# Impulse on a thrown ball

1. Dec 26, 2004

### UrbanXrisis

A 0.2kg ball is thrown at 15m/s, 45 degrees below the horizontal. The ball is hit by at bat at 40m/s, 30 degrees above the horizontal. What is the impulse delivered to the ball?

So, impulse is change in momentum. The initial momentum is .2*15=3Ns
How would I find the final momentum and how does the angles come into play?

2. Dec 26, 2004

### dextercioby

1.Make a drawing (diagram) in which place the horizontal and vertical directions along with the 2 velocity vectors.
2.Chose a system of coordinates with 2 axis orthogonal on each other.Project on these 2 axis the 2 velocity vectors.Compute the initial and the final comonents of the momentum vectors.Add/subtract carefully the vectors and find the final answer.

Daniel.

3. Dec 26, 2004

### UrbanXrisis

how is that possible when one angle is above the horizontal and the other is below? I'm not quite sure how I should place the two diagrams for comparison

and how would I find final velocity?

Last edited: Dec 26, 2004
4. Dec 26, 2004

### UrbanXrisis

5. Dec 26, 2004

### dextercioby

1.The drawing is pretty awkward,as you imagined it like an Arab,from right to left.It's not important.As I said,pick a system of axis and compute this vector:
$$\Delta\vec{p}=\vec{p_{f}}-\vec{p_{i}}$$
2.If your choise of coordonates is smart,then u should get the result pretty easily.

Daniel.

6. Dec 26, 2004

### UrbanXrisis

So I do have to split the 40m/s and the 15m/s into it's x and y componets. With that, I have to subtract x-axis momentum of 40m/s from the x-axis momentum of 15m/s. Then subtract the y-axis momentum of 40m/s from the y-axis momentum of 15m/s. The resultant should give me what I'm looking for right?

7. Dec 26, 2004

### dextercioby

Yes.To me it looks like those components add instead of subtract,as they correspond to a vector subtraction of vectors which have different dirrections (the support is the same,but point one at another).

Daniel.

8. Dec 26, 2004

### UrbanXrisis

40 m/s at 30 degrees
x-componet = 34.6 m/s to the right
y-componet= 20 m/s upwards

15 m/s at 45 degrees
x-componet = -10.6 m/s to the left
y-componet= -10.6 m/s downwards

so I add the vectors up?
x-componet = 24 m/s to the right
y-componet= 9.4 m/s upwards

resultant=25.8 m/s

9. Dec 26, 2004

### UrbanXrisis

P=mv=.2*-15m/s=-3Ns
P=mv=.2*25.8N/s=5.2Ns

Impulse is 8.2Ns

is this correct?

10. Dec 26, 2004

### dextercioby

Here's how i see it.Consider the initial state of the ball being pictured in the left of the screen and having a velocity vector $\vec{v}_{i}$ and moving towards the right part of the screen.In the final state it is hit by a bat and therefore moves to the left again with a velocity $\vec{v}_{f}$.Pick the Oxy system with the direction of 0x from left to right and with direction of 0y upwards.
U will have
$$\Delta \vec{p}=\vec{p_{f}}-\vec{p_{i}}$$
$$\vec{p_{f}}=m(-v_{f,x}\vec{i}+v_{f,y}\vec{j})$$
$$\vec{p_{i}}=m(v_{i,x}\vec{i}-v_{i,y}\vec{j})$$
Therefore:
$$\Delta \vec{p}=m[-(v_{f,x}+v_{i,x})\vec{i}+(v_{f,y}+v_{i,y})\vec{j}]$$

I'll let u plug in the numbers.

Daniel.

11. Dec 26, 2004

### UrbanXrisis

so you're saying: Impulese = mass * (resultant force in y + resultant force in x)

is that correct?

12. Dec 27, 2004

### dextercioby

What do you mean resulting force?? U probably meant "velocity". I gave you the final vector decomposed on the coordinate axis.Square it and find its modulus and then plug in the numbers.It's not a big deal... :tongue2:

Daniel.