Final answer:
Using conservation of energy, a ball treated as a spherical shell rolling up a hill will reach a height of 1.00 m, while if it slides without rolling, it will reach a height of 0.63 m.
Step-by-step explanation:
To answer the student's question about a ball rolling up a hill, we need to use energy conservation principles. First, let's consider case (a), where the ball is treated as a spherical shell, which means that not only the translational kinetic energy but also the rotational kinetic energy should be taken into account. As the ball rolls up without slipping, the initial kinetic energy is converted into potential energy at the highest point.
The initial kinetic energy (translational + rotational) of the ball is given by:
KE = ½ mv² + ½ Iω², where m is the mass, v is the velocity, I is the moment of inertia for a spherical shell (I = (2/3)mr²), and ω is the angular velocity (ω = v/r).
The potential energy at the height h is given by:
PE = mgh
Setting initial kinetic energy equal to potential energy yields:
KE_initial = PE_at_height
½ mv² + ½ (2/3)mr²(v/r)² = mgh
We can solve for h:
h = ½ v²/g + (1/3)v²/g = (5/6)v²/g
Substituting v = 3.53 m/s and g = 9.81 m/s² gives:
h = (5/6)*3.53²/9.81 ≈ 1.00 m (approximately)
For case (b), if the ball is sliding instead of rolling, it does not have rotational kinetic energy, so only translational kinetic energy is considered.
KE = ½ mv²
½ mv² = mgh
h = v²/(2g)
Substituting v = 3.53 m/s and g = 9.81 m/s² gives:
h = 3.53²/(2*9.81) ≈ 0.63 m (approximately)
Therefore, the correct answers are:
- (a) 1.00 m (for the ball treated as a spherical shell rolling up the hill)
- (b) 0.63 m (for the same ball if it slides up the hill without rolling)