Elastic Collisions in One Dimension — Question 9

Question 9

Center-frame reversal simplifies two moving masses

A $3\,\mathrm{kg}$ cart moving at $4\,\mathrm{m/s}$ meets a $1\,\mathrm{kg}$ cart moving at $-2\,\mathrm{m/s}$. Their collision is elastic and one dimensional.

1. Use the center frame to find both final laboratory velocities.

2. Check momentum and kinetic energy independently.

Question 9 - Solution

Center-frame reversal simplifies two moving masses

(a) Derivation and calculation

Compute the center velocity and incoming center-frame velocities: $$V_C=\frac{3(4)+1(-2)}4=2.5\ \mathrm{m/s},\qquad (u_1',u_2')=(1.5,-4.5).$$ Reverse and transform back: $$\boxed{v_1=2.5-1.5=1,\qquad v_2=2.5+4.5=7\ \mathrm{m/s}}.$$

(b) Result and interpretation

Both invariants agree before and after: $$P_i=12-2=10=3(1)+7=P_f\ \mathrm{kg\,m/s},$$ $$K_i=\tfrac12[3(16)+4]=26=\tfrac12[3(1)+49]=K_f\ \mathrm J.$$ The outgoing separation speed is $7-1=6$ m/s, equal to the incoming approach speed.

Physical check

The lighter object gains the larger velocity change.