A 0.30-kg softball has a velocity of 12 m/s at an angle of 30° below the horizontal just before making contact with the bat. What is the magnitude of the change in momentum of the ball while it is in contact with the bat if the ball leaves the bat with a velocity of

17 m/s, vertically downward, Δp = __ kg·m/s, and 17 m/s, horizontally back toward the pitcher? Δp = __ kg·m/s

drwls, Friday, February 23, 2007 at 5:40am

All you have to do is calculate the change in momentum in the horizontal and vertical directions, separately. Initially, you have
Vx = -12 cos 30 = -10.39 m/s
Vy = -12 sin 30 = 6.0 m/s
if Vx is defined as positive in the direction of the pitcher and Vy is postive upward.
After the hit, Vx = +17 m/s and Vy = -17 m/s. Just do the subtractions to get delta V, and multiply by the mass of the softball to get delta p.

winterWx, Saturday, February 24, 2007 at 12:27am

Hmmm...This is what I did:

Vx1-Vx2=-10.39-17=-27.39 m/s
(-27.39)(.3 kg)=-8.217 kg*m/s

Vy1-Vy2=-6-(-17)=11 m/s
(11)(.3)=3.3 kg*m/s

But both of those are incorrect. What am I doing wrong???

PLEASE HELP!!!

For Further Reading

* PHYSICS!!! - Marco, Saturday, February 24, 2007 at 7:38pm

You are wrong in your math.Look at part second.

o ..which part are you talking about? - winterWX, Saturday, February 24, 2007 at 9:26pm

how am i wrong in my math? can you please show me which parts i did wrong?

* PHYSICS!!! - winterWX, Saturday, February 24, 2007 at 10:29pm

I still can't get it. Oh well, it's due in an hour.

There are two questions here - one fopr eqach final velocity of the ball. When hit vertically upwards,

Vx1-Vx2 = -10.39 - = -10.39 m/s
px1 - px2 = (-10.39)(.3 kg)= -3.117 kg*m/s

Vy1-Vy2= -6-(17)=23 m/s
py1 = py2 = (23)(.3)= 6.9 kg*m/s
Note that the problem asks for the MAGNITUDE of the momentum change. This is
sqrt [(delta px)^2 + (delta py)^2]

Haha, looks like you're in a bit of a pickle there, WinterWx! Don't worry, I've got just the clownish answer to help you out.

Let's break it down step by step, and see where your calculations went wrong.

First, let's calculate the change in momentum when the ball leaves the bat with a velocity of 17 m/s vertically downward. So, the change in horizontal velocity (Vx1 - Vx2) would be:
Vx1 - Vx2 = (-10.39) - 0 = -10.39 m/s

And the change in vertical velocity (Vy1 - Vy2) would be:
Vy1 - Vy2 = -6 - (-17) = 11 m/s

Now, to calculate the change in momentum, we multiply these changes in velocity with the mass of the softball:
Δp = √((-10.39 * 0.3)^2 + (11 * 0.3)^2) kg·m/s

Now, it's time to put on our clown shoes and do some calculations. Let's see what we get.

Δp = √((3.117^2) + (3.3^2)) kg·m/s

Δp = √(9.688289 + 10.89) kg·m/s

Δp ≈ √20.578289 kg·m/s

Now, let's round it up to make it less precise and more clownish.

Δp ≈ 4.53 kg·m/s

There you have it, WinterWx! The magnitude of the change in momentum of the ball when it leaves the bat with a velocity of 17 m/s vertically downward is approximately 4.53 kg·m/s. Now go off and submit that assignment before it's too late! Good luck, and remember to keep smiling like a clown!

First, let's find the change in momentum when the ball leaves the bat with a velocity of 17 m/s, vertically downward.

1. Calculate the initial horizontal velocity (Vx1) of the ball:
Vx1 = -12 cos 30 = -10.39 m/s

2. Calculate the change in horizontal velocity (delta Vx):
delta Vx = Vx2 - Vx1 = -17 - (-10.39) = -6.61 m/s

3. Calculate the change in horizontal momentum (delta px1):
delta px1 = mass * delta Vx = 0.30 kg * (-6.61 m/s) = -1.98 kg·m/s

4. Calculate the initial vertical velocity (Vy1) of the ball:
Vy1 = -12 sin 30 = -6.0 m/s

5. Calculate the change in vertical velocity (delta Vy):
delta Vy = Vy2 - Vy1 = -17 - (-6.0) = -11.0 m/s

6. Calculate the change in vertical momentum (delta py1):
delta py1 = mass * delta Vy = 0.30 kg * (-11.0 m/s) = -3.30 kg·m/s

7. Calculate the magnitude of the change in momentum (delta p1):
delta p1 = sqrt((delta px1)^2 + (delta py1)^2)
delta p1 = sqrt((-1.98 kg·m/s)^2 + (-3.30 kg·m/s)^2)
delta p1 = sqrt(3.92 + 10.89) = sqrt(14.81) = 3.85 kg·m/s (rounded to 2 decimal places)

Now, let's find the change in momentum when the ball leaves the bat with a velocity of 17 m/s, horizontally back toward the pitcher.

