What is the Velocity of the Center of Mass in a Proton-Carbon Collision?

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In a proton-carbon collision, the velocity of the center of mass can be calculated using the formula Mv = mv1 + mv2, where M is the total mass of the system. For a proton of mass m and a stationary carbon nucleus of mass 12m, the total mass is 13m. The initial velocity of the proton is 790 m/s, while the carbon nucleus has a velocity of 0. By substituting these values into the equation, the velocity of the center of mass is determined to be 790 m/s in the positive x direction, reflecting the influence of the proton's motion. Understanding this concept is crucial for analyzing the dynamics of two-particle systems.
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A proton of mass m undergoes a head-on elastic collision with a stationary carbon nucleus of mass 12×m. The initial velocity of the proton is 790 m/s in the positive x direction. Find the velocity of the center of mass of the system.

I am really confused as to how to find the 'velocity of the center of mass'. I thought using the equation Mv = mv1 + mv2 would work but nope it surely didnt. I am not sure if I am missing an equation or what but I can't seem to figure it out.
Thanks for your help and time!
 
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envscigrl said:
A proton of mass m undergoes a head-on elastic collision with a stationary carbon nucleus of mass 12×m. The initial velocity of the proton is 790 m/s in the positive x direction. Find the velocity of the center of mass of the system.

I am really confused as to how to find the 'velocity of the center of mass'. I thought using the equation Mv = mv1 + mv2 would work but nope it surely didnt. I am not sure if I am missing an equation or what but I can't seem to figure it out.
Thanks for your help and time!

I do not see any problem with that equation.
V = (m1v1+m2v2)/(m1+m2) = (790*m+0*12m)/(13m) = 60.8 m/s in the positive x direction is really the velocity of the center of mass.

ehild
 


The velocity of the center of mass of a system can be calculated using the equation Mv = mv1 + mv2, where M is the total mass of the system, v is the velocity of the center of mass, m1 and m2 are the masses of the individual particles, and v1 and v2 are their respective velocities. In this case, the total mass of the system is M = m + 12m = 13m.

We know the initial velocity of the proton, v1 = 790 m/s in the positive x direction. Since the carbon nucleus is stationary, v2 = 0. Plugging these values into the equation, we get:

13m * v = m * 790 m/s + 12m * 0
13m * v = 790m * s
v = 790 m/s

Therefore, the velocity of the center of mass of the system is 790 m/s in the positive x direction. This makes sense intuitively, as the proton is much lighter than the carbon nucleus, so the center of mass will be closer to the proton and will have a higher velocity in the same direction as the initial velocity of the proton.

I hope this helps clarify the concept of center of mass and how to calculate its velocity in a two-particle system. It's important to remember that the center of mass is a point that represents the overall motion of the system, and its velocity is the average velocity of all the particles in the system.
 
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