A sphere of mass $m$, moving with velocity $V$, enters a hanging bag of sand and stops. If the mass of the bag is $M$ and it is raised by height $h$, then the velocity of the sphere was
A sphere of mass $m$, moving with velocity $V$, enters a hanging bag of sand and stops. If the mass of the bag is $M$ and it is raised by height $h$, then the velocity of the sphere was
$\frac{{M + m}}{m}\sqrt {2gh} $
$\frac{M}{m}\sqrt {2gh} $
$\frac{m}{{M + m}}\sqrt {2gh} $
$\frac{m}{M}\sqrt {2gh} $
A sphere of mass $m$, moving with velocity $V$, enters a hanging bag of sand and stops. If the mass of the bag is $M$ and it is raised by height $h$, then the velocity of the sphere was
By the conservation of linear momentum
Initial momentum of sphere
= Final momentum of system
$mV = (m + M){v_{{\rm{sys}}{\rm{.}}}}$…(i)
If the system rises up to height h then by the conservation of energy
$\frac{1}{2}(m + M)v_{{\rm{sys}}{\rm{.}}}^{\rm{2}} = (m + M)gh$…(ii)
==> ${v_{{\rm{sys}}{\rm{.}}}} = \sqrt {2gh} $
Substituting this value in equation (i)
$V = \left( {\frac{{m + M}}{m}} \right)\;\sqrt {2gh} $
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