Explaining the Equation: ½ mu2 = ½ mv2 + ½ ms2

[Disputationem]

Homework Statement

I have no idea how this equation (½ mu^2 = ½ mv^2 + ½ ms^2 ) came to be. I understand that work done is equal to final kinetic energy minus initial (w = KE_v - KE_u) but I have no idea how the "1/2 ms^2" portion came to be. Is there anyone that can help me on explaining why the equation is in this manner?

Relevant Equations

½ mu^2 = ½ mv^2 + ½ ms^2 , w = KE_v - KE_u.

My attempt is that I think that, since the equation was derived in an interaction between particles (proving that the angle between the subsequent paths of both particles must be 90 degrees), that acceleration (in m/s^2) is basically negligable since it is so small that it could be rounded off to 0. However, I'm unsure about the validity of this theory, so I would like an explanation for why ½ mu2 = ½ mv2 + ½ ms2 is the way it is.

 

[anuttarasammyak]

May I understand your case that energy and velocity of the body we are watching is before collision
KE_u=\frac{1}{2}mv^2
and after collision
KE_v=\frac{1}{2}mu^2
and s is speed of another body of same mass that I assume it has been at still but starts moving collided by the body mentioned above ?

 

[Steve4Physics]

The final terms

Homework Statement:: I have no idea how this equation (½ mu^2 = ½ mv^2 + ½ ms^2 ) came to be. I understand that work done is equal to final kinetic energy minus initial (w = KE_v - KE_u) but I have no idea how the "1/2 ms^2" portion came to be. Is there anyone that can help me on explaining why the equation is in this manner?
Relevant Equations:: ½ mu^2 = ½ mv^2 + ½ ms^2 , w = KE_v - KE_u.

My attempt is that I think that, since the equation was derived in an interaction between particles (proving that the angle between the subsequent paths of both particles must be 90 degrees), that acceleration (in m/s^2) is basically negligable since it is so small that it could be rounded off to 0. However, I'm unsure about the validity of this theory, so I would like an explanation for why ½ mu2 = ½ mv2 + ½ ms2 is the way it is.

Aplologies if I'm wrong, but from your wording, it seems like you believe the term "½ ms^2” is a value of acceleration. (a=0.5 m/s²).

This would not make sense because you can’t add an acceleration to an energy.

To make sense, ‘s’ needs to be the symbol for the speed of some object. (It's a bad choice of a symbol for speed though!)

For example, the equation would be valid in this situation:

Particle A (mass m, speed u) collides elastically with particle B (mass m, stationary). After the collision A's speed is v and B’s speed is s.

Since kinetic energy is conserved in elastic collisions:
½mu² = ½mv² + ½ms²

 

[jbriggs444]

[moved my response here from the other thread]

I never understood why the equation was ½ mu2 = ½ mv2 + ½ ms2 in the first place. I understand the mu and mv portion, but I don't quite understand the ms part of the equation. What is it derived from?

The equation comes from conservation of energy. $u$ is the initial velocity of the incoming proton. The initial velocity of the other proton is zero. So the total kinetic energy is $\frac{1}{2}mu^2$.

[In the other thread, context is supplied. This is an elastic impact of an arriving proton with an equally massive proton that begins at rest]

After the collision, the incoming proton moves away with velocity $v$ and the other proton moves away with velocity $s$. So the total kinetic energy is $\frac{1}{2}mv^2 + \frac{1}{2}ms^2$.

By conservation of energy, initial energy is equal to final energy so $\frac{1}{2}mu^2=\frac{1}{2}mv^2 + \frac{1}{2}ms^2$.The equation also looks very much like a statement of the Pythagorean theorem.

If one looks at conservation of momentum then the momentum of the incoming proton $m\vec{u}$ must be equal to the sum of the momenta of the two departing protons, $m\vec{v} + m\vec{s}$. We can divide out the $m$ and get $\vec{u}=\vec{v} + \vec{s}$.

If the two departing protons move away at right angles to one another then the magnitude of the vector sum is given by the Pythagorean theorem so that $|\vec{u}|^2=|\vec{v}|^2+|\vec{s}|^2$. One can multiply by $\frac{1}{2}m$ to recover the original equation: $\frac{1}{2}mu^2=\frac{1}{2}mv^2 + \frac{1}{2}ms^2$ and all is well.

Conversely, if the departing velocities do not have the Pythagorean relationship with the arrival velocity then the angle between the departure velocities will not be a right angle.

 

[rude man]

Looks like you tried to say that in a collision the sum of k.e. before the collision = sum of k.e. after.
But you have to add other sources of energy in addition to k.e.
Say a bomb goes off at the collision.. Obviously, k.e. before collision $\neq$ k.e. after collision:
k.e. after = k.e before + energy of explosion going to k.e.

Note that conservation of momentum is not affected by the explosion since the explosion is not an externally applied force.

 

[Disputationem]

The final terms

Aplologies if I'm wrong, but from your wording, it seems like you believe the term "½ ms^2” is a value of acceleration. (a=0.5 m/s²).

This would not make sense because you can’t add an acceleration to an energy.

To make sense, ‘s’ needs to be the symbol for the speed of some object. (It's a bad choice of a symbol for speed though!)

For example, the equation would be valid in this situation:

Particle A (mass m, speed u) collides elastically with particle B (mass m, stationary). After the collision A's speed is v and B’s speed is s.

Since kinetic energy is conserved in elastic collisions:
½mu² = ½mv² + ½ms²

Thank you so much! I was so confused about why they were mentioning displacement instead of acceleration all of a sudden and now I know!

 

[Disputationem]

[moved my response here from the other thread]

The equation comes from conservation of energy. $u$ is the initial velocity of the incoming proton. The initial velocity of the other proton is zero. So the total kinetic energy is $\frac{1}{2}mu^2$.

[In the other thread, context is supplied. This is an elastic impact of an arriving proton with an equally massive proton that begins at rest]

After the collision, the incoming proton moves away with velocity $v$ and the other proton moves away with velocity $s$. So the total kinetic energy is $\frac{1}{2}mv^2 + \frac{1}{2}ms^2$.

By conservation of energy, initial energy is equal to final energy so $\frac{1}{2}mu^2=\frac{1}{2}mv^2 + \frac{1}{2}ms^2$.The equation also looks very much like a statement of the Pythagorean theorem.

If one looks at conservation of momentum then the momentum of the incoming proton $m\vec{u}$ must be equal to the sum of the momenta of the two departing protons, $m\vec{v} + m\vec{s}$. We can divide out the $m$ and get $\vec{u}=\vec{v} + \vec{s}$.

If the two departing protons move away at right angles to one another then the magnitude of the vector sum is given by the Pythagorean theorem so that $|\vec{u}|^2=|\vec{v}|^2+|\vec{s}|^2$. One can multiply by $\frac{1}{2}m$ to recover the original equation: $\frac{1}{2}mu^2=\frac{1}{2}mv^2 + \frac{1}{2}ms^2$ and all is well.

Conversely, if the departing velocities do not have the Pythagorean relationship with the arrival velocity then the angle between the departure velocities will not be a right angle.

Your reply is complex but it makes a bundle of sense! Thanks a mil!