Question

: A sand bag of mass $M$ is suspended by a rope. A bullet of mass $m$ is fired towards the sand bag with speed $v$ and gets embedded in it. The loss of kinetic energy is:
(A) $0$
(B) $\dfrac{1}{2}m{v^2} \times \dfrac{1}{{M + m}}$
(C) $\dfrac{1}{2}m{v^2} \times \dfrac{M}{{M + m}}$
(D) $\dfrac{1}{2}m{v^2} \times \dfrac{m}{{M + m}}$

Answer 1

First of all, we will find the momenta before and after hitting. Then we will find the velocity of the embedded system. We will then calculate the loss of kinetic energy followed by manipulation accordingly to obtain the result.

Complete step by step answer:
In the given problem, the supplied data are as follows:
The mass of the sand bag is $M$ which is suspended by a rope.
The mass of the bullet is $m$ which is fired towards the sand bag with velocity $v$ .
We are asked to find the loss of kinetic energy.
This problem is based on the conservation of linear momentum. It states that whenever a body collides with another body, and gets attached to it, they begin to move again at some velocity. In such a case, the momentum before and after collision remains the same. However, the mass of the embedded system increases due to combination and the velocity of combination decreases as compared to the velocity of the first object.
Let $v'$ be the final speed of the embedded system
The momentum of the bullet is $mv$ .
The momentum of the embedded system i.e. the bullet and the sand bag is equal to $\,(m\, + \,M)\,v'$ .
Initial momentum is equal to the final momentum.

$$mv\, = \,(m\, + \,M)\,v' \\\ v' = \dfrac{{mv}}{{m + M}} \\\$$

Loss of K.E can be written as the difference between the initial kinetic energy and the final kinetic energy, which can be written mathematically as follows:

$$= \,{\text{Initial}}\,\,{\text{K}}{\text{.E}} - {\text{Final}}\,\,{\text{K}}{\text{.E}} \\\ = \dfrac{1}{2}\,m{v^2} - \dfrac{1}{2}\,(m + M)\,{{v'}^2} \\\ = \dfrac{1}{2}\left[ {m{v^2} - (m + M){{\left( {\dfrac{{mv}}{{m + M}}} \right)}^2}} \right] \\\ = \dfrac{1}{2}\left[ {m{v^{2\,}} - \dfrac{{{m^2}{v^2}}}{{{{(m + M)}^2}}} \times (m + M)} \right] \\\$$

Again, by simplifying further, we get:

$$\Delta K.E = \dfrac{1}{2}m{v^2}\left[ {1 - \dfrac{m}{{m + M}}} \right] \\\ \Delta K.E = \dfrac{1}{2}m{v^2}\left[ {\dfrac{{m + M - m}}{{m + M}}} \right] \\\ \Delta K.E = \dfrac{1}{2}m{v^2} \times \dfrac{M}{{m + M}} \\\$$

Hence, the loss in the kinetic energy is given by
$\dfrac{1}{2}m{v^2} \times \dfrac{M}{{m + M}}$ .
The correct option is (C).

Note: While solving this problem, you need to have some knowledge on the conservation of linear momentum. Always use this principle whenever you see that bodies are colliding or like that. It does not depend whether the colliding bodies are too big or too small, momentum will always be conserved.