An 8.0-g bullet is shot into a 4.0-kg block, at rest on a frictionless horizontal surface (see thefigure). the bullet remains lodged in the block. the block moves into an ideal massless spring andcompresses it by 8.7 cm. the spring constant of the spring is 2400 nm. the initial velocity of the
bullet is closest to

The initial velocity of the bullet is 1066.63 m/s

Explanation:

The problem involves two different events: A collision between two objects and the stretching of a spring

Since we know the final condition of the system, we start from the end and back to the initial condition

The system of the bullet+block with mass= 4Kg+8gr = 4.008 Kg stretches the spring a distance of x=0.087 m

The work done when stretching a spring of constant k is

This work is equivalent to the change of potential energy of the spring, which is the result of the transformation of the kinetic energy of the total mass of the system

Solving for v' (the velocity of the system after the collision took place)

This is the resulting velocity after the bullet lodged into the block. That event complies with the law of conservation of linear momentum

Since the block is initially at rest ()

Solving for :

The initial velocity of the bullet is 1066.63 m/s

well, you would divide 135 by 33 to get your answer. then that is how many seconds it would take. boardman should fly about 132 m or 4.090909090909091 seconds.