Question

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 will be

• A. $\frac{M +m}{m} \sqrt{2 g h}$
• B. $\frac{M}{m}\sqrt{2gh}$
• C. $\frac{m}{M+m}\sqrt{2gh}$
• D. $\frac{m}{M}\sqrt{2gh}$

Answer 1

Answer: (A)

By the conservation of linear momentum
Initial momentum of sphere = Final momentum of system
$mV=\left(m +M\right)v_{\text{sys}}$ (i)
If the system rises up to height h then by the conservation of energy
$\frac{1}{2}\left(m +M\right)v^{2}_{\text{sys}}$
$=\left(m +M\right)gh$ (ii)
$\Rightarrow v_{\text{sys}}=\sqrt{2gh}$
Substituting this value in equation (i)
$V=\left(\frac{m +M}{m}\right)\sqrt{2gh}$