JEE Main 2021 · 17 March · morning · Single correct · +4 / −1
A boy is rolling a 0.5 kg ball on the frictionless floor with the speed of 20 ms⁻¹. The ball gets deflected by an obstacle on the way. After deflection it moves with 5% of its initial kinetic energy. What is the speed of the ball now?
- (a) 14.41 ms⁻¹
- (b) 19.0 ms⁻¹
- (c) 4.47 ms⁻¹
- (d) 1.00 ms⁻¹
Answer and solution
Answer (c) 4.47 ms⁻¹
Solution
Physics sir · 3 steps
Kinetic energy goes as speed squared, so 5% of the energy leaves only √0.05 of the speed.
- Initial energy. K₁ = 1/2mv² = 1/2 × 0.5 × 20² = 100 J.
- Energy after deflection. Only 5% remains, so K₂ = 0.05 × 100 = 5 J. The mass does not change.
v₂ = √(2K₂/m) = √((2 × 5)/0.5) = √20 m s⁻¹
- So. v₂ = √20 ≈ 4.47 m s⁻¹. You get the same result from 20 × √0.05 = 20 × 0.2236. The answer is (c).
The idea to keep. When a fraction of kinetic energy remains with the same mass, the speed scales by the square root of that fraction: v₂ = v₁√(K₂/K₁).
Answer from the NTA final answer key. Solution by PrepWiser.