Answer:
The kinetic energy of the two curling stones is 320 J
Step-by-step 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