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Two equal spheres B and C, each of mass m, are in contact on a smooth horizontal table. A third sphere A of same size as that of B or C but mass m//2 impinges symmetrically on them with a velocity u and is itself brought to rest. The coefficient of restitution between the two spheres A and B (or between A and C) is

Answer» <html><body><p>`1//3`<br/>`1//4`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>//3`<br/>`3//4`</p>Solution :`u=`velocity of sphere `A` before impact. As the spheres are identical, the triangle `ABC` formed by joining their centres is equilateral. The spheres `B` and `C` will move in <a href="https://interviewquestions.tuteehub.com/tag/direction-1696" style="font-weight:bold;" target="_blank" title="Click to know more about DIRECTION">DIRECTION</a> `AB` and `AC` after impact making can <a href="https://interviewquestions.tuteehub.com/tag/angle-875388" style="font-weight:bold;" target="_blank" title="Click to know more about ANGLE">ANGLE</a> of `30^(@)` with the origina lines of motion of <a href="https://interviewquestions.tuteehub.com/tag/ball-891849" style="font-weight:bold;" target="_blank" title="Click to know more about BALL">BALL</a> `A`. <br/> Let `v` be the speed ofthe balls `B` and `C` after impact. <br/> Momentum conservationgives <br/> `(m/2)u=mvcos30^(@) +mvcos30^(@)` <br/> `u=2sqrt(3)vimpliesv=u/(2sqrt(3))`.......i <br/> From Newton's experimental law, for an oblique collision, we have take components along normal, i.e. along `AB` for balls `A` and `B`. <br/> `v_(B)-v_(A)=-e(u_(B)-u_(A))` <br/> `v-0=-e(0-ucos30^(@))` <br/> `v=eucos30^(@)`.........ii <br/> Combining <a href="https://interviewquestions.tuteehub.com/tag/eqn-973463" style="font-weight:bold;" target="_blank" title="Click to know more about EQN">EQN</a> i and ii we get `e=1/3` <br/> Loss in `KE =1/2 m/2 u^(2)-2(1/2mv^(2))` <br/> `=1/4mu^(2)-m(u/(2sqrt(3)))^(2)=1/6mu^(2)`</body></html>


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