Answer:
The kinetic energy of the ball 1 is 2.06 J
Explanation:
The kinetic energy of a rolling object K = 1/2Iω² + 1/2mv² where I is its rotational inertia, ω its angular speed, m its mass and v = its velocity of center of mass.
Let m₁, I₁, v₁, d₁ represent the mass, rotational inertia, speed and diameter of solid ball 1. and Let m₂, I₂, v₂, d₂ represent the mass, rotational inertia, speed and diameter of solid ball 2.
Since both objects are spheres, I =2/5mr²
Let r₁ = radius of ball 1 and r₂ = radius of ball 2. Since d₂ = 2d₁
⇒ 2r₂ = 4r₁ ⇒ r₂ = 2r₁
Now, the the kinetic energy of sphere 1 is
K₁ = 1/2I₁ω₁² + 1/2m₁v₁² ω₁ = v₁/r₁ which is the angular speed of solid ball 1.
K₁ = 1/2(2/5mr²)v₁²/r₁² + 1/2m₁v₁²
K₁ = 1/5m₁v₁² + 1/2m₁v₁²
K₁ = 7/10m₁v₁²
Also, the the kinetic energy of sphere 2 is
K₂ = 1/2I₂ω₂² + 1/2m₂v₂² ω₂ = v₂/r₂ which is the angular speed of solid ball 2.
K₂ = 1/2(2/5m₂r₂²)v₂²/r₂² + 1/2m₂v₂²
K₂ = 1/5m₂v₂² + 1/2m₂v₂²
K₂ = 7/10m₂v₂²
Now, m₁ = m₂/2 and v₁ = v₂/3
Substituting these into K₁, we have
K₁ = 7/10(m₂/2)(v₂/3)²
K₁ = 7/10 × 1/18m₂v₂²
K₁ = (1/18)(7/10m₂v₂²)
K₁ = K₂/18
K₂ = 37.0 J/18
K₂ = 2.06 J
So, the kinetic energy of the ball 1 is 2.06 J
