Collision and Impulse — Question 5

Question 5

The impulse vector of an oblique reflection

A $0.20\,\mathrm{kg}$ puck hits a smooth stationary wall parallel to the $y$ axis. Its velocity changes from $(6,2)$ to $(-4,2)\,\mathrm{m/s}$.

1. Find the wall impulse and the kinetic-energy change.

2. Explain why an unchanged tangential component is consistent with a smooth wall, but does not imply an elastic collision.

Question 5 - Solution

The impulse vector of an oblique reflection

(a) Derivation and calculation

Apply the vector impulse theorem: $$\boxed{\vec J=m(\vec v_f-\vec v_i)=0.20[(-4,2)-(6,2)]=(-2,0)\ \mathrm{N\,s}}.$$ $$\boxed{\Delta K=\tfrac12(0.20)[(-4)^2+2^2-6^2-2^2]=-2.0\ \mathrm J}.$$

(b) Result and interpretation

The tangential impulse vanishes: $$J_y=m(v_{f,y}-v_{i,y})=0,\qquad |v_{f,x}|=4<6=|v_{i,x}|.$$ A smooth wall supplies only normal impulse. It need not return all deformation energy, so smoothness does not establish elasticity.

Physical check

The tangential kinetic energy is unchanged; all 2 J of loss comes from the normal component.