# Force, Impulse, and trajectory help request

1. Mar 25, 2014

### kahunah

1. The problem statement, all variables and given/known data

A baseball with a mass of 0.125 kg is moving horizontally at 32.0 m/s [E], when it is stuck by a bat for 0.00200 seconds. The velocity of the ball just after the collision is 52.0 m/s [W 20 N].

2. Relevant equations

p = m Δv
Fnet = ∆p/∆t

there are also these equations

Fnetx ∆t = m(v2x – v1x)
Fnety ∆t = m(v2y – v1y)

3. The attempt at a solution

i've gone over this a few times but nothing looks right to me...

Given
m for mass of baseball = 0.125 kg
v1 for the velocity of the baseball before the collision = 32.0 m/s [E]
v2 for the velocity of the baseball after the collision = 52.0 m/s [W 20 N]
Δt for duration of time the baseball struck the bat = 0.00200 s

a. Find the impulse experienced by the ball. (6 marks)

Let [E] be positive

To solve for the horizontal x-component of v2

v2x = 52.0 m/s * cos (20)

v2x = 48.86 m/s

To solve for the vertical y-component of v2

v2y = 52.0 m/s * sin (20)

v2y = 17.79 m/s Since v2x is in the [W] direction, let v2x be negative

To calculate the impulse, if

p = m Δv

Then,

Δv = √((v2x – v1x)2 – (v2y – v1y)2)

Δv = √((-48.86 m/s – 32.0 m/s)2 – (17.79 m/s – 0 m/s)2)

Δv = √((-80.86 m/s)2 – (17.79 m/s)2)

Δv = √((6538.34 m/s) – (316.48 m/s))

Δv = √(6221.87 m/s)

Δv = 78.88 m/s

So,

p = m Δv

p = (0.125 kg)(78.88 m/s)

p = 8.86 kgm/s

b. Find the average net force of the ball. (2 marks)

Fnet = ∆p/∆t

Fnet = (8.86 kgm/s)/(0.00200 s)

Fnet = 4430 N <--- that's WAY too much force

I also know there's an angle in there somewhere that i SHOULD be using force to calculate for the direction of the impluse

tan θ = Fnety/Fnetx?

2. Mar 25, 2014

### dauto

Why is it way too much force?

3. Mar 25, 2014

### paisiello2

I assume that the velocity given in the problem is always horizontal.

And 4430 N is not a lot of force to be acting over a small amount of time. So I think you did it correctly.

4. Mar 25, 2014

### kahunah

Well theres a vector for velocity of the ball after its hit. 52 m/s w20n so that would tell me that the impulse needs a direction to it. Perhaps Fnet as well.

5. Mar 25, 2014

### paisiello2

Good point. You know what one of the components of the vector must be for the impulse.

And the net force and impulse must have the same direction

6. Mar 25, 2014

### kahunah

i know i'm just lost :S

would it make sense to ensure that

Fnetx ∆t = m(v2x – v1x)
Fnety ∆t = m(v2y – v1y)

yeild FnetTotal and that

Fnettotal = ∆p/∆t?

7. Mar 25, 2014

### paisiello2

Yes, that makes sense. However, you already did that math as part of your answer to a).

8. Mar 25, 2014

### kahunah

so what would i use to calculate the direction of the impulse and net force?

9. Mar 26, 2014

### Rellek

Hey there!

We know that Newton's third law is a vector equation, so we can see that $$\vec{F}=m\vec{a}$$

Also, by the definition of acceleration, $$a = \frac{d\vec{v}}{dt}$$

So, now substituting the definition of acceleration and integrating, you can get this equation: $$\int_{v_{1}}^{v_{2}} m\,d\vec{v} = \int \vec{F}\,dt$$

Which can be simplified further after integrating to: $$m\vec{v_{2}} = \int \vec{F}\,dt + m\vec{v_{1}}$$

Where $\int \vec{F}\,dt$ is defined as impulse.

So impulse is a vector quantity, and if you know the time that your objects were in contact (which you do), you can find the original force vector. In your case, it will actually be slightly larger than the force you found, which you thought was too big in the first place.

In my opinion though, this problem would be most easily solved by keeping your velocities in the standard cartesian vector form, and then finding the magnitude and the angle of your force/impulse after the vector components have been found.

10. Mar 26, 2014

### Rellek

The only place you went wrong is here.

$$|\Delta \vec{v}| = \sqrt{(\Delta v_{x})^2 + (\Delta v_{y})^2}$$

where
$$\Delta v_{x} = v_{x2} - v_{x1}$$
$$\Delta v_{y} = v_{y2} - v_{y1}$$

And note $$\tan^{-1} \frac{\Delta v_{y}}{\Delta v_{x}}$$ is going to be your angle of impulse AND force.

Last edited: Mar 26, 2014
11. Mar 26, 2014

### paisiello2

I don't see where he went wrong other than maybe the notation was slightly confusing.

You already figured out the x- and y-components of the impulse to use for the direction in step a).

12. Mar 26, 2014

### Rellek

13. Mar 26, 2014

### kahunah

so i went wrong in that i subtracted

$$|\Delta \vec{v}| = \sqrt{(\Delta v_{x})^2 - (\Delta v_{y})^2}$$

$$|\Delta \vec{v}| = \sqrt{(\Delta v_{x})^2 + (\Delta v_{y})^2}$$???