A particle of mass m is projected with a speed u from the ground at an angle θ=π/3 w.r.t. horizontal (x-axis). When it has reached its maximum height, it collides completely inelastically with another particle of the same mass and velocity ûi. The horizontal distance covered by the combined mass b…

Question

A particle of mass m is projected with a speed u from the ground at an angle θ=π/3 w.r.t. horizontal (x-axis). When it has reached its maximum height, it collides completely inelastically with another particle of the same mass and velocity ûi. The horizontal distance covered by the combined mass before reaching the ground is

Accepted Answer

(2) (3√3)/8(u²)/g. At the top the first particle moves horizontally at u cos 60°=/u2; sticking with the second: V=(u/2+u)/2=(3u)/4. Height of the top: H=(u² sin² 60°)/(2g)=(3u²)/(8g); fall time t=√(2H)/g=(√3 u)/(2g). Distance =Vt=(3u)/4×(√3 u)/(2g)=(3√3)/8(u²)/g.

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