back toward the pitcher. The momentum is calculated using the formula P = M * V, and the magnitude of the change in momentum is determined using vector subtraction and the Pythagorean Theorem.

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  • Basic physics concepts related to collisions and motion
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Zach Lunch
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Homework Statement


A 0.33 kg softball has a velocity of 14 m/s at an angle of 50° below the horizontal just before making contact with the bat. What is the magnitude of the change in momentum of the ball while it is in contact with the bat if the ball leaves the bat with a velocity of (a) 15 m/s, vertically downward, and (b) 15 m/s, horizontally back toward the pitcher?

Homework Equations


→P=M*→V

The Attempt at a Solution


Find the Vector for momentum before hitting the bat .33<14*cos 50, 14*sin 50>
Subtract vector for each case. a) subtract .33<0,-15> b.) subtract .33<-15,0>
Then use Pythagorean Theorem to get magnitude sqrt(x^2 + y^2) for each case.
 
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Zach Lunch said:

Homework Statement


A 0.33 kg softball has a velocity of 14 m/s at an angle of 50° below the horizontal ...
Find the Vector for momentum before hitting the bat .33<14*cos 50, 14*sin 50>

The angle is 50° below the horizontal. The vertical component of the original momentum has to be negative.

ehild
 

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