A bullet of mass m moving with velocity v strikes a block of mass M at rest and gets embeded into it. The kinetic energy of the composite block will be

The correct option is (mv²/2)x(m/m+M).

To find the kinetic energy of the composite block after the collision, we can follow these steps:

Step 1: Analyze the system before the collision
Let:
- Mass of the bullet = $m$
- Velocity of the bullet before collision = $v$
- Mass of the block at rest = $M$
- Velocity of the block before collision = 0
The initial linear momentum of the system ($P_i$) is equal to the momentum of the moving bullet:
$P_i=mv$

Step 2: Analyze the system after the collision
Since the bullet gets embedded in the block, this is a completely inelastic collision. After the collision, the bullet and the block stick together and move as a single composite body.
- Total mass of the composite block = $m+M$
Let the velocity of this composite block immediately after the collision be $V$.
The final linear momentum of the system ($P_f$) is:
$P_f=(m+M)V$

Step 3: Apply the conservation of linear momentum
Since there are no external forces acting on the system in the horizontal direction, linear momentum is conserved:
$P_i=P_f$
Substituting the momentum expressions:
$mv=(m+M)V$
Solving for the common velocity $V$ gives:
$V=\frac{mv}{m+M}$

Step 4: Determine the kinetic energy of the composite block
The kinetic energy of the composite block ($K_f$) after the collision is given by:
$K_f=\frac{1}{2}(m+M)V^2$
Substitute the expression for $V$ derived in Step 3:
$K_f=\frac{1}{2}(m+M){(\frac{mv}{m+M}}^2$
Simplify the term:
$K_f=\frac{1}{2}(m+M)\times \frac{m^2{v}^2}{{(m+M)}^2}$
Cancel $(m+M)$ from the numerator and denominator:
$K_f=\frac{m^2{v}^2}{2(m+M)}$
This expression can be rearranged to match the format of the options:
$K_f=\left(\frac{m{v}^2}{2}\right)\times \left(\frac{m}{m+M}\right)$