Impulse / 2-D Collisions / Need Clarification?

• julianwitkowski
In summary, the baseball experienced a change in velocity of -80.9 m/s after it was struck by the bat.
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?

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

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?

basheer uddin said:
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.

Very Much Appreciated

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 ?

julianwitkowski said:
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?

haruspex said:
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, vE=-48.9 m/s and ΔvN=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 ΔvE. Now I feel comfortable with this problem.

Thanks for everyones input!

1. What is impulse in physics?

Impulse is a physical quantity that measures the change in momentum of an object. It is equal to the force applied to an object multiplied by the time interval over which the force acts. Mathematically, it can be represented as I = FΔt.

2. How is impulse related to collisions?

In collisions, impulse is the force applied to an object over a specific time interval, causing a change in its momentum. The impulse-momentum theorem states that the impulse of an object is equal to the change in its momentum. Therefore, in a collision, the impulse experienced by each object involved is equal and opposite, resulting in a conservation of momentum.

3. What is a 2-D collision?

A 2-D collision is a type of collision that occurs in a two-dimensional space, where the objects involved have motion in two perpendicular directions. In 2-D collisions, the conservation of momentum still holds, but the calculations become more complex as they involve vectors and trigonometry.

4. Can you explain the difference between elastic and inelastic collisions?

In an elastic collision, both momentum and kinetic energy are conserved, meaning that the total energy of the system before and after the collision remains the same. In an inelastic collision, only momentum is conserved, and some kinetic energy is lost in the form of heat, sound, or deformation of the objects involved.

5. Why is it important to study impulse and collisions?

Impulse and collisions are crucial concepts in physics as they help us understand and predict the behavior of objects in motion. They are used in a variety of fields, including engineering, sports, and transportation, to design and analyze systems and improve their efficiency and safety. Additionally, these concepts play a significant role in developing technologies such as airbags, seatbelts, and protective gear.

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