Does Kinetic Energy increase or decrease with mass ?

Question

In an expression for Kinetic energy KE=(mv^2)/2 KE is directly proportional to mass.

On the other hand, using linear momentum p=mv, expression for KE becomes KE=p²/(2m)

which is inversely proportional to mass.

Which of those two expressions is right? If both are right, is KE directly or inversely proportional to mass?

Can you give some similar examples from different areas of physics?

Answer 1

BOTH expressions are correct but at different conditions.

When we use KE=mv²/2 and say that KE is directly proportional to m we assume that the other parameter (speed) stays the same. So, between two bodies moving with the same speed the more massive one would have larger KE

However, when we analyze the mass dependence of KE using the expression KE=p²/2m

we must assume two bodies have same linear momentum p . So, between two bodies with same linear momentum the less massive one would have larger KE.

An example of the latter is firing a gun. Suppose the gun's body has a mass M while the bullet - mass m. Suppose the gun and the bullet inside it were at rest before the firing. Therefore, before filing, the linear momenta of the gun and the bullet are both zero P_G=0 P_b=0 (so, the total linear momentum of a system "gun+bullet" P_tot=P_G+ P_b=0).

Now, the gun is fired. The bullet and the gun move in opposite directions : the bullet moves forward and the gun recoils but with what speeds?

That is instantly decided by the two of the great Conservation Laws that control every single event in the Universe: conservation of Linear Momentum and Conservation of Energy

Namely, the speeds of the gun and the bullet after firing must be such that the total linear momentum of the "gun+bullet" P_tot=P_G+ P_b would NOT change (i.e. remain zero), another words, P_tot=P_G+ P_b =0 ---> P_b=-P_G=P . Let's stress it: We do NOT know how large P would be : the Linear Momentum Conservation leaves that one completely open: it would allow ANY P as long as the bullet and the gun would both have that P (with opposite signs of course).

How to compare KE of the gun and the bullet? The situation suggests how to do it:

KE_G=P²/(2M_G) ; KE_b=P²/(2m_b)

We see here that b/c both the gun and the bullet must have same (square of) linear momenta P²,

KE must be inversely proportional to mass, and because the bullet's has normally a much smaller mass than the gun, m_b<<m_G its KE is much larger than that of the gun : KE_b>>KE_G

With same bullet's mass and the cartridge's charge, the heavier the body of the gun the smaller is the energy of recoil

[BTW: what is the Energy Conservation's role here? Answer: to decide the size of the P . Namely, suppose a certain amount Q of chemical energy stored in the cartridge is channeled toward moving of the bullet and the gun, then Q=KE_b+KE_G=P²/(2m_b)+P²/(2M_G)=P²(1/(2m_b)+1/(2M_G)) so, P is determined by Q at a given masses of the bullet and the gun]

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The answer to the second part of my question.

Potential Energy U stored in a capacitor with capacitance C with a charge Q and the voltage V allows for two expressions:

U=CU²/2 and U=Q²/2C

So, is U directly or inversely proportional to Capacitance C ?

The answer could be obtained with a very similar reasoning

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When looking on relationship between two variables, we need to clarify the condition of comparisons.

Dr. Ariel B.

Answer 2

You're correct that KE = p²/(2m), but p = mv.

So p²/(2m) = (mv)²/(2m) = (m²v²)/(2m) = (mv²)/2

so KE is still directly proportional to mass.