A bag P (mass M) hangs by a long thread and a bullet (mass m) comes horizontally with velocity v and gets caught in the bag. Then for the combined (beg + bullet) system the

The correct option is: Kinetic energy is m²V²/2(M+m)

Let's analyze the problem step-by-step using the principles of conservation of momentum.

Step 1: Understand the system before the collision
- The mass of the bullet is $m$ and it moves horizontally with velocity $V$.
- The mass of the bag is $M$ and it is initially hanging at rest, so its velocity is 0.

Step 2: Apply the conservation of linear momentum
Since there is no external horizontal force acting on the combined bullet-bag system during the collision, horizontal linear momentum is conserved.
Let $V^{\prime }$ be the common velocity of the combined system (bullet + bag) immediately after the collision.

The initial momentum of the system is:
$P_{\text{initial}}=mV+M\left(0\right)=mV$
The final momentum of the combined system is:
$P_{\text{final}}=(M+m)V^{\prime }$
Equating the initial and final momentum:
$mV=(M+m)V^{\prime }$
Solving for the post-collision velocity $V^{\prime }$:
$V^{\prime }=\frac{mV}{M+m}$

Step 3: Calculate the kinetic energy of the combined system
The kinetic energy ($K$) of the combined system of mass $(M+m)$ immediately after collision is:
$K=\frac{1}{2}(M+m){\left(V^{\prime }\right)}^2$
Substitute the value of $V^{\prime }$ from Step 2 into the kinetic energy equation:
$K=\frac{1}{2}(M+m){\left(\frac{mV}{M+m}\right)}^2$
Simplify the expression:
$K=\frac{1}{2}(M+m)\frac{m^2{V}^2}{{(M+m)}^2}$
$K=\frac{m^2{V}^2}{2(M+m)}$

Therefore, the kinetic energy of the combined system is $\frac{m^2{V}^2}{2(M+m)}$.