Kinetic Energy — Question 5

Question 5

Relative motion survives the frame that minimizes total kinetic energy

Use classical translational kinetic energy $K=\tfrac12 m|\vec v|^2$. Two 1 kg carts have ground velocities +1 and +5 m/s. An observer moves at constant velocity u. Define the pair kinetic energy as the sum of the two individual kinetic energies; no collision is considered.

1. Express the pair kinetic energy in terms of u.

2. Find its minimum and the observer velocity at that minimum.

3. Can any inertial observer make both carts have zero kinetic energy simultaneously?

Question 5 - Solution

Relative motion survives the frame that minimizes total kinetic energy

(a) Transform both cart velocities.

The governing relation is $$\boxed{K(u)=\tfrac12m_1(v_1-u)^2+\tfrac12m_2(v_2-u)^2}$$

$$K_{\mathrm{pair}}(u)=\tfrac12(1-u)^2+\tfrac12(5-u)^2=(u-3)^2+4\ \mathrm J,$$ using u in m/s.

(b) Read the completed square.

The minimum is 4 J at $$u=3\ \mathrm{m/s}.$$ In that frame the velocities are $-2$ and $+2\ \mathrm{m/s}$, each contributing 2 J.

(c) Require both nonnegative terms to vanish.

Zero total kinetic energy would require $u=1$ and $u=5$ at the same time. No such frame exists. A frame shift removes shared motion but cannot remove the carts’ relative velocity.