Questions about x and y components of its change in momentum

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SUMMARY

The discussion centers on the change in momentum of a baseball with a mass of 0.135 kg, initially traveling at 30.0 m/s along the x-axis and striking a fence at a 45-degree angle. The user incorrectly calculated the change in momentum components, specifically ∆Px and ∆Py. The correct approach reveals that the y component of velocity remains equal to the x component post-collision, leading to a reversal of the x component while the y component remains unchanged at 30.0 m/s.

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A 135 grams baseball, moving along the x-axis with speed 30.0 m/s, strikes a fence at a 45 degree angle and rebounds along the y-axis with unchanged speed.
Give the y and y components of its change in momentum --> ∆Px and ∆Py

are the following correct?
M= 0.135 kg
Vx = 30.0 m/s
θ=45
Vy=30.0 m/s

I calculated ∆Px = M*Vx*cos 45
0.135 * 30.0 * cos 45 = 2.9 kg m/s
∆Py = M*Vy*sin 0
= 0.135*30.0*sin 0 = 0

but I got it wrong..
can anyone help me?

thanks..
 
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The 30 m/s is already the x component of the velocity. If it strikes the fence at 45 degrees, then the y component is equal to the x component.

By bouncing backward at 45 degrees, then the x component is completely reversed; you can see this because it was going to the right, and then after the collision, to the left.

THe vertical momentum... does it change? before the collision the vertical component is down. How about after the collision?
 

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