Kinetic Energy — Question 1

Question 1

Equal kinetic energies need not imply equal speeds

Use classical translational kinetic energy $K=\tfrac12 m|\vec v|^2$. Cart A has mass 2 kg and speed 6 m/s. Cart B has speed 3 m/s and the same kinetic energy. Treat both carts as particles.

1. Find the kinetic energy and the mass of B.

2. If both speeds increase by 2 m/s, compare their new kinetic energies.

3. If instead both speeds are doubled, does the equal-energy relation survive? Explain.

Question 1 - Solution

Equal kinetic energies need not imply equal speeds

(a) Use the equal-energy constraint.

The governing relation is $$\boxed{K=\tfrac12mv^2,\qquad K_A=K_B\Longrightarrow m_Av_A^2=m_Bv_B^2}$$

$$K_A=\tfrac12(2)6^2=36\ \mathrm J.$$ Hence $$36=\tfrac12 m_B3^2$$ gives $$m_B=8\ \mathrm{kg}.$$

(b) Square each new speed separately.

$$K_A^{\prime}=\tfrac12(2)8^2=64\ \mathrm J,\qquad K_B^{\prime}=\tfrac12(8)5^2=100\ \mathrm J.$$ Equal additive speed changes do not give equal changes in kinetic energy.

(c) Apply a common multiplicative change.

Doubling speed multiplies every kinetic energy by four, so both become 144 J. The distinction is between adding the same speed and multiplying speeds by the same factor.