bullet fired into block on spring

[candyq27]

1. The problem statement, all variables and given/known data
A 1.01x10^-2 kg bullet is fired horizontally into a 2.49kg wooden block attached to one end of a massless, horizontal spring (k=831 N/m). The other end of the spring is fixed in place, and the spring is unstrained initially. The block rests on a horizontal, frictionless surface. The bullet strikes the block perpendicularly and quickly comes to a halt with in it. As a result of this completely inelastic collision, the spring is compressed along its axis and causes the block/bullet to oscillate with an amplitude of 0.20m. What is the speed of the bullet?


2. Relevant equations

1/2mvf^2 + 1/2Iwf^2 +mghf+ 1/2kxf^2 = 1/2mvo^2 + 1/2Iwo^2 +mgho+ 1/2kxo^2
Aw = v
Aw^2 = a

3. The attempt at a solution

I'm not sure what to do w/ this equation. Do I use energy laws to figure it out? If I do then I'm not sure how to change the equation to find vf. I originally tried w= sqrt(k/m+M) and got 18.2m/s but that's wrong

[Doc Al]

Think of this problem as having two parts:
(1) The collision of bullet with block. What's conserved?
(2) The compression of the spring. What's conserved?

[candyq27]

Well for the collision of the bullet with the block u have vom1=(m1+m2)v, but i don't know the intial speed so i'm not sure
and for the compression of the spring...? all I know is Hooke's Law is F=-kx, and U=1/2kx^2

[Doc Al]

Well for the collision of the bullet with the block u have vom1=(m1+m2)v, but i don't know the intial speed so i'm not sure
Good. Momentum is conserved during the collision.

and for the compression of the spring...? all I know is Hooke's Law is F=-kx, and U=1/2kx^2
Good. After the bullet collides with the block, and the spring is compressed, what's conserved?

[candyq27]

when they collide its an inelastic collision so the kinetic energy isn't conserved, so i'm not sure

[Doc Al]

when they collide its an inelastic collision so the kinetic energy isn't conserved, so i'm not sure
You are right: During the collision, mechanical energy is not conserved (but momentum is). But what about after the collision, when the block/bullet moves and compresses the spring?

[candyq27]

potential energy?

[candyq27]

well since its oscillating then the w is conserved and the energy is also conserved

[Doc Al]

After the collision, total mechanic energy is conserved. What are the two kinds of energy relevant here?

[candyq27]

1/2kx^2 + ?

[Doc Al]

1/2kx^2 + ?
Yes, spring potential energy is one kind. What's the other? (Hint: It's not conserved in an inelastic collision.)

[candyq27]

kinetic 1/2mv^2 ?

[Doc Al]

Of course!

[candyq27]

so then do i set everything equal? i'm still a little lost
is it (m1+m2)v = 1/2kx^2 + 1/2mv^2 ? but then how do i know the original v?

[Doc Al]

so then do i set everything equal? i'm still a little lost
is it (m1+m2)v = 1/2kx^2 + 1/2mv^2 ? but then how do i know the original v?

Momentum and energy are two completely different things--you can't set them equal.

Instead, get two equations--one describing conservation of momentum during the collision, the other describing conservation of energy after the collision--and solve them together to determine the initial speed.

What are those two equations?

[candyq27]

well the momentum is m1v=(m1+m2)v so it's m1v-(m1+m2)v=0
and the energy is 0=1/2kx^2 + 1/2mv^2
so setting them equal i get...
0.0101v-2.5001v= 166.2 + 1.25005v^2
but then i'm confused

[Doc Al]

well the momentum is m1v=(m1+m2)v so it's m1v-(m1+m2)v=0
Use different symbols for V (bullet) and Vf (block, after the collision).
and the energy is 0=1/2kx^2 + 1/2mv^2
The energy is certainly not zero! The block moves after the bullet hits it. Initially it has only KE; when the block fully compresses the spring, it has only spring potential energy.
so setting them equal i get...
Stop trying to set them equal. :smile:

Instead, write the equation for conservation of energy and use it to figure out the speed of the block after the bullet hits it. That's your first step.

[candyq27]

well if i just use energy and i use 1/2kx^2 + 1/2mv^2 i get v=11.5m/s
and then using momentum i do m1v=(m1+m2)v2 so it's 0.0101v= (2.5001)(11.5m/s) so v=2854.2, ok thats def wrong

[Doc Al]

well if i just use energy and i use 1/2kx^2 + 1/2mv^2 i get v=11.5m/s
How did you arrive at this answer?

[candyq27]

i plugged in numbers and solved for v, i'm confused because i don't know what the intial is

[Doc Al]

I know you must have plugged numbers into something! I want to know the exact equation you used, and what values you used for m, k, and x.

[candyq27]

I did 1/2kx^2 = 1/2mv^2
so 1/2(831N/m)(0.2m)^2 = 1/2(2.5001kg)v^2 so v=11.5m/s

[Doc Al]

I did 1/2kx^2 = 1/2mv^2
so 1/2(831N/m)(0.2m)^2 = 1/2(2.5001kg)v^2 so v=11.5m/s
Correct equation; Correct values. Good! But redo your calculation for v.

[candyq27]

my reply didn't post...but i recalculated and got v=3.65m/s so then when i plug that into my other equation i get 902.5 m/s, that still seems really large

[Doc Al]

Well, it is a speeding bullet! (That's not an unusual speed for a bullet.)

[candyq27]

ok good, thanks so much!