A 12.0-g bullet is fired horizontally into a 114-g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 146 N/m. The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of 75.5 cm, what was the speed of the bullet at impact with the block

The speed of the bullet-block is .

Explanation:

The expression of the conservation of energy to the given system is as follows;

Here, are the initial kinetic energy and final kinetic energy and .

Here, m is the mass of the bullet, M is the mass of the wooden block, k is the spring constant, x is the distance, v is the speed of the bullet and V is the speed of the bullet with the block.

Calculate the speed of the bullet-block.

Convert the distance from cm to m.

x=75.5 cm

x=0.755 m

Put m= 12 g, M=114 g, v=0, x= 0.755 m and in the above expression according to the given values in the problem.

Therefore, the speed of the bullet-block is .

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

0.55x745.7j and a calculator