Calculating Impulse Force for a Pitched Ball: How to Find the Final Velocity?

In summary, a baseball with a mass of 0.1417 kg was struck by a bat, resulting in a maximum vertical height of 44.196m. Using this information, the initial and final velocities of the ball can be calculated using equations of motion. The change in momentum can then be found by subtracting the final momentum from the initial momentum. This change in momentum can be further divided by the time of contact between the ball and bat to determine the magnitude and direction of the impulse force. In this case, the impulse force was found to be 2286.57N at an angle of 52.6 degrees west of north.
  • #1
lab-rat
44
0

Homework Statement


A fast ball is pitched at a batter who hits a pop fly vertically into the air. The ball was moving horizontally at 38.45 m/s when it was hit by the bat. As it left the bat it was moving vertically and reached a maximum height of 44.196m. The mass of a baseball is 0.1417 kg. If the ball was in contact with the bat for 3/1000 of a second, determine the magnitude and direction of the impulse force with which it was struck.


I need to find the change in momentum. So I need to find the initial and final velocity. The only thing left that I can't do is find the final velocity. How do I calculate it when all that I know is the vertical height?
 
Physics news on Phys.org
  • #2
lab-rat said:

Homework Statement


A fast ball is pitched at a batter who hits a pop fly vertically into the air. The ball was moving horizontally at 38.45 m/s when it was hit by the bat. As it left the bat it was moving vertically and reached a maximum height of 44.196m. The mass of a baseball is 0.1417 kg. If the ball was in contact with the bat for 3/1000 of a second, determine the magnitude and direction of the impulse force with which it was struck.


I need to find the change in momentum. So I need to find the initial and final velocity. The only thing left that I can't do is find the final velocity. How do I calculate it when all that I know is the vertical height?

For vertical motion, there is a constant downwards acceleration of g [we are told to take g= 9.8, 9.81 or 10 depending what level Physics you are taking]

From that , and the height reached, you can calculate the initial velocity.
 
  • #3
but I don't have the time and the formula is v=vo + at
 
  • #4
or s=vot + a(1/2 a) t2
 
  • #5
lab-rat said:
or s=vot + a(1/2 a) t2

ever heard of V2 = Vo2 + 2as
 
  • #6
OK, but the acceleration would be negative, so how can I find the square root?
 
  • #7
lab-rat said:
OK, but the acceleration would be negative, so how can I find the square root?

Yes, it is negative, but you have to move it to the other side of the equation [whence it will become positive] since you are trying to find Vo.

V = 0 because it has finally stopped going up when it reaches maximum height.
 
  • #8
Ohh, I thought I was looking for V. Vo = 29.43 j m/s

So I found Pi= 5.448i kgm/s
and Pf= 4.17j kgm/s

I need to find the change in momentum so Pf-Pi, how do I go about subtracting those vectors?
 
  • #9
lab-rat said:
Ohh, I thought I was looking for V. Vo = 29.43 j m/s

So I found Pi= 5.448i kgm/s
and Pf= 4.17j kgm/s

I need to find the change in momentum so Pf-Pi, how do I go about subtracting those vectors?

A pythagorus solution is what you are after.

The change in momentum has to get rid of a lot of horizontal momentum, and produce a lot of vertical momentum at the same time, so it will have an up component, and a "back towards the pitcher" component
 
  • #10
So Vo is 29.43 m/s, and the back towards pitcher component is -29.43 m/s correct?

And I would just need to find the horizontal velocity ?
 
  • #11
lab-rat said:
So Vo is 29.43 m/s, and the back towards pitcher component is -29.43 m/s correct?

And I would just need to find the horizontal velocity ?

No. The ball arrived at 38+ m/s, then went up at 29+ m/s, so the change is 38+ back, and 29+ up.
 
  • #12
OK, so what would the variation of momentum be ?

P = mv

I have (4.17 j - 5.45 i) kgm/s But my problem is that I can't subtract j's and i's.

I need either two horizontal vectors or two vertical vectors

Then I just need to divide the change in momentum by 3/1000
 
  • #13
OK so I used the pythagorea theorem but that gave me the resulting momentum when I only need the difference in momentum.

It's probably really simple, I'm just not seeing it right now...
 
  • #14
I ended up just doing (4.17j-5.448i) / 3/1000
Fnet=(1390j-1816i) N

does that work?
 
  • #15
lab-rat said:
I ended up just doing (4.17j-5.448i) / 3/1000
Fnet=(1390j-1816i) N

does that work?

NO.

This question is basically a test of whether you can add and subtract vectors. Looks like you can't.
It is done by drawing arrows which forms triangles which you use trigonometry and/or pythagorus to solve. Google search if you can't find it in your own text.
 
  • #16
lab-rat said:
OK, so what would the variation of momentum be ?

P = mv

I have (4.17 j - 5.45 i) kgm/s But my problem is that I can't subtract j's and i's.

I need either two horizontal vectors or two vertical vectors

Then I just need to divide the change in momentum by 3/1000

The bit I highlighted is a real problem. You have to learn how to do that.
 
  • #17
I actually do know how to subtract vectors, but my professor had said that we did not have time to cover that section, so we wouldn't see vector addition/subtraction.
So I assumed we wouldn't have any assignments on it, and thought there was another way to do this problem.

Anyways, I ended up doing it anyway and found 2286.57N (N52.6degresW)
 

1. What is a vector and how is it different from a scalar?

A vector is a mathematical quantity that has both magnitude and direction. This is different from a scalar, which only has magnitude and no direction. For example, velocity is a vector because it includes both the speed (magnitude) and the direction (e.g. north or south).

2. How is the magnitude of a vector determined?

The magnitude of a vector is determined by calculating the length of the vector using the Pythagorean theorem. This involves finding the square root of the sum of the squared components of the vector (e.g. x and y). The magnitude is always a positive value.

3. What is an impulse force?

An impulse force is a type of force that is applied over a very short period of time. It is often described as a sudden, strong force that causes a change in an object's momentum. This can be seen in sports, such as when a baseball player hits a ball with their bat.

4. How is the direction of a vector represented?

The direction of a vector is typically represented using an arrow pointing in the direction of the vector. The length of the arrow represents the magnitude of the vector, and the direction of the arrow represents the direction of the vector. Alternatively, the direction can also be represented using angles or coordinates.

5. How is the impulse-momentum theorem related to vectors and impulse force?

The impulse-momentum theorem states that the change in an object's momentum is equal to the impulse force applied to it. This means that the greater the impulse force, the greater the change in momentum. Since impulse force is a vector, this theorem also involves considering the direction of the force and its impact on the object's momentum.

Similar threads

  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
1K
Replies
1
Views
568
  • Introductory Physics Homework Help
Replies
24
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
3K
  • Introductory Physics Homework Help
Replies
17
Views
3K
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
996
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
2
Replies
57
Views
672
Back
Top