JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2020
- Hard
- MCQ
A particle of mass \(\mathrm{m}\) is projected with a speed \(u\) from the ground at an angle \(\theta=\frac{\pi}{3}\) 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 \(u \hat i\). The horizontal distance covered by the combined mass before reaching the ground is
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Answer & Solution
Correct Answer
(C) \(\frac{3 \sqrt{3}}{8} \frac{u^{2}}{g}\)
Step-by-step Solution
Detailed explanation
By momentum conservation,
\(\frac{\mathrm{mu}}{2}+\mathrm{mu}=2 \mathrm{mv}^{\prime}\)
\(v^{\prime}=\frac{3 v}{4}\)
Range after collision \(=\frac{3 \mathrm{v}}{4} \sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}\)
\(=\frac{3 v}{4} \sqrt{\frac{2 \cdot u^{2} \sin ^{2} 60^{\circ}}{g 2 g}}\)
\(=\frac{3}{4} \frac{\sqrt{3}}{2} \cdot \frac{u^{2}}{g}=\frac{3 \sqrt{3} u^{2}}{8 g}\)
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