Impulse on a thrown ball

In summary, the impulse delivered to the ball is 8.2 Ns based on the change in momentum from the initial velocity of 3 Ns to the final velocity of 5.2 Ns after being hit by the bat. The x and y components of the initial and final velocities are used to calculate the change in momentum. The impulse is found by squaring the final vector and finding its modulus.
  • #1
UrbanXrisis
1,196
1
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?
 
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  • #2
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
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?
 
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  • #4
http://home.earthlink.net/~urban-xrisis/pic001.jpg [Broken]

so basically find the x and y componets and subtract right?
 
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  • #5
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:
[tex] \Delta\vec{p}=\vec{p_{f}}-\vec{p_{i}} [/tex]
2.If your choise of coordonates is smart,then u should get the result pretty easily.

Daniel.
 
  • #6
dextercioby said:
pick a system of axis and compute this vector:
[tex] \Delta\vec{p}=\vec{p_{f}}-\vec{p_{i}} [/tex]

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
UrbanXrisis said:
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?

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

Impulse is 8.2Ns

is this correct?
 
  • #10
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 [itex] \vec{v}_{i} [/itex] 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 [itex] \vec{v}_{f} [/itex].Pick the Oxy system with the direction of 0x from left to right and with direction of 0y upwards.
U will have
[tex] \Delta \vec{p}=\vec{p_{f}}-\vec{p_{i}} [/tex]
[tex] \vec{p_{f}}=m(-v_{f,x}\vec{i}+v_{f,y}\vec{j}) [/tex]
[tex] \vec{p_{i}}=m(v_{i,x}\vec{i}-v_{i,y}\vec{j}) [/tex]
Therefore:
[tex] \Delta \vec{p}=m[-(v_{f,x}+v_{i,x})\vec{i}+(v_{f,y}+v_{i,y})\vec{j}] [/tex]

I'll let u plug in the numbers.

Daniel.
 
  • #11
so you're saying: Impulese = mass * (resultant force in y + resultant force in x)

is that correct?
 
  • #12
What do you mean resulting force?? :confused: U probably meant "velocity". :wink: 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.
 

1. What is impulse on a thrown ball?

Impulse on a thrown ball is the change in momentum that occurs when the ball is thrown. It is a force that is applied to the ball for a specific amount of time, resulting in a change in the ball's velocity.

2. How is impulse related to force?

Impulse is related to force through the equation J = FΔt, where J is impulse, F is force, and Δt is the time interval over which the force is applied. This means that the greater the force applied to the ball, or the longer the force is applied, the greater the impulse will be.

3. Can the impulse on a thrown ball be negative?

Yes, the impulse on a thrown ball can be negative. This occurs when the force applied to the ball is in the opposite direction of its motion, causing a decrease in the ball's momentum. For example, when a catcher catches a ball thrown by a pitcher, the catcher applies a negative impulse to the ball to stop it.

4. How does the mass of the ball affect the impulse?

The mass of the ball does not directly affect the impulse. However, a larger mass will result in a smaller change in velocity for the same amount of impulse, as shown by the equation J = mΔv, where m is the mass of the ball and Δv is the change in velocity.

5. Why is it important to consider impulse when throwing a ball?

It is important to consider impulse when throwing a ball because it determines the amount of force and time needed to achieve a desired change in the ball's velocity. This can affect the accuracy and distance of the throw. In sports such as baseball or football, understanding and controlling impulse can also help prevent injuries to the player's arm or shoulder.

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