Relation Between KE and Momentum

Derivation

Momentum: $p = mv \implies v = \frac{p}{m}$

Kinetic energy:

$KE = \frac{1}{2}mv^2 = \frac{1}{2}m\left(\frac{p}{m}\right)^2 = \frac{p^2}{2m}$

$\boxed{KE = \frac{p^2}{2m}}$

Why this form is useful

In many problems — explosions, collisions, radioactive decay — momentum is conserved and is easier to work with than velocity. This formula connects momentum directly to kinetic energy without needing to find velocity first.

Comparing KE of bodies with equal momentum

Two bodies of masses $m_1$ and $m_2$ have the same momentum $p$:

$KE_1 = \frac{p^2}{2m_1}, \quad KE_2 = \frac{p^2}{2m_2}$

$\frac{KE_1}{KE_2} = \frac{m_2}{m_1}$

The lighter body has more kinetic energy.

A bullet and a gun recoiling with equal and opposite momenta (from an explosion): the bullet has far more kinetic energy — it does the damage.

Comparing KE of bodies with equal KE

Two bodies with the same KE:

$\frac{p^2}{2m_1} = \frac{p^2}{2m_2}...$

Wait — if KE is equal: $\frac{p_1^2}{2m_1} = \frac{p_2^2}{2m_2}$

$\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}}$

The heavier body has more momentum when both have equal KE.

Explosion from rest

A body at rest explodes into two fragments of masses $m_1$ and $m_2$.

By conservation of momentum: $p_1 = p_2 = p$ (equal and opposite)

Ratio of kinetic energies:

$\frac{KE_1}{KE_2} = \frac{p^2/(2m_1)}{p^2/(2m_2)} = \frac{m_2}{m_1}$

The lighter fragment carries more KE. The total KE released equals the chemical energy of the explosion.

Stopping distance

A vehicle of mass $m$ moving at speed $v$ has momentum $p = mv$ and KE $= \frac{p^2}{2m}$.

Braking force $F$ does work over stopping distance $d$:

$Fd = \frac{p^2}{2m} \implies d = \frac{p^2}{2mF} = \frac{mv^2}{2F}$

For same braking force: heavier vehicle needs more distance to stop (larger $m$ → larger $d$).

For same momentum: lighter vehicle stops in shorter distance (smaller $m$ → smaller $d = \frac{p^2}{2mF}$).