A 14.3 g bullet is fired into and embeds itself
in a 2.76 kg block attached to a spring with a
spring constant of 18.4 N/m and whose mass
How far is the spring compressed if the
bullet has a speed of 247 m/s just before it
strikes the block, and the block slides on a
Answer in units of m.
.5(m)(v^2) + .5 (k)(x^2) = .5(k)(A^2)
velocity = (omega)(amplitude)
momentum = m1v1
The Attempt at a Solution
Here's my process but its not right because I got square root of a negative number:
I took the v=wA and solved for Amplitude. But to get the velocity of the wood, I did momentum of bullet = momentum of block. I plugged in my velocity(v2) into the formula and solved for amplitude. I then used the .5(m)(v^2) + .5 (k)(x^2) = .5(k)(A^2) and solved for x. To find the velocity, i assumed the velocity was velocity of block + bullet, in the equation, I set up another momentum equation: momentum of bullet + momentum of block = momentum of bullet + block and solved for velocity of block+ bullet. I then used this new velocity to find the distance, x. The only problem is that I got a square root of a negative number, which is obviously wrong. What am i doing wrong?