Problem — Kinetic energy and doubled speed

Kinetic energy and doubled speed. Explain why doubling a car's speed quadruples its kinetic energy, but only doubles its momentum.

Solution

The two quantities depend on speed differently:

$E_k = \tfrac{1}{2}mv^2 \qquad p = mv$

Kinetic energy is quadratic in $v$: if $v \to 2v$ at constant mass,

$E_k' = \tfrac{1}{2}m(2v)^2 = 4\cdot\tfrac{1}{2}mv^2 = 4\,E_k$

so $E_k$ quadruples. Momentum, on the other hand, is linear in $v$:

$p' = m(2v) = 2\,mv = 2\,p$

so $p$ doubles.

$\ev{E_k \propto v^2 \ \Rightarrow\ \times 4; \qquad p \propto v \ \Rightarrow\ \times 2}$

Practical consequence. Stopping a car means dissipating all of its $E_k$ through the brakes. Going from $50$ to $100\;\mathrm{km/h}$ doubles the speed but quadruples the energy to be disposed of: the braking distance (at constant deceleration, $E_k = F\,d$) becomes four times longer. This is the physical reason why high speed is so dangerous.

Collegamenti

Argomenti: Lavoro ed energia Concetti: Energia cinetica · Quantità di moto Competenze: Impostazione simbolica