# Vector diagram for momentum

• shan
In summary, the white ball, with a mass of 0.35kg and a velocity of 5.0ms-1, collides with a colored ball and is deflected through an angle of 33º with a velocity of 4.3ms-1. To calculate the change in momentum, a vector triangle can be used with the cosine rule to find the magnitude of the change in momentum. The direction of the force exerted by the white ball on the colored ball can be found using the angle between the final momentum and the change in momentum. The force is in the direction of the change in momentum, which can be determined by adding 33º to the angle opposite the final momentum in the vector triangle.

#### shan

The situation: In a game of pool, the white ball of mass 0.35kg traveling at 5.0ms-1 collides with a colored ball and is deflected through an angle of 33º with a velocity of 4.3ms-1.

The question is a carry on (from questions about momentum and impulse) and is: What is the direction of the force the white ball exerts on the colored ball?

The answer is 87.8+33º to the original direction. I found the 87.8º but I don't understand why I have to add 33º to it. Can someone explain?

Ok... the whole question is:

The game of pool requires the player to use the cue to hit the white ball so that it rolls towards and hits the coloured ball. The direction of the white ball must be such that it hits the coloured ball at an angle which makes the coloured ball roll into one of the pockets. The white ball will be deflected but must not roll into a pocket.

The white ball, of mass 0.35kg, is hit and rolls towards the coloured ball at a velocity of 5.0 ms-1. It collides and is deflected through an angle of 33&deg; with a velocity of 4.3ms-1. Frictional forces acting to slow the ball can be assumed to be negligible.

[There's a diagram of the white ball traveling left, hitting the coloured ball and being deflected 33&deg; upwards]

Q.1
Calculate the size of the change in momentum of the white ball.
-> p(initial) = 0.35 x 5 = 1.75
p(final) = 0.35 x 4.3 = 1.505
Using vector equation of p(final) - p(initial), the vector triangle looks like:
A triangle with it's apex on the bottom. The top side is p(initial) to the right. The left side is p(final) going up. The right side is change in p going up. p(final) makes a 33&deg; with a horizontal line at the bottom.
Using cosine rule: change in p = 1.75&sup2; + 1.505&sup2; - 2 x 1.75 x 1.505 x cos33 = 0.95385

Q.2 What is the direction of the force the white ball exerts on the coloured ball during the collision?
-> Using the previous vector triangle, &theta; is the angle between p(final) and change in p:
sin&theta;/1.75 = sin33/0.95385
&theta; = 87.756 to the original direction

Hope that helps.

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Excellent. Now we're cooking.

You did fine, but you are mixing up angles. First let's find the angle that the change in momentum (of the white ball) makes with respect to the original direction.

If I understand your triangle properly, you have: a horizontal arrow to the right (the original momemtum), another arrow dipping down at 33 degrees to the first (the final momentum), and the third connecting the two (the change in momentum). Note that the change vector points from the head of the original to the head of the final. For some reason you calculated the angle opposite the original momentum; instead, calculate the angle opposite the final momentum (call it theta). The angle that the change in momentum makes with respect to the original direction of the white ball will be 180 - theta.

Of course, the force that the colored ball exerted on the white ball will be in the direction of the change in momentum of the white ball. By Newton's third law, the force on the colored ball is equal and opposite.

Let me know if this helps.

That said, to keep the angles straight I would have approached the problem using an x & y coordinate system and then calculating components in each direction. I wouldn't have used the triangle rules. (Not that there's anything wrong with that!)

Maybe I should use an actual diagram...

http://www.boomspeed.com/mewmeow/momen.jpg [Broken]

The top picture is what is happening/happened. The other one is the vector triangle. Could you explain what you were talking about again? ^^;

Hmmm x & y coordinate system... We've only been taught the triangle way to do things

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Originally posted by shan

The top picture is what is happening/happened. The other one is the vector triangle. Could you explain what you were talking about again? ^^;

Your triangle is fine, but you must interpret it correctly. It's a picture of Final - Intial = Change. I think you are forgetting that what you labeled as initial is really minus the initial. Now find the angle that the change makes with the original direction (which is horizontal to the left). You found the angle of the triangle opposite the initial vector: you have to add 33 degrees to it to get the total angle.

Bottom line: find the angle between the vectors not just the sides of the triangle.

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Yay I get it now! Thank you so much

## 1. What is a vector diagram for momentum?

A vector diagram for momentum is a graphical representation of the direction and magnitude of an object's momentum. It is used to visually understand the motion and momentum of an object.

## 2. How is momentum represented in a vector diagram?

Momentum is represented by an arrow on the vector diagram. The direction of the arrow indicates the direction of the momentum, while the length of the arrow represents the magnitude of the momentum.

## 3. What is the relationship between velocity and momentum in a vector diagram?

Velocity and momentum are directly related in a vector diagram. The direction of the momentum arrow is the same as the direction of the velocity arrow, and the length of the momentum arrow is proportional to the speed of the object.

## 4. How is the law of conservation of momentum shown in a vector diagram?

The law of conservation of momentum states that the total momentum of a closed system remains constant. In a vector diagram, this is shown by the momentum arrows before and after a collision or interaction remaining the same length and direction, as long as there are no external forces acting on the system.

## 5. What are some real-world applications of vector diagrams for momentum?

Vector diagrams for momentum are commonly used in fields such as physics, engineering, and sports. For example, they can be used to analyze the motion and impact of objects in car crashes, the trajectory of a thrown ball, or the forces acting on a rocket during launch.