Inelastic Collisions in One Dimension — Question 8

Question 8

Known external impulse corrects a sticking calculation

A $2\,\mathrm{kg}$ cart at $5\,\mathrm{m/s}$ sticks to a stationary $3\,\mathrm{kg}$ cart. During contact a brake gives the two-cart system a measured impulse $-1.0\,\mathrm{N\,s}$.

1. Find the common final speed and total kinetic-energy change.

2. Can the energy loss be assigned entirely to deformation in the collision?

Question 8 - Solution

Known external impulse corrects a sticking calculation

(a) Derivation and calculation

Apply the impulse balance to the full pair: $$P_f=P_i+J_{\mathrm{ext}}=2(5)-1=9\ \mathrm{kg\,m/s}.$$ $$\boxed{V=9/(2+3)=1.8\ \mathrm{m/s}}.$$

(b) Result and interpretation

The observed kinetic-energy change is $$K_i=\tfrac12(2)(5)^2=25\ \mathrm J,\qquad K_f=\tfrac12(5)(1.8)^2=8.1\ \mathrm J.$$ $$\boxed{\Delta K=-16.9\ \mathrm J}.$$ The brake impulse $\int F\,dt$ does not determine its work $\int F\,dx$. Therefore these data do not separately determine brake work and internal heating.

Physical check

Using momentum conservation without the external impulse would incorrectly predict 2 m/s.