# Conservation of momentum - Collision of pool balls

• crazedbeat
In summary, The green ball has a velocity of 15 cm/s north and the orange ball has a velocity of 15 cm/s west. The two balls never collide because they have different directions of momentum.
crazedbeat
Hi,

I am having trouble recalling my concepts in momentum. I just remember its conserved

Plus, I am horrible with directions, so need help understanding this problem, as well as the underlying concepts.

Two pool balls are rolling along a pool table. The orange pool ball has a velocity of 15 cm/s west and the green ball a velocity of 15 cm/s north 30° east. What are the velocities of the balls after the collision?

What I am doing is as follows:

15*m + 15*m = m(15 + 15)
which doesn't take me anywhere

Do I need to resolve into x and y components?

So any help on this question would be greatly appreciated. I will try and return the help by helping others. (i.e. pay it forward!)

Last edited:
Yeah, you need to resolve the velocities into components.

Okay, so if I resolve them, what exactly does North 30 degrees east mean? Is it north or east or north east? If its north east, what's the 30 degrees about?

How do I draw the diagram? The way I am drawing it right now, they never appear to collide?

The oragne ball will be a vector in the negative x direction magnitude 15cm. The green ball has 15cm north+30 degrees east. I would interpret this to be 30 degrees east of north, which would be 60 degrees from the x axis. At the point of collision, the two vectors will be nose-to-nose, so you can start from that point and draw the vectors from there. You can break down the green ball's vectors by making a triangle and using sin and cosin to find each component.

Then conserve momentum in each direction.

remember the i component is (magnitude)cos(angle from + x axis) and the j component is (magnitude)sin(angle from + x axis) then all you have to do is add them, and convert them back into angles and magnitudes (tan^-1 j/i is angle from + x axis, magnitude is sqrt (i^2 + j^2))

Or you can just draw a vector diagram and work it out with sin and cos rules.

So then would this be correct?

Momentum is conserved

x direction:
$$mv_{1} + mv_{2} = 2mv_{x}$$
$$10 + 10cos(60) = 2v_{x}$$
$$15 = 2v_{x}$$
$$7.5 = v_{x}$$

y direction:
It remains the same $$10sin(60)$$ for $$m_{2}$$

I just want to know if I did this correctly :) After this, I would simply have to find the direction and composite velocity for the green ball.

## 1. What is conservation of momentum?

Conservation of momentum is a fundamental principle of physics that states that the total momentum of a closed system remains constant over time. In other words, the total momentum before a collision is equal to the total momentum after the collision.

## 2. How does conservation of momentum apply to the collision of pool balls?

In the collision of pool balls, the total momentum of the system (the two balls) is conserved. This means that the sum of the initial momentums of the two balls before the collision is equal to the sum of their final momentums after the collision. This is true for both elastic and inelastic collisions.

## 3. What factors can affect the conservation of momentum in a pool ball collision?

The conservation of momentum in a pool ball collision can be affected by factors such as the mass and velocity of the balls, the angle and direction of the collision, and any external forces acting on the system (such as friction).

## 4. What is the difference between elastic and inelastic collisions in terms of conservation of momentum?

In elastic collisions, the total kinetic energy of the system is conserved in addition to the total momentum. This means that the two colliding objects bounce off each other without any loss of energy. In inelastic collisions, the total kinetic energy is not conserved as some energy is lost in the form of heat, sound, or deformation.

## 5. Is conservation of momentum always applicable in real-life collisions?

In general, conservation of momentum is applicable in all collisions, including real-life collisions, as long as the system is closed and there are no external forces acting on it. However, in some situations, other factors such as air resistance or friction may cause small deviations from the principle of conservation of momentum.

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