# A Linear Momentum Problem

1. Jan 5, 2008

### rabar789

1. The problem statement, all variables and given/known data

A 10.0g bullet is fired horizontally into a 106 g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring of constant 155 N/m. If the bullet-block system compresses the spring by a maximum of 78.0 cm, what was the velocity of the bullet at impact with the block?

2. Relevant equations

p=mv
where p is momentum, m is mass (kg), and v is velocity (m/s)

(1/2 m1 v1^2 + 1/2 m2 v2^2)initial = (1/2 m1 v1^2 + 1/2 m2 v2^2)final + 1/2kx^2
Where m is mass (kg), v is velocity (m/s), k is the spring constant (N/m), and x is the distance of compression (meters).

(1/2 m1 v1^2 + 1/2 m2 v2^2)initial = (1/2m3 v3^2)final + 1/2kx^2
Where m is mass (kg), v is velocity (m/s), k is the spring constant (N/m), x is the distance of compression (meters), m3 is the sum of the masses, and v3 is the combined, new velocity. Not even sure if this one makes sense, I kind of combined some equations.

3. The attempt at a solution

To sum up my attempt, I plugged all of my variables into my "combined" equation. Thus I had two variables, v1 and v3^2. I decided to make everything equal to v1 and plug that equation into the original "combined" equation. (I can't really copy that onto here because it had a gigantic square root symbol.) So then I solved for v3 and got a ridiculously low answer, 1.0217m/s. According to my text book, which has the same problem with different variables, the answer is 237m/s.

Can someone help me out? Also, is this considered a perfectly inelastic collision?

2. Jan 5, 2008

### Staff: Mentor

Yes, so energy is not conserved during the collision. But what else is?

After the collision, then you can use energy conservation.

3. Jan 5, 2008

### rabar789

Momentum is, also. My main problem is that I can't set up a good equation (or two, if need be) or get rid of the double unknowns.

4. Jan 5, 2008

### Staff: Mentor

Think of the situation as having two phases:
(1) The collision of bullet and block
(2) The compressing of the spring after the collision

Set up equations for each phase. Each phase uses a different conservation law. Hint: You might want to analyze phase 2 first and work backwards.

5. Jan 6, 2008

### rabar789

OK I tried this.

(1/2*m1*v1^2 + 1/2*m2*v2^2)final = ½*k*x^2
And solved for v2
I got 97.11 m/s, understanding that the final velocity of the bullet is 0 because it’s lodged into the block.
Then I plugged it into (1/2*m1*v1^2 + 1/2*m2*v2^2)initial = (1/2*m1*v1^2 + 1/2*m2*v2^2)final + ½*k*x^2
And solved for v1 initial.
I got 137.317 m/s. This seems much more reasonable. Am I close?

6. Jan 7, 2008

### physixguru

bro..u r missing the equilibrium condition of the spring....i.e. when the spring is in its natural state...think about it and then proceede..:p

also..never go on answers..they may be wrong ..

have self belief and proceed...if ur concepts are right..u may even deny the text book's answer..

7. Jan 7, 2008

### Staff: Mentor

The left side of this equation is incorrect. After the collision, the bullet and block move together with a single speed. Their KE = 1/2 (m1+m2)V^2. Find that speed.
The final speed of the bullet (immediately after the collision) is not zero--it's the same speed as the block.