A 20 kg curling stone is sliding is sliding in a positive direction at 4 m/s. A second curling stone is sliding at the same speed but in the opposite direction. What is the kinetic energy of the two stones?

The kinetic energy of the two curling stones is 320 J

Explanation:

Given;

mass of the first curling stone, m₁ = 20 kg

velocity of the first curling stone, v₁ = 4 m/s

velocity of the second curling stone, v₂ = - 4m/s

assuming the second curling stone to have equal mass with the first stone = 20 kg

The kinetic energy of the first curling stone is given by;

K.E₁ = ¹/₂m₁v₁²

K.E₁ = ¹/₂(20)(4)²

K.E₁ = 160 J

The kinetic energy of the second stone is given by;

K.E₂ = ¹/₂m₂v₂²

K.E₂ = ¹/₂ (20) (-4)²

K.E₂ = 160 J

Thus, the kinetic energy of the two curling stones is given by;

K.E = K.E₁ + K.E₂

K.E = 160 J + 160 J

K.E = 320 J

Therefore, the kinetic energy of the two curling stones is 320 J