Impulse / 2-D Collisions / Need Clarification?

[julianwitkowski]

Homework Statement

A baseball with a mass of 0.152 kg is moving horizontally at 32.0 m/s [E], when it is struck by a bat for 0.00200 seconds. The velocity of the ball just after the collision is 52.0 m/s [W 20⁰ n].

a) Find the impulse experienced by the ball?

b) Find the average net force of the ball?

Homework Equations

J = F ∙ ∆t
F ∙ ∆t = m ∙ ∆v

The Attempt at a Solution

I've done my work like it says in the textbook and it's easy to do when I see examples of similar problems but I still haven't completely wrapped my head around it and I'm not sure why it's right...

Can you please check it over and describe what's going on?

Thank you,
Julian

 

[basheer uddin]

can you please state a bit more clearly about the point which is troubling you?
why do you think it is not right ?
what is the doubt?

 

[julianwitkowski]

can you please state a bit more clearly about the point which is troubling you?
why do you think it is not right ?
what is the doubt?

I know its right because I read it in the textbook, but I wouldn't know what to do if it was a test, and I really want to ace my exam.

I guess I just don't really understand about the x and y component of velocity and why the hypotenuse of that right triangle is Δv.

What is it the change in velocity because it's not equaling the Δd/Δt of the ball the way I would assume it would? ... Here, this is my problem...

My assumption would be Δv = Δd/Δt 32/1 - 0/0.002- -52/1 = /1 which I know is just wrong, but that's the silly thing that comes to mind when I think about this with no external reference. I know this sounds stupid its just, I guess I'm a noob.

Thanks for your help though!
Very Much Appreciated

 

[julianwitkowski]

Also... A triangle with the hypotenuse being the baseballs velocity of 52 m/s leaves an x-axis value of 48.9 m/s W and 17.8 m/s E for the y axis. How does the 48.9 relate to Δv ?

 

[haruspex]

Also... A triangle with the hypotenuse being the baseballs velocity of 52 m/s leaves an x-axis value of 48.9 m/s W and 17.8 m/s E for the y axis. How does the 48.9 relate to Δv ?

Sounds like you're not drawing the correct triangle.
Set an origin at O. Draw a line E of that to a point A to represent the original velocity vector. Now draw line W 20⁰ N from O to a point B to represent the final velocity vector. The change in velocity is the vector you have to add to OA to get OB; i.e. it is represented by the line from A to B. You want to calculate the length of that line.
How far W of A is B? How far N of A is B?

 

[julianwitkowski]

Sounds like you're not drawing the correct triangle.
The change in velocity is the vector you have to add to OA to get OB; i.e. it is represented by the line from A to B. You want to calculate the length of that line.
How far W of A is B? How far N of A is B?

I realized that the baseball velocity triangle v=52 m/s, v_E=-48.9 m/s and Δv_N=17.8 m/s is the v₁ of v₁ - v₂ in Δv --> -48.9 m/s - 32 m/s = -80.9 m/s which is Δv_E. Now I feel comfortable with this problem.

Thanks for everyones input!