# Homework Help: I thought I understood

1. Oct 18, 2004

### doxigywlz

Basically, I understand this question but I keep coming up with the wrong answer!! Please show me where I am going wrong!

A 12.0-g bullet is fired horizontally into a 100-g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 150 N/m. The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of 80.0cm, what was the speed of the bullet at impact with the block?

Book answer: 273 m/s

My work:
Conservation of Mechanical Energy
(KE + PEg + PEs)i = (KE + PEg + PEs)f
1/2(12)(Vi^2) = 1/2(112)(Vf^2) + 1/2(150)(80^2)
soo..... 6(Vi^2) = 56(Vf^2) + 480,000
and from Conservation of Momentum:
12V1i =112V2f

0= 56Vf^2 - 6 Vi^2 + 480,000
plugging in for Vi.. I get 87 x 6 (I think I'm wrong here..)
so 46Vf2 = 480,000
Vf = 32.0 m/s

12Vi = 112(32)
Vi = 299 m/s

2. Oct 18, 2004

### Staff: Mentor

During the collision of bullet with block, energy is NOT conserved. But momentum is. After the collision, the combined "bullet + block" will have a final speed, and thus a kinetic energy. Energy is conserved during the compression of the spring.

3. Oct 18, 2004

### doxigywlz

are you telling me that the first part of the kinetic energy is equal to zero? sorry,i'm a little confused....

4. Oct 18, 2004

### Staff: Mentor

I don't know what you mean. Treat this problem as having two related parts:
(1) the collision: use conservation of momentum
(2) the spring compression: use conservation of energy

5. Oct 18, 2004

### doxigywlz

i hate to be a bother... but isn't that what i did?

6. Oct 18, 2004

### Staff: Mentor

No. You tried to apply conservation of energy during the collision--that won't work.
For the collision, only use conservation of momentum to relate Vi to Vf.

7. Oct 18, 2004

### doxigywlz

okay, i think i see what you mean. so... will it be m1v1 + m2v2 = (m1 + m2) vf ??

8. Oct 18, 2004

### Staff: Mentor

Right. (v2 = 0)

9. Oct 18, 2004

### doxigywlz

okay, so then i have
V1i = (112/12) Vf

and then
(12)(V1i)^2 = 12 V1f^2 + 100 V2f^2 + (150)(80)^2??

10. Oct 18, 2004

### Staff: Mentor

Right.
I have no idea what you are doing here. Instead, apply conservation of energy to the system after the collision. Hint: After the collision, the only speed involved is Vf.

11. Oct 18, 2004

### doxigywlz

i'm really confused.... i don't get it. i tried what you said but i'm still not getting the right answer.

12. Oct 18, 2004

### doxigywlz

ooookkkkaaayyy.... doesn't look like i'm getting gonna get anywhere with this problem 2nite.

13. Oct 18, 2004

### Pyrrhus

I thought you understood the problem??

Well ok I'll try to explain..

Imagine a Bullet..... It collides with a wooden block and stays inside the block, What type of collision this is Doxi??? Of course perfectly inelastic (Well an ideal case), so energy is NOT conserved (look up elastic collision), but Momentum is conserved.

So let's start with our first equation.

$$m_{bullet}v_{bullet} + m_{box}v_{box} = (m_{bullet} + m_{box})v_{both}$$

we know the box initially is at rest so.

$$m_{bullet}v_{bullet} = (m_{bullet} + m_{box})v_{both}$$

But we have 2 unknowns!!!!

But WAIT!!!!!!!! Mechanical Energy is conserved for our bullet+box system!!!!

Conservation of Mechanical Energy [Conservative System]
$$\Delta K + \Delta \Omega = 0$$

$$K + \Omega = K_{o} + \Omega_{o}$$

so, we analyse two points we can use, we can use the point when the box+bullet system get in contact with the spring (spring at equilibrium) and the point where the spring is at its max extend and there's no speed for our system.

$$\frac{1}{2}(m_{bullet} + m_{box})v_{both}^2 = \frac{1}{2}kx_{max}^2$$

Note: i was trying to be funny, because sometimes thinking "funny" works

Last edited: Oct 18, 2004
14. Oct 19, 2004

### doxigywlz

thanks, by the way... that helped a lot