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 u hati. The horizontal distance covered by the combined mass before reaching the ground is :

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 u hati. The horizontal distance covered by the combined mass before reaching the ground is : (A) (3√3/8) (u2/g) (B) 2√2 (u2/g) (C) (3√2/4) (u2/g) (D) (5/8) (u2/g)

Tardigrade Admin · Accepted Answer

By momentum conservation, (mu/2) + mu = 2mv' v' = (3v/4) Range after collision = (3v/4)√(2H/g) = (3v/4) √(2⋅ u2 sin2 60°/g2g) =(3/4) (√3/2). (u2/g) = (3√3u2/8g)

$\frac{3 3}{8} \frac{u ^{2}}{g}$

$22 \frac{u ^{2}}{g}$

$\frac{3 2}{4} \frac{u ^{2}}{g}$

$\frac{5}{8} \frac{u ^{2}}{g}$

By momentum conservation,
$\frac{m u}{2} + m u = 2 m v^{'}$
$v^{'} = \frac{3 v}{4}$
Range after collision $= \frac{3 v}{4} \frac{2 H}{g}$
$= \frac{3 v}{4} \frac{2 \cdot u ^{2} s i n ^{2} 6 0 ^{\circ}}{g 2 g}$
$= \frac{3}{4} \frac{3}{2} . \frac{u ^{2}}{g} = \frac{3 3 u ^{2}}{8 g}$

833​​gu2​

22​gu2​

432​​gu2​

85​gu2​

By momentum conservation, 2mu​+mu=2mv′ v′=43v​ Range after collision =43v​g2H​​ =43v​g2g2⋅u2sin260∘​​ =43​23​​.gu2​=8g33​u2​

$\frac{3 3}{8} \frac{u ^{2}}{g}$

$22 \frac{u ^{2}}{g}$

$\frac{3 2}{4} \frac{u ^{2}}{g}$

$\frac{5}{8} \frac{u ^{2}}{g}$

By momentum conservation,
$\frac{m u}{2} + m u = 2 m v^{'}$
$v^{'} = \frac{3 v}{4}$
Range after collision $= \frac{3 v}{4} \frac{2 H}{g}$
$= \frac{3 v}{4} \frac{2 \cdot u ^{2} s i n ^{2} 6 0 ^{\circ}}{g 2 g}$
$= \frac{3}{4} \frac{3}{2} . \frac{u ^{2}}{g} = \frac{3 3 u ^{2}}{8 g}$