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JEE Mains · Physics · STD 11 - 5. work,energy,power and collision

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

  1. A \(\frac{3 \sqrt{2}}{4} \frac{u^{2}}{g}\)
  2. B \(2 \sqrt{2} \frac{\mathrm{u}^{2}}{\mathrm{g}}\)
  3. C \(\frac{3 \sqrt{3}}{8} \frac{u^{2}}{g}\)
  4. D \(\frac{5}{8} \frac{\mathrm{u}^{2}}{\mathrm{g}}\)
Verified Solution

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 \(=\frac{3 \mathrm{v}}{4} \sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}\) \(=\frac{3 v}{4} \sqrt{\frac{2 \cdot u^{2} \sin ^{2} 60^{\circ}}{g 2 g}}\)…
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