Momentum and Kinetic Energy in Collisions — Question 8

Question 8

An energy claim can violate the momentum floor

A system of fixed mass $4\,\mathrm{kg}$ has momentum magnitude $12\,\mathrm{kg\,m/s}$. After an isolated collision, a report claims its total kinetic energy is $12\,\mathrm J$.

1. Test the report without knowing individual masses or velocities.

2. If the reported energy is correct, what is the largest possible final momentum magnitude?

Question 8 - Solution

An energy claim can violate the momentum floor

(a) Derivation and calculation

The fixed-momentum lower bound is $$K_f\geq\frac{P^2}{2M}=\frac{12^2}{2(4)}=18\ \mathrm J.$$ $$\boxed{K_f=12\ \mathrm J\text{ is incompatible with isolation and }P=12.}$$

(b) Result and interpretation

If the energy really is 12 J, then $$P^2\leq2MK=2(4)(12)=96,$$ $$\boxed{|P|\leq\sqrt{96}=4\sqrt6=9.80\ \mathrm{kg\,m/s}}.$$ The proposed energy would require a smaller total momentum.

Physical check

The bound applies even when additional internal relative motion is present.