Kinetic Energy — Question 2

Question 2

A right-angle turn can leave kinetic energy unchanged

Use classical translational kinetic energy $K=\tfrac12 m|\vec v|^2$. A 2 kg particle changes velocity from $(3,4)$ m/s to $(-4,3)$ m/s in a fixed perpendicular x-y frame.

1. Find initial and final kinetic energies.

2. Find the velocity-change vector and its magnitude.

3. Is $\tfrac12m|\Delta\vec v|^2$ the kinetic-energy change? Explain numerically.

Question 2 - Solution

A right-angle turn can leave kinetic energy unchanged

(a) Compute the squared norms.

The governing relation is $$\boxed{K=\tfrac12m(v_x^2+v_y^2),\qquad\Delta K=\tfrac12m(|\vec v_f|^2-|\vec v_i|^2)}$$ Both velocity norms squared are $$3^2+4^2=25\ \mathrm{m^2/s^2},$$ giving $$K_i=K_f=25\ \mathrm J,\qquad \Delta K=0.$$

(b) Subtract vectors first.

$$\Delta\vec v=(-7,-1)\ \mathrm{m/s},$$ so $$|\Delta\vec v|=\sqrt{50}=5\sqrt2\ \mathrm{m/s}.$$ The velocity changes despite unchanged speed.

(c) Distinguish two quadratic expressions.

$$\tfrac12m|\Delta\vec v|^2=50\ \mathrm J,$$ not zero. In general $$\Delta K=\tfrac12m(|\vec v_f|^2-|\vec v_i|^2);$$ squaring a vector difference adds different cross terms.