Final answer:
To calculate the speed of object A at the end of the ramp, we use conservation of energy and the equation v = sqrt(2gh). Plugging in the values for object A, we can calculate the speed. To calculate the speed of object B, we consider its moment of inertia and use the equation v = sqrt(5/7gh), since the hollow cylinder is rolling without slipping.
Step-by-step explanation:
Part (a) To calculate the speed of object A at the end of the ramp, we can use conservation of energy. The potential energy at the top of the ramp is equal to the kinetic energy at the bottom of the ramp. The potential energy is given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. The kinetic energy is given by (1/2)mv^2, where v is the speed. Setting the two equal and solving for v, we get:
Part (a) v = sqrt(2gh)
Plugging in the values for object A: m = 175 g = 0.175 kg and h = 21.5 cm = 0.215 m, we can calculate the speed.
Part (b) To calculate the speed of object B, we need to consider its moment of inertia. The moment of inertia of a hollow cylinder is given by I = (1/2)mR^2, where m is the mass and R is the radius. Using conservation of energy again, the potential energy is mgh and the kinetic energy is (1/2)mv^2 + (1/2)Iw^2, where w is the angular velocity. Setting the two equal and solving for v, we get:
Part (b) v = sqrt(2gh - (1/2)Iw^2)
Since the hollow cylinder is rolling without slipping, the linear speed v is related to the angular velocity w by v = Rw. Substituting this into the equation and simplifying, we get:
Part (b) v = sqrt(5/7gh)
Plugging in the values for object B: m = 175 g = 0.175 kg and R = 29 cm = 0.29 m, we can calculate the speed.