解析：设v ₁ =1.0× 10 ⁶ m /s；v ₂ =2.0× 10 ⁸ m /s

由相对论的动能表达式，得

E _k1 = -m ₀ c ² =4.55×10 ^-19 J

E _k2 = m ₀ c ² =27.9×10 ^-15 J

由经典力学公式E _k = m ₀ v ² ，得

E _k1 = m ₀ v ₁ ² =4.55×10 ^-19 J

E _k2 = m ₀ v ₂ ² =18.2×10 ^-15 J.

答案：(1)E _k1 =4.55×10 ^-19 J

E _k2 =27.9×10+ J

(2)E _k1 =4.55×10 ^-19 J

E _k2 =18.2×10 ^-15 J

解析：设v ₁ =1.0× 10 ⁶ m /s；v ₂ =2.0× 10 ⁸ m /s

由相对论的动能表达式，得

E _k1 = -m ₀ c ² =4.55×10 ^-19 J

E _k2 = m ₀ c ² =27.9×10 ^-15 J

由经典力学公式E _k = m ₀ v ² ，得

E _k1 = m ₀ v ₁ ² =4.55×10 ^-19 J

E _k2 = m ₀ v ₂ ² =18.2×10 ^-15 J.

答案：(1)E _k1 =4.55×10 ^-19 J

E _k2 =27.9×10+ J

(2)E _k1 =4.55×10 ^-19 J

E _k2 =18.2×10 ^-15 J