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A sphere of radius a and mass m rolls along a horizontal plane with constant speed v₀. It encounters an inclined plane at angle θ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?

Asked in JEE Main 25th Feb 2nd Shift 2021 · Rolling on an incline

• (1) 2/5(v₀²)/(g sin θ)
• (2) (v₀²)/(2g sin θ)
• (3) (v₀²)/(5g sin θ)
• (4) (7v₀²)/(10g sin θ)

Step-by-step solution

Why the other options are wrong

• (1) 2/5 is the sphere’s shape factor on its own. What the energy balance needs is 1+2/5=7/5, and it then has to be halved: 7/(10).
• (2) (v₀²)/(2g sin θ) is the distance a *sliding* body would travel, from 1/2mv₀²=mgs sin θ. It leaves the rotational energy out, and a rolling sphere carries 40% more energy than that.
• (3) A check with numbers: take v₀=7 m/s, g=10 m/s² and θ=30°. The sphere carries 7/(10)v₀²=34.3 J/kg, and g sin θ=5 m/s², so it travels (34.3)/5=6.86 m. Option (c) gives 0.98 m, seven times too little.

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