A mass m travels in a straight line with velocity v0 perpendicular to a uniform stick of mass m and length L, which is initially at rest. The distance from the centre of the stick to the path of the travelling mass is h (see figure). Now the travelling mass m collides elastically with the stick, and the centre of the stick and the mass m are observed to move with equal speed v after the collision. Assuming that the travelling mass m can be treated as a point mass, it follows that the distance h must be

Engineering

Physics

Conservation of Angular Momentum

Question

A mass m travels in a straight line with velocity v₀ perpendicular to a uniform stick of mass m and length L, which is initially at rest. The distance from the centre of the stick to the path of the travelling mass is h (see figure). Now the travelling mass m collides elastically with the stick, and the centre of the stick and the mass m are observed to move with equal speed v after the collision. Assuming that the travelling mass m can be treated as a point mass, it follows that the distance h must be

$\frac{L}{2}$

$\frac{L}{\sqrt{6}}$

$\frac{L}{\sqrt{3}}$

$\frac{L}{4}$

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

mv₀ = 2mv
v = $\frac{v_{0}}{2}$
mv₀h = mvh + $\frac{{mℓ}^{2}}{12} ω$
⟹ $\frac{v_{0}}{2}$ h = $\frac{ℓ^{2}}{12}$ ω
e = 1 = $\frac{v + ωh - v}{v_{0} - 0}$ ⟹ v₀ = ωh
$\frac{{ωh}^{2}}{2}$ = $\frac{{ωℓ}^{2}}{12}$ ⟹ h = $\frac{ℓ}{\sqrt{6}}$

Please subscribe our Youtube channel to unlock this solution.