A 5.00-g bullet is shot through a 1.00-kg wood block suspended on a string 2.00 m long. The center of mass of the block rises a distance of 0.38 cm. Find the speed of the bullet as it emerges from the block if its initial speed is 450 m/s.

The initial speed of the bullet is$$" width="9" height="19" role="math">u=450m/s

The height raised by the block is $h=0.38cm=0.0038m$

The length of the string is $l=2m$

The mass of the wooden block is: $M=1kg$

The mass of the bullet is: $M=5g=0.005kg$

The velocity of the wooden block is calculated by using raised height and then applying momentum conservation for the emerging speed of a bullet.

The speed of the wooden block after the collision is given as:

$v_w=\sqrt{2gh}$

Here, g is the gravitational acceleration.

Substitute all the values in the above equation.

$$\begin{gathered} v_w=\sqrt{2\left(9.8m/s\right)\left(0.0038m\right)} \\ {v}_w=0.0745m/s \end{gathered}$$

The emerging speed of bullet after the collision is given as:

$mu=M{v}_w+mv$

Substitute all the values in the above equation.

$$\begin{gathered} \left(0.005kg\right)\left(450m/s\right)=\left(1kg\right)\left(0.0745m/s\right)+\left(0.005kg\right)v \\ v=435m/s \end{gathered}$$

Therefore, the emerging speed of bullet after collision is .