Conservation of Linear Momentum — Question 5

Question 5

A capture speed sets an unknown mass ratio

A moving cart of mass $m$ at speed $u=6\,\mathrm{m/s}$ captures a stationary cart of unknown mass $M$. They subsequently move together at $2\,\mathrm{m/s}$. External horizontal impulse is negligible.

1. Find $M/m$ without using energy conservation.

2. What fraction of the initial kinetic energy remains, and why can energy conservation not be applied to kinetic energy alone?

Question 5 - Solution

A capture speed sets an unknown mass ratio

(a) Derivation and calculation

The stationary cart initially contributes no momentum: $$m(6)+M(0)=(m+M)(2).$$ $$6m=2m+2M\Rightarrow\boxed{M/m=2}.$$

(b) Result and interpretation

The retained kinetic-energy fraction is $$\boxed{\frac{K_f}{K_i}=\frac{(m+M)(2)^2}{m(6)^2} =\frac{3m(4)}{36m}=\frac13}.$$ The remaining two thirds becomes internal energy through capture and deformation. Total energy remains conserved, but translational kinetic energy does not.

Physical check

The combined mass triples while its speed falls to one third, giving one third the initial energy.