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Hi
ans: 3vcosx
At the top of its path, vel is vcos x only in horizontal direction , so after explosion to retrace its path , the velocity of one of the equal masses should be -vcos x . therefore now if v conserve the momentum, then 2mvcosx = -mvcosx + mv' . calculate v' .
hi vaibhav,
dis question is quite straight forward
as velocity at d highest point is v cosx
just apply momentum conservation
m(v cosx) = m/2 (-ucosx ) + m/2 v1, [ as it is divided into parts of equal mass n one part retraces to d cannon itself so its velocity will be vcosx but wid a negative sign]
we hav to find v1
here simply u get v1= 3v cosx
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