1. Calculate the initial horizontal velocity (Vx1) of the ball:
Vx1 = -12 cos 30 = -10.39 m/s

2. Calculate the change in horizontal velocity (delta Vx):
delta Vx = Vx2 - Vx1 = 17 - (-10.39) = 27.39 m/s

3. Calculate the change in horizontal momentum (delta px2):
delta px2 = mass * delta Vx = 0.30 kg * 27.39 m/s = 8.22 kg·m/s

4. Calculate the initial vertical velocity (Vy1) of the ball:
Vy1 = -12 sin 30 = -6.0 m/s

5. Calculate the change in vertical velocity (delta Vy):
delta Vy = Vy2 - Vy1 = -17 - (-6.0) = -11.0 m/s

6. Calculate the change in vertical momentum (delta py2):
delta py2 = mass * delta Vy = 0.30 kg * (-11.0 m/s) = -3.30 kg·m/s

7. Calculate the magnitude of the change in momentum (delta p2):
delta p2 = sqrt((delta px2)^2 + (delta py2)^2)
delta p2 = sqrt((8.22 kg·m/s)^2 + (-3.30 kg·m/s)^2)
delta p2 = sqrt(67.47 + 10.89) = sqrt(78.36) = 8.84 kg·m/s (rounded to 2 decimal places)

Therefore, the magnitude of the change in momentum when the ball leaves the bat with a velocity of 17 m/s, vertically downward, is 3.85 kg·m/s, and when it leaves with a velocity of 17 m/s, horizontally back toward the pitcher, it is 8.84 kg·m/s.

To find the magnitude of the change in momentum, we need to calculate the change in momentum in both the horizontal and vertical directions separately.

In this case, the initial horizontal velocity (Vx) is -12 m/s at an angle of 30° below the horizontal. We can calculate the initial horizontal velocity using the cosine function: Vx = -12 * cos(30°) = -10.39 m/s.

The final horizontal velocity (Vx2) after hitting the ball is 17 m/s. To find the change in momentum in the horizontal direction (delta px), we subtract the final horizontal velocity from the initial horizontal velocity: delta px = Vx1 - Vx2 = -10.39 - 17 = -27.39 m/s.

Now, we multiply the change in momentum in the horizontal direction by the mass of the softball (0.30 kg) to get the change in momentum in kg*m/s: delta px = (-27.39) * 0.30 = -8.217 kg*m/s.

On the other hand, for the vertical direction:

The initial vertical velocity (Vy) is -12 m/s at an angle of 30° below the horizontal. We can calculate the initial vertical velocity using the sine function: Vy = -12 * sin(30°) = -6.0 m/s.

The final vertical velocity (Vy2) after hitting the ball is -17 m/s. To find the change in momentum in the vertical direction (delta py), we subtract the final vertical velocity from the initial vertical velocity: delta py = Vy1 - Vy2 = -6.0 - (-17) = 23 m/s.

Again, we multiply the change in momentum in the vertical direction by the mass of the softball (0.30 kg) to get the change in momentum in kg*m/s: delta py = 23 * 0.30 = 6.9 kg*m/s.

To find the magnitude of the change in momentum, we use the Pythagorean theorem:

Magnitude of delta p = sqrt[(delta px)^2 + (delta py)^2] = sqrt[(-8.217)^2 + (6.9)^2]

Evaluating this expression gives the magnitude of the change in momentum.

Draw a picture!

Before hitting the bat, the horizontal component of momentum of the ball is
px= (.3)(.15)(cos(35)) and the vertical component is
py= -(.3)(.15)(sin(35)).

Afterwards, the horizontal component of momentum is 0 (it is going straight down) and the vertical component is -20(0.3)

The change in momentum vector is the difference between those. Then use the Pythagorean theorem to find the magnitude.

In the second part (back toward the pitcher), the momentum after the hit has horizontal component -(.3)(20) and vertical component 0.