A particle of mass m is projected with a speed u from the ground at an angle $\theta = {\pi \over 3}$ 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 $u\widehat i$ . The horizontal distance covered by the combined mass before reaching the ground is:
Solution
By momentum conservation,
<br><br>${{mu} \over 2}$ + mu = 2mV
<br><br>$\Rightarrow$ V = ${{3u} \over 4}$
<br><br>H<sub>max</sub> = ${{{u^2}{{\sin }^2}60^\circ } \over {2g}}$ = ${{{u^2} \times {3 \over 4}} \over {2g}}$ = ${{3{u^2}} \over {8g}}$
<br><br>Time taken = $\sqrt {{{2{H_{\max }}} \over g}}$ = $\sqrt {{2 \over g} \times {{3{u^2}} \over {8g}}}$ = ${{\sqrt 3 } \over 2}{u \over g}$
<br><br>Horizontal distance traveled = ut
<br><br>= ${{3u} \over 4}.{{\sqrt 3 } \over 2}{u \over g}$ = ${{3\sqrt 3 } \over 8}{{{u^2}} \over g}$
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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