A uniform horizontal rod of length ℓ falls vertically from height h on two identical blocks placed symmetrically below the rod as shown in figure. The coefficients of restitution are e1 and e2. The maximum height through which the centre of mass of the rod will rise after bouncing off the blocks is

Engineering

Physics

Acceleration

Newtons Second Law of Motion

Collision

Question

A uniform horizontal rod of length ℓ falls vertically from height h on two identical blocks placed symmetrically below the rod as shown in figure. The coefficients of restitution are e₁ and e₂. The maximum height through which the centre of mass of the rod will rise after bouncing off the blocks is

$\frac{4 h}{({e_{1}}^{2} + {e_{2}}^{2})}$

$\frac{{(e_{1} + e_{2})}^{2} h}{4}$

$\frac{{(e_{1} + e_{2})}^{2} h}{2}$

$\frac{h}{(e_{1} + e_{2})}$

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A uniform rod falls vertically and strikes two identical blocks symmetrically. The coefficients of restitution are e₁ and e₂. The velocity of the center of mass (COM) after collision is determined by the average of the impulses from both ends. The vertical rebound velocity v of the COM is given by v = (e₁ + e₂)√(2gh)/2, since the impact velocity is √(2gh). The maximum height H is found using v² = 2gH.

Substituting: H = v²/(2g) = [(e₁ + e₂)² * 2gh] / (8g) = (e₁ + e₂)²h/4

Final Answer: $\frac{{(e_{1} + e_{2})}^{2} h}{4}$