# Is kinetic energy of center of mass of a system is equal to KE of of system?

• vkash
In summary, the kinetic energy of the center of mass of a system is not equal to the kinetic energy of the system. The total kinetic energy of a system is the sum of the kinetic energies of all particles or objects in the system. In order to solve a problem involving a movable block and wedge, it is necessary to consider the kinetic energy of both the block and the wedge, as well as the potential energy. By conserving energy, the velocity of the center of mass before collision can be found and substituted into the equation to solve for the minimum value of u.
vkash
Is kinetic energy of center of mass of a system is equal to KE of of system??

Does Kinetic energy(KE) of center of mass is equal to kinetic energy of system?
I think no...
Example take two identical particle moving with velocity v and -v. kinetic energy is $$mv^2$$ but KE of center of mass is zero>
have a look at this example.
A block of mass m is pushed towards towards a movable wedge of mass nm. and height h, with a velocity v. All the surfaces are smooth. Find the minimum value of u for which block will reach top of height.
one of the equation placed(in book) to solve this question is
Work done =change in Kinetic energy
$$-mgh=\frac{(m+nm)V^2}{2}-\frac{mu^2}{2}$$ (v=velocity of center of mass before collision)
(where this latex code is wrong)
I think this equation is incorrect. Because final Kinetic energy can't be written as kinetic energy of center of mass.
Am I correct??

Secondly i failed to solve this question. Any hint.,

Hey! First post here so maybe I can be of help and not give wrong info haha. To start with your initial question, no. The kinetic energy of the center of mass is not equal to the kinetic energy of the system. Kinetic energy of the system is the summation of all of the particles/objects in that system. So if you have two objects the total kinetic energy = 0.5(m1)(v1)^2 + 0.5(m2)(v2)^2

So yeah, for your actual problem. You have kinetic energy at the start and all of that is turning into potential energy when it goes up the block (if the initial speed is a minimum). Considering that energy is conserved, initial energy (kinetic) = final energy (potential)

Don't think I'm supposed to give away too much so maybe I should leave it at that and come back if you get it from there? Good luck!

vkash said:
Does Kinetic energy(KE) of center of mass is equal to kinetic energy of system?
I think no...
Example take two identical particle moving with velocity v and -v. kinetic energy is $$mv^2$$ but KE of center of mass is zero>
have a look at this example.

one of the equation placed(in book) to solve this question is
Work done =change in Kinetic energy
$$-mgh=\frac{(m+nm)V^2}{2}-\frac{mu^2}{2}$$ (v=velocity of center of mass before collision)

You are right, the KE is the sum of the KE-s of all participants, but the whole KE is equal to the sum of the KE of the CM and the KE of all parts relative to the CM, that is Ʃ1/2 mi (v-VCM)2.

First the total KE is that of the block. If it goes up along the wedge it pushes it and slows down with respect to it. If it just reaches the top, its relative velocity with respect to the wedge is zero, the whole system moves together as a single body, with the velocity of the CM.
As there are internal forces only, the velocity of the CM does not change.

Edit: As there are only internal forces in the horizontal direction, the horizontal velocity of the CM does not change.

Find the velocity of the center of mass before collision and substitute for V into the equation. Solve for u.

(About Latex: Do not use "[bold]" inside it. )

ehild

Last edited:

Hi Syrianto,

welcome at PF.

Syrianto said:
So yeah, for your actual problem. You have kinetic energy at the start and all of that is turning into potential energy when it goes up the block (if the initial speed is a minimum). Considering that energy is conserved, initial energy (kinetic) = final energy (potential)

That is not right. The energy of the whole system conserves. When the block slides upward along the wedge, it pushes it and the wedge starts to move. It accelerates till the block has velocity relative to it. At the end the whole thing moves together as a single body. The final KE is not zero.

ehild

ehild said:
You are right, the KE is the sum of the KE-s of all participants, but the whole KE is equal to the sum of the KE of the CM and the KE of all parts relative to the CM, that is Ʃ1/2 mi (v-VCM)2.

First the total KE is that of the block. If it goes up along the wedge it pushes it and slows down with respect to it. If it just reaches the top, [highlight]its relative velocity with respect to the wedge is zero, the whole system moves together as a single body, with the velocity of the CM.[highlight]
As there are internal forces only, the velocity of the CM does not change.

Find the velocity of the center of mass before collision and substitute for V into the equation. Solve for u.

(About Latex: Do not use "[bold]" inside it. )

ehild

thanks!
since there is no horizontal force so there will no change in velocity of center of mass in horizontal direction.

Syrianto said:
Hey! First post here so maybe I can be of help and not give wrong info haha. To start with your initial question, no. The kinetic energy of the center of mass is not equal to the kinetic energy of the system. Kinetic energy of the system is the summation of all of the particles/objects in that system. So if you have two objects the total kinetic energy = 0.5(m1)(v1)^2 + 0.5(m2)(v2)^2

So yeah, for your actual problem. You have kinetic energy at the start and all of that is turning into potential energy when it goes up the block (if the initial speed is a minimum). Considering that energy is conserved, initial energy (kinetic) = final energy (potential)

Don't think I'm supposed to give away too much so maybe I should leave it at that and come back if you get it from there? Good luck!

welcome to PF!

You got it little wrong actually wedge is also movable so it is required to calculate the change in KE of that wedge...

vkash said:
thanks!
since there is no horizontal force so there will no change in velocity of center of mass in horizontal direction.

You are right, I meant that there are no horizontal external forces.

ehild

Ohh, sorry for putting out bad info then D=

When he said all surfaces were smooth I assumed that it meant the block + floor, and the ramp part of the wedge. If the bottom of the wedge was firmly planted to the ground would my statement be correct?

## 1. What is kinetic energy of center of mass of a system?

Kinetic energy of center of mass of a system is the energy possessed by the system as a whole due to its motion. It is a measure of the total mechanical energy of the system.

## 2. How is the kinetic energy of center of mass of a system calculated?

The kinetic energy of center of mass of a system is calculated by multiplying the mass of the system by the square of its velocity, and then dividing by 2. This can be expressed as KE = 1/2 * m * v^2.

## 3. Is the kinetic energy of center of mass of a system always equal to the KE of the system?

Yes, the kinetic energy of center of mass of a system is always equal to the KE of the system. This is because the center of mass represents the average motion of all the particles in the system, and the KE takes into account the motion of all the particles in the system.

## 4. How does the kinetic energy of center of mass of a system affect its overall motion?

The kinetic energy of center of mass of a system determines the overall motion of the system. A higher KE of center of mass means the system has a greater ability to do work and a larger velocity, while a lower KE of center of mass means the system has less energy and a slower velocity.

## 5. Can the kinetic energy of center of mass of a system change?

Yes, the kinetic energy of center of mass of a system can change if there is a change in the velocity or mass of the system. This can happen due to external forces acting on the system or internal interactions between the particles within the system.

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