# Conservation of energy

1. Oct 27, 2008

### steelydan8821

1. The problem statement, all variables and given/known data
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

2. Relevant equations
U= 1/2Kx^2
K = 1/2mv^2

3. 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

2. Oct 27, 2008

### Staff: Mentor

Good!
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.

3. Oct 27, 2008

### steelydan8821

The momentum is conserved. Thanks Dr Al