Magnitude and direction of impulse of baseball

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SUMMARY

The discussion focuses on calculating the magnitude and direction of the impulse imparted on a baseball after being hit by a player. The baseball has a mass of 0.145 kg and an initial velocity of -35 m/s. After being struck, the ball travels at a speed of 51 m/s at a 45-degree angle. The impulse is calculated using the formula Δp = mv_f - mv_i, with separate calculations for the x and y components of the impulse, leading to the conclusion that the impulse imparted to the bat is equal and opposite to that imparted to the ball, in accordance with Newton's Third Law.

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dorian_stokes
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Homework Statement


A baseball player hits a baseball (m = 0.145 kg) as shown in the figure below. The ball is initially traveling horizontally with speed of 35 m/s. The batter hits a fly ball as shown, with a speed vf = 51 m/s. What is the magnitude and direction of the impulse imparted on the ball. The initial velocity is going along the x-axis in the negative direction and the final velocity is going at a 45 degree angle in the first quadrant.


Homework Equations





The Attempt at a Solution

First I had the initial velocity at -35. The final velocity I had the final velocity* cos theta. I don't know where to go from here.
 
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dorian_stokes said:
First I had the initial velocity at -35. The final velocity I had the final velocity* cos theta. I don't know where to go from here.
You need to consider both x (horizontal) and y (vertical) components of the velocity. What's the y component of the final velocity?

Once you have that, how would you compute the impulse?
 
The y component is final velocity* sin theta. I have the x component for both final a initial velocities and the y component for the final velocity. The y component for the initial will be just the -35m/s. How do I add them together to get the magnitude?
 
dorian_stokes said:
The y component is final velocity* sin theta.
OK.
I have the x component for both final a initial velocities and the y component for the final velocity.
OK.
The y component for the initial will be just the -35m/s.
No. (It's thrown horizontally.)
How do I add them together to get the magnitude?
Once you get the final components of the impulse, then you can find the magnitude using Pythagorus (like you would with any other vector).

But before you do anything, how will you calculate the impulse? How does it relate to the velocities?
 
Impulse is dp=mv_f-mv_i. I have the final velocity* cos theta+ final velocity *sin theta+ initial velocity. Is this right?
 
dorian_stokes said:
Impulse is dp=mv_f-mv_i.
OK. Impulse = Δp. But you need to find the x and y components of the Impulse separately:
Impulsex = m(Vfx -Vix)
Impulsey = m(Vfy - Viy)
 
Thanks I got the answer. I knew how to get the magnitude but got stuck with how to set it up after finding the components.
 
I have one more problem it asks for the direction of the impulse imparted to the bat. I have the magnitude but don't know how would I find the direction.
 
dorian_stokes said:
I have one more problem it asks for the direction of the impulse imparted to the bat. I have the magnitude but don't know how would I find the direction.
The impulse that the ball imparts to the bat is equal and opposite to the impulse the bat imparts to the ball. (This is from Newton's 3rd law.)
 

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