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 is raised by height h then the velocity of the sphere was

Answer 1

$\frac{m+M}{m}\sqrt{2gh}$

Answer 2

The problem involves two distinct physical processes: an inelastic collision followed by upward motion under gravity.

1. Conservation of Linear Momentum (Collision Phase): When the sphere of mass $m$ moving with velocity $v$ collides with the hanging bag of mass $M$ (initially at rest), momentum is conserved. Let $V$ be the common velocity of the sphere and the bag immediately after the collision. Initial momentum = Momentum of sphere + Momentum of bag $p_i = mv + M(0) = mv$ Final momentum = Momentum of combined mass $(m+M)$ $p_f = (m+M)V$ By conservation of linear momentum: $mv = (m+M)V$ From this, we get the velocity of the combined system after the collision: $V = \frac{mv}{m+M}$

2. Conservation of Mechanical Energy (Upward Motion Phase): After the collision, the combined mass $(m+M)$ moves upwards with velocity $V$ and reaches a maximum height $h$. During this upward motion, the kinetic energy of the system is converted into gravitational potential energy. Initial kinetic energy (at the start of upward motion) = $\frac{1}{2}(m+M)V^2$ Final potential energy (at height $h$) = $(m+M)gh$ By conservation of mechanical energy (assuming no energy loss due to air resistance after the collision): $\frac{1}{2}(m+M)V^2 = (m+M)gh$ Simplifying this equation, we get: $\frac{1}{2}V^2 = gh$ $V^2 = 2gh$ $V = \sqrt{2gh}$

3. Equating and Solving for $v$: Now, we equate the two expressions for $V$ obtained from momentum and energy conservation: $\frac{mv}{m+M} = \sqrt{2gh}$ To find the initial velocity of the sphere, $v$, we rearrange the equation: $v = \frac{m+M}{m}\sqrt{2gh}$