Final answer:
The linear speed of the ball when it leaves the incline is approximately 5.16 m/s.
Step-by-step explanation:
To find the linear speed of the ball when it leaves the incline, we need to use conservation of energy. The gravitational potential energy at the top of the incline is converted into both kinetic energy and rotational kinetic energy at the bottom. Therefore, we can express the conservation of energy equation as:
mgh = 0.5mv^2 + 0.5Iw^2
Where m is the mass of the ball, g is the acceleration due to gravity, h is the initial height, v is the linear speed, I is the moment of inertia, and w is the angular velocity. Since the ball is hollow, its moment of inertia can be calculated as I = (2/3) * mr^2, where r is the radius of the ball.
Plugging in the given values, we have:
(0.434 kg)(9.8 m/s^2)(2 m) = 0.5(0.434 kg)v^2 + 0.5((2/3)(0.434 kg)(0.446 m)^2)w^2
Simplifying the equation, we can solve for v:
v = ±sqrt(2gh - (4/3)(r^2)w^2)
Since the ball is rolling down the incline, its angular velocity can be calculated using w = v/r. Substituting this into the equation, we get:
v = ±sqrt(2gh - (4/3)(v^2)[(0.446 m)/r])
Now we can solve for v:
v^2 + (4/3)(v^2)[(0.446 m)/r] = 2gh
v^2(1 + (4/3)(0.446 m)/r) = 2gh
v^2 = (2gh) / (1 + (4/3)(0.446 m)/r)
v = ±sqrt((2gh) / (1 + (4/3)(0.446 m)/r))
Plugging in the values, we get:
v = ±sqrt((2)(9.8 m/s^2)(2 m) / (1 + (4/3)(0.446 m)/0.446 m))
v ≈ ±5.16 m/s