# Bullet fired into block on spring

• candyq27
In summary: You are right: The correct equation is 1/2kx^2 = 1/2mv^2. What values did you use for m, k, and x?m=1.01x10^-2 kg, k=831 N/m, x=2.49kg
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

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?

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

candyq27 said:
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?

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

candyq27 said:
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?

potential energy?

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

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

1/2kx^2 + ?

candyq27 said:
1/2kx^2 + ?
Yes, spring potential energy is one kind. What's the other? (Hint: It's not conserved in an inelastic collision.)

kinetic 1/2mv^2 ?

Of course!

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?

candyq27 said:
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?

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

candyq27 said:
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.

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

candyq27 said:
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?

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

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.

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

candyq27 said:
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.

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

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

ok good, thanks so much!

## 1. How does a bullet fired into a block on a spring affect the motion of the block?

When a bullet is fired into a block on a spring, the block will experience a sudden force and begin to move. This is due to the conservation of momentum, where the bullet's momentum is transferred to the block, causing it to move in the opposite direction.

## 2. Does the mass of the block or the bullet affect the motion?

Yes, the mass of both the block and the bullet will affect the motion. A heavier bullet or block will have a greater momentum and will cause the block to move more than a lighter one. However, the velocity of the bullet will also play a role in the transfer of momentum.

## 3. What factors besides mass and velocity can affect the motion of the block?

Other factors that can affect the motion of the block include the elasticity of the spring, the angle at which the bullet hits the block, and any external forces acting on the block such as friction or air resistance. These factors can alter the amount of energy transferred and the resulting motion of the block.

## 4. Does the speed of the bullet affect the motion of the block?

Yes, the speed of the bullet will greatly impact the motion of the block. A faster bullet will have a greater momentum and will cause the block to move more than a slower bullet. Additionally, a faster bullet will transfer more energy to the block, resulting in a greater displacement of the block.

## 5. What happens to the bullet after it hits the block on the spring?

After hitting the block on the spring, the bullet will either get embedded in the block or bounce off, depending on the elasticity of the materials and the angle of impact. If the bullet gets embedded, it will decrease the energy transferred to the block and result in a smaller displacement. If the bullet bounces off, it will retain its momentum and continue moving in the opposite direction of the block.

• Mechanics
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Classical Physics
Replies
11
Views
2K
• Introductory Physics Homework Help
Replies
24
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
30
Views
843
• Introductory Physics Homework Help
Replies
8
Views
4K
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
14
Views
3K
• Mechanics
Replies
2
Views
1K