# Kinetic Energy of 10 0.50-kg Balls in Symmetrical Pattern

• tony873004
In summary: If you cancel out the velocities, you would end up with 0 for the kinetic energy, which we know is not the case.In summary, the system described consists of 10 0.50-kg balls moving radially outward from a common center in a symmetrical pattern, each at a speed of 12 m/s. The kinetic energy of the system can be calculated using the formula E_k= (1/2)mv^2 and is equal to 360 J. The velocities of the balls cannot be cancelled out as it would impose a direction requirement on the kinetic energy, which is a scalar. Therefore, the kinetic energy of the system cannot be 0.
tony873004
Gold Member
A system consists of 10 0.50-kg balls, each moving radially outward from a common center in a symmetrical pattern, all at a speed of 12 m/s. What is the kinetic energy of the system?

My instinct is to say 0, since it it symmetrical and every ball moving one direction will be canceled by a ball of the same mass and speed moving the opposite direction.

But energy is a scalar, right? And scalars only have magnitude, not direction, right?

$$e_k=10*\frac{1}{2}mv^2$$

$$e_k=10*\frac{1}{2}0.50kg *(12m/s)^2$$

$$e_k=360 J$$

My guess is 0, but I'm not sure why.
I think if they all are radiating outward from a common center, then they must have been motionless at some point during which the potential enegry and kinetic energy is zero.
Energy is a scalar though and they don't cancel each other out, I'd still like to know why its not 0 in a definitive fashion though.

No, your "second thought" was correct. E= (1/2)mv2 where you can think of v2 as meaning either the dot product or the square of the magnitude of the v vector.

In any case, kinetic energy, unlike momentum, IS a scalar, not a vector, and is always positive. The kinetic energys of the two objects do not cancel, they add.

whozum said:
I think if they all are radiating outward from a common center, then they must have been motionless at some point during which the potential enegry and kinetic energy is zero.

No, if they are radiating outward, they are moving. If they were "motionless at some point" there must have been some force acting to cause them to move. That force does work and contributes energy.

I'm guessing that the velocities cancel. Since they're vectors and not scalars, I can cancel them before plugging them into the K equation. Then one of my factors is 0 which makes the whole answer 0.

I think.

Well take the balls to be physical objects. If they are all moving outwards at t=t1, then isn't it feasible to say at some t<t1 they were at the same point? If so, motionless?

Its kind of like the big bang.

I posted before I saw your post HallsofIvy. Can't I cancel the v's first?

Thinking about it, I guess it couldn't be 0. I could always spin down this system with a generator and light a light bulb for a moment.

Cancelling out the velocities would impose a direction requirement on the kinetic energy, which we already know it doesn't have.

## 1. What is kinetic energy?

Kinetic energy is the energy possessed by an object due to its motion.

## 2. How is kinetic energy calculated?

The kinetic energy of an object can be calculated using the formula: KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.

## 3. What is the significance of the symmetrical pattern in this scenario?

The symmetrical pattern in this scenario ensures that each ball has the same mass and velocity, making it easier to calculate the total kinetic energy of the system.

## 4. How does the number of balls affect the kinetic energy?

The kinetic energy of the system will increase with the number of balls, as the total mass and velocity will also increase.

## 5. Can the kinetic energy of this system be changed?

Yes, the kinetic energy of this system can be changed by altering the mass or velocity of the balls, or by changing the number of balls in the system.

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