Two particles of equal mass m- have respective initial velocities u- and u -2 - They collide completely inelastically- The energy lost in the process is-

Since there is no external force we can apply conservation of momentum along x direction: P_f=P_i ⇒ mu+mu2=2mv_x ⇒ v_x=3u4 Since there is no external force w ...

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Standard XII

Physics

Collision Examples I

Two particles...

Question

Two particles of equal mass $m$ , have respective initial velocities $u^i$ and $u (\frac{^i +^j}{2}) .$ They collide completely inelastically. The energy lost in the process is:

\frac{1}{3} m u^{2}

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\frac{1}{8} m u^{2}

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\frac{3}{4} m u^{2}

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\sqrt{\frac{2}{3}} m u^{2}

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Solution

The correct option is B $\frac{1}{8} m u^{2}$

Conservation of momentum along x-direction:

$m u + \frac{m u}{2} = 2 m v_{x} \Rightarrow v_{x} = \frac{3 u}{4}$

Conservation of momentum along y-direction:

$0 + \frac{m u}{2} = 2 m v_{y} \Rightarrow v_{y} = \frac{u}{4}$

$(K . E .)_{i} = \frac{1}{2} m u^{2} + \frac{1}{2} m [{(\frac{u}{2})}^{2} + {(\frac{u}{2})}^{2}]$

$= \frac{1}{2} m u^{2} + \frac{m u^{2}}{4} = \frac{3 m u^{2}}{4}$

$(K . E .)_{f} = \frac{1}{2} (2 m) (v_{x})^{2} + \frac{1}{2} (2 m) (v_{y})^{2}$

$= \frac{1}{2} 2 m [{(\frac{3 u}{4})}^{2} + {(\frac{u}{4})}^{2}] = \frac{5}{8} m u^{2}$

$∴ Loss of K E = (K E)_{i} - (K E)_{f}$

$= m u^{2} (\frac{6}{8} - \frac{5}{8}) = \frac{m u^{2}}{8}$

Hence, $(B)$ is the correct answer.

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Two particles of equal mass m, have respective initial velocities u^i and u(^i+^j2). They collide completely inelastically. The energy lost in the process is:

Two particles of equal mass $m$ , have respective initial velocities $u^i$ and $u (\frac{^i +^j}{2}) .$ They collide completely inelastically. The energy lost in the process is: