Bullet fired into block on spring

[candyq27]

Homework Statement

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?

Homework 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

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.

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 that's 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!