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A ball of mass m moving with a speed v makes a head on collision with an identical ball at rest. The kinetic energy after collision of the balls is three fourth of the original kinetic energy. The coefficient of restitution is

Answer» <html><body><p>`1/2`<br/>`1/3`<br/>`1/(sqrt2)`<br/>`1/(sqrt3)`</p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/MTG_NEET_GID_PHY_XI_C04_E01_127_S01.png" width="80%"/> <br/> Applying the principle of conservation of linear momentum , we get <br/>`mv = mv_1 + mv_2 " or " v = v_1 + v_2 "" ….(i)` <br/>By <a href="https://interviewquestions.tuteehub.com/tag/defination-947003" style="font-weight:bold;" target="_blank" title="Click to know more about DEFINATION">DEFINATION</a> of coefficient of restitution <br/>`e = (v_2 - v_1)/(u_1 - u_2) = (v_2 - v_1)/(v - 0)" or " v_2 - v_1 = ev "" .....(<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>)` <br/> As KE after <a href="https://interviewquestions.tuteehub.com/tag/collision-922060" style="font-weight:bold;" target="_blank" title="Click to know more about COLLISION">COLLISION</a> = `3/4 KE` before collision <br/>`:. 1/2 m (v_1^2 + v_2^2) = 3/4 xx 1/2 mv^2 " or " v_1^2 + v_2^2 = 3/4 v^2 "".....(iii)` <br/> Squaring eq. (i), we get `v_1^2 + v_2^2 + 2v_1 v_2 = v^2 "".........(<a href="https://interviewquestions.tuteehub.com/tag/iv-501699" style="font-weight:bold;" target="_blank" title="Click to know more about IV">IV</a>)` <br/>Subtracting (iiii), from <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a> (iv), we get <br/>`2v_1 v_2 = 1/4 v^2"".....(v)` <br/> Squaring (ii), we get <br/> `v_2^2 + v_1^2 - 2v_1 v_2 = e^2 v^2 "" ....(vi)` <br/>Using (iii) and (v) in (vi), we get <br/> `3/4 v^2 - 1/4 v^2 = e^2 v^2 , 1/2 v^2 = e^2 v^2 or e = 1/(sqrt2)`.</body></html>


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