Find kinetic energy from center of mass.

[Sneakatone]

two automobiles each of mass 1200kg traveling a the same direction. Speed of one automobile is 29.0 m/s and the other is 19 m/s. Regard these automobiles as a system of two particles.

a)what is translational energy of center of mass?
I found velocity at center of mass which is 24 m/s and plugged it in K=mv^2 and ended up with 345600 J.

b) what is the total kinetic energy?
(0.5*1200*29^2)+(0.5*1200*19^2)=721200

c) what is the kinetic energy in a reference frame moving with the center of mass?
Im guessing that you subtract part b with part a.

 

[Sunil Simha]

I think you have to find the velocities of each of the objects in the CM reference frame, plug them in mv²/2 and add them.

 

[Sneakatone]

504600+216600=721200
which is what I have for part b
so would the reference frame be zero for part c?

 

[Sunil Simha]

so would the reference frame be zero for part c?

I don't understand your statement. How can a reference frame be 0?

 

[Sneakatone]

I thought the total-translational energy gives kinetic energy in a reference frame (part c).
But I guess not, I think I should just use k=.5mv^2 but with velocity of the center of mass.

 

[Sunil Simha]

In part c, since the frame is moving with the center of mass(I'm assuming 0 relative velocity here between the frame and the CM as the question doesn't say otherwise) it is essentially the frame of the center of mass and thus the kinetic energy should be the same as part b)

Does your source say the same?

 

[Sneakatone]

source dosent have a name.

so are you saying that part c would be zero?

 

[Sunil Simha]

In part c, since the frame is moving with the center of mass(I'm assuming 0 relative velocity here between the frame and the CM as the question doesn't say otherwise) it is essentially the frame of the center of mass and thus the kinetic energy should be the same as part b)

Does your source say the same?

Sorry. My bad. Strike that.

so are you saying that part c would be zero?

I'm not saying it is zero.
Find the velocities of the objects relative to the center of mass frame. Then add their respective kinetic energies.

504600+216600=721200
which is what I have for part b
so would the reference frame be zero for part c?

This is wrong. Get the velocities w.r.t. the CM frame.

 

[Sneakatone]

(19*1200+29*1200)/2400=24m/s
if I plugg it in 0.*1200*24=345600 J is that correct?

 

[Sunil Simha]

The velocity of the bodies w.r.t. the center of mass frame of reference is v_body-v_CM (Vector difference). Now try solving part c)

P.S. If velocity of a body is v_{Body in A} in one frame of reference, say frame A and there is a frame B which is moving at velocity v_B with respect to frame A, then the velocity of the body in frame B is
v_{Body in B} = v_{Body in A} - v_B.