What is the initial velocity of the bullet when it is shot into the block?

[steelydan8821]

Homework Statement

An 8 g bullet is shot into a 4 kg block, at rest on a frictionless horizontal surface. The bullet remains lodged in the block. The block moves into a spring and compresses it by 8.7cm. the force constant of the spring is 2400N/m. The initial velocity of the bullet is closest to
A. 1190m/s B. 1020m/s C. 1110m/s D. 1150m/s E. 1070m/s

Homework Equations

U= 1/2Kx^2
K = 1/2mv^2

The Attempt at a Solution

I calculated the potential energy from the spring compressing using U=1/2(2400)(0.087^2) and set that equal to the kinetic energy of the block moving with the bullet lodged into it K=1/2(4.008)V^2. solved for v then set that kinetic energy equation equal to the kinetic energy of the bullet K=1/2(0.008)v^2 but the value of v for the bullet I get is much much smaller than any of the choices
1/2(2400)(0.087^2)=1/2(4.008)v^2=1/2(0.008)v^2

 

[Doc Al]

The Attempt at a Solution

I calculated the potential energy from the spring compressing using U=1/2(2400)(0.087^2) and set that equal to the kinetic energy of the block moving with the bullet lodged into it K=1/2(4.008)V^2.

Good!

solved for v then set that kinetic energy equation equal to the kinetic energy of the bullet K=1/2(0.008)v^2 but the value of v for the bullet I get is much much smaller than any of the choices
1/2(2400)(0.087^2)=1/2(4.008)v^2=1/2(0.008)v^2

There's your error. The KE of the bullet does not equal the KE of block + bullet. KE is not conserved during the collision. But something else is.

 

[steelydan8821]

The momentum is conserved. Thanks Dr Al