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A shell fired a cannon with speed v m/s at angle thetawith horizontal explodes into three pieces of equal masses at the highest point of trajectory. One piece falls down vertically while the other retraces its path. What is speed of the third piece? |
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Answer» SOLUTION :If .m. is mass of shell, at highest point, initial momentum =`m(vcostheta)`, along HORIZONTAL given, momentum of one fragment `=m/3 (-v cos theta)` momentum of SECOND along horizontal = 0 If `V_(1)` is speed of third fragment, from momentum CONSERVATION `=-m/3 v cos theta + m/3 v_(1) + 0 =m(v cos theta), v_(1) =4V cos theta` |
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