What is the Impulse of a Baseball Bat on a 0.15 kg Ball Traveling at 6.0 m/s?

[redshift]

A 0.15 kg ball traveling horizontally at 6.0 m/s is hit by a bat, after which it is traveling at 8.0 m/s perpendicular to the original direction.
I'm supposed to determine the impulse that the ball receives from the bat.

Since cos90 is zero, would this just be the mass multiplied by the initial velocity (i.e., 0.9 Ns)?

 

[arildno]

Use the relation:
$$\vec{I}=m\vec{v}_{f}-m\vec{v}_{i}$$
Remember that the impulse, initial velocity, and final velocity are vector quantities.

 

[redshift]

Plugging in the numbers...
I = 0.15kg(8.0m/s)(cos90) - 0.15kg(-6.0m/s)
I = 0.9 Ns

I guess this shows my initial answer is correct. But this doesn't seem right as I would get the same result for any velocity at which the ball leaves the bat. But since the velocity is a vector quantity, it seems to me that the final velocity must be multiplied by cos90 to reflect its x-component.

 

[Doc Al]

Plugging in the numbers...

This is not correct. As arildno explained, you must treat momentum as a vector.

Start by writing the x and y components of the momentum of the ball before and after the collision with the bat.

 

[redshift]

Okay, I think I get it now. The ball, in order to head in a perpendicular direction to its original direction, must have been struck at an angle LESS THAN 90 degrees. I'm guessing it's 45 degrees. Therefore, I = 0.15kg(8.0m/s)(cos45) - 0.15kg(-6.0m/s) = 1.74 Ns

 

[Doc Al]

Okay, I think I get it now. The ball, in order to head in a perpendicular direction to its original direction, must have been struck at an angle LESS THAN 90 degrees. I'm guessing it's 45 degrees. Therefore, I = 0.15kg(8.0m/s)(cos45) - 0.15kg(-6.0m/s) = 1.74 Ns

No. No need to guess, since you are given all the information needed to calculate the change in momentum.

I'll call the original direction the x direction; perpendicular to it, the y direction.
Thus the initial momentum of the ball is:
$$m\vec{v}_{i} = (.15)(6.0)\hat{x} + 0\hat{y}$$
And the final momentum is:
$$m\vec{v}_{f} = 0\hat{x} + (.15)(8.0)\hat{y}$$

Now find the impulse (change in momentum) by subtracting these two vectors.

 

[redshift]

I'll have to meditate on this for a while.
I understand the division into x and y components of the respective momentums but am confused as to the substraction operation.

 

[Doc Al]

I'll have to meditate on this for a while.
I understand the division into x and y components of the respective momentums but am confused as to the substraction operation.

Meditation is good for the soul.

You subtract the vectors by subtracting each component separately. The final answer will be a vector with both an x and y component. You'll need to find the magnitude of that vector. (And perhaps the direction, as well.)

 

[arildno]

redshift:
I believe you are confusing the concepts of "work" and "impulse" in this problem.

"Work" is a scalar concept (i.e represented by a single number), that is related to the change of kinetic energy.

However, "impulse" is a vector concept, and is related to the change of momentum.

Do not confuse the momentum of a particle with its kinetic energy!

 

[redshift]

Just a heads up that the penny eventually dropped last night (with a little help from a glass of oloroso sherry. Sherry and physics seem to go well together). I know, I know, you guys practically solved it for me.

 

[arildno]

I prefer a cup of black coffee..