A wooden block of mass M is suspended by a cord and is at rest. A bullet of mass m, moving with a velocity v pierces through the block and comes out with a velocity v / 2 in the same direction. If there is no loss in kinetic energy, then upto what height the block will rise

The correct option is m²v²/8M²g.

To find the height to which the wooden block rises, we can break the problem down into two parts:
1. Applying the law of conservation of linear momentum during the collision to find the velocity of the block immediately after the bullet exits.
2. Applying the law of conservation of mechanical energy to determine the height the block rises after the collision.

Step 1: Conservation of Linear Momentum
Let the mass of the bullet be m and its initial velocity be v.
Let the mass of the wooden block be M, which is initially at rest (initial velocity is 0).
After the bullet pierces the block, it emerges with a velocity of
$\frac{v}{2}$
in the same direction.
Let the velocity acquired by the block immediately after the collision be V.

According to the law of conservation of linear momentum:
Initial Momentum = Final Momentum

$mv+M\left(0\right)=m\left(\frac{v}{2}\right)+MV$

Simplifying the equation to solve for V:

$mv=\frac{mv}{2}+MV$

$MV=mv-\frac{mv}{2}$

$MV=\frac{mv}{2}$

$V=\frac{mv}{2M}$

Step 2: Conservation of Mechanical Energy for the Block
After the collision, the block swings upward. As it rises, its kinetic energy is converted into gravitational potential energy. Let h be the maximum height the block reaches.
At the maximum height, the velocity of the block becomes 0.

By conservation of energy for the block:
Kinetic Energy immediately after collision = Potential Energy at maximum height

$\frac{1}{2}M{V}^2=Mgh$

We can cancel M from both sides:

$gh=\frac{V^2}{2}$

$h=\frac{V^2}{2g}$

Now, substitute the expression for V from Step 1 into this equation:

$h=\frac{1}{2g}{\left(\frac{mv}{2M}\right)}^2$

$h=\frac{1}{2g}\left(\frac{m^2{v}^2}{4M^2}\right)$

$h=\frac{m^2{v}^2}{8M^{2}g}$