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In Fig. there are n identical suspended with wires of equal length. The spheres are almost in contact with each other. Sphere 1 is pulled aside and released. If sphere 1 strikes sphere 2 with velocity u. find an expression for velocity v_(n) of the nth sphere immediately after being struck by the one adjacent to it. The coefficient of restitution for all the impacts is e. |
Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/let-11597" style="font-weight:bold;" target="_blank" title="Click to know more about LET">LET</a> velocities of <a href="https://interviewquestions.tuteehub.com/tag/sphere-1222094" style="font-weight:bold;" target="_blank" title="Click to know more about SPHERE">SPHERE</a> `1` and `2` 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> be `v_(1)` and `v_(2)`, respectively, then <br/> `(v_(2)-v_(1))/u=e` and `mv_(2)+mv_(1)="mu"` <br/> From above equations `v_(2)=((1+e)u)/2` <br/> Now sphere `2` will collide with sphere `3` and affter collision velocity of `3` can be formed as <br/> `v_(3)=((1+e)/2)^(2)u` <br/> Hence, similarly velocity of `nth` sphere can be formed as <br/> `v_(<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>)=((1+e)/2)^(n-1)u`</body></html> | |