A sphere of mass 2 kg and radius 0.5 m is rolling with an initial speed of 1 ms⁻¹ goes up an inclined plane which makes an angle of 30° with the horizontal plane, without slipping. How long will the sphere take to return to the starting point A?

Solution

Physics sir · 4 steps

Rolling cuts the slope's pull to 5/7 of g sin θ, and the trip down mirrors the trip up.

1. Acceleration on incline. For a body rolling without slipping, static friction makes the acceleration along the incline a = (g sin θ)/(1 + k²/R²). For a solid sphere k²/R² = 2/5, so mass and radius drop out.

   a = (g sin θ)/(1+2/5) = 5/7× 9.8 × 1/2 = 3.5 m s⁻²

2. Same both ways. Friction points up the slope while the sphere climbs and also while it rolls back down. So the acceleration is 3.5 m s⁻² down the slope the whole time, and the motion is symmetric, just like a ball thrown straight up.
3. Time up and back. Time to stop is t = v/a = 1/3.5 = 0.286 s. Coming back down takes the same time, so the total is 2 × 0.286 = 0.571 s.
4. So. The sphere returns to A after about 0.57 s, which is option (d).

The idea to keep. For rolling without slipping on an incline, use a = (g sin θ)/(1+k²/R²). Mass and radius cancel, and the up and down trips take equal times.

Answer from the NTA final answer key. Solution by PrepWiser.