# Homework Help: Ratio of momentum

1. Nov 23, 2008

### hansel13

1. The problem statement, all variables and given/known data
There's actually 2 problems that I'm totally stuck on:

A. Bullets from two revolvers are fired with the same velocity. The bullet from gun #1 is twice as heavy as the bullet from gun #2. Gun #1 weights three times as much as gun #2. The ratio of momentum imparted to gun #1 to that imparted to gun #2 is ?

B. Two bodies, A and B, have equal kinetic energies. The mass of A is nine times that of B. The ratio of the momentum of A to that of B is ?

2. Relevant equations
KE = 1/2*m*v2
P = m*v

3. The attempt at a solution
A. I figured if the gun weighed 3 times gun #2, and the bullet weighed twice as heavy, the answer would be 6:1?

B.
And for A: 1/2*9m*v2
So for B: 1/2*m*(3v)2

So for momentum:
A = 9m*v
B = m*3v

So the ratio is: 3:1?

I'm pretty sure Problem A is wrong, just not sure what to do. However, I think I did Problem B correctly, but I'm not positive.

2. Nov 24, 2008

### Kurdt

Staff Emeritus
What was your calculation for A?

3. Nov 24, 2008

### hansel13

So, was my answer to B correct?

Wasn't sure where to start..
M1g = Mass of Gun #1
M2g = Mass of Gun #2
M1b = Mass of Bullet #1
M2b = Mass of Bullet #2

M1g = 3*M2g
M1b = 2*M2b

So,
3*M2g = Mass of Gun #1
M2g = Mass of Gun #2
2*M2b = Mass of Bullet #1
M2b = Mass of Bullet #2

P = mv.

So
Gun #1: 3*M2g*2*M2b*V = 6mv
Gun #2: M2g*M2b*V = 2mv

So the ratio is 6:2 (3:1). Still doesn't seem right.

4. Nov 24, 2008

### Kurdt

Staff Emeritus
Part B looks fine. For part A your momentum equations are slightly off. The sum of the gun and bullets momentum for each system should be equal to the momentum beforehand.

5. Nov 24, 2008

### hansel13

Not sure if I follow, do you mean write out the momentum formulas first then substitute, Like the following?:
Gun #1: (M1g+M1b)*V = 2mv
Gun #2: (M2g+M2b)*V = 2mv

6. Nov 24, 2008

### Kurdt

Staff Emeritus
What is the momentum of the gun and bullet before it is fired?

7. Nov 24, 2008

### hansel13

Nothing because the velocity will be 0.

8. Nov 24, 2008

### Kurdt

Staff Emeritus
Ok, so the momentum of the bullet and the gun after being fired must cancel each other. Remember the bullet and gun don't necessarily travel at the same speed.

9. Nov 24, 2008

### hansel13

ah I see..

3*M2g + 2*M2b = 0
So,
3M2g = -2*M2b
M2g = -2/3*M2b

AND

M2g + M2b = 0
So,
M2g = -M2b

-2*M2b = Mass of Gun #1
-M2b = Mass of Gun #2
2*M2b = Mass of Bullet #1
M2b = Mass of Bullet #2

For Gun #1: P = -2*M2b * 2*M2b * V = -4mv
For Gun #2: P = -M2b * M2b * V = mv

So ratio is 4:1?

Last edited: Nov 24, 2008
10. Nov 24, 2008

### hansel13

Well 4:1 isn't one of the options haha, so I must have done something wrong.

For Gun #1: P = 2*M2b + 2*M2b * V = 4mv
For Gun #2: P = M2b + M2b * V = 2mv

Now I'm kinda guessing, but 2:1 ratio was one of the options and it would make more sense to add their masses...

11. Nov 24, 2008

### Kurdt

Staff Emeritus
Why are you only using one velocity? Both bullets have the same velocity, the question tells you that, but the two velocities for the guns are different. remember that momentum is a mass multiplied by a velocity.

12. Nov 24, 2008

### hansel13

Sorry, It's easier when I apply actual data into a scenario, I'm not good with understanding problems like these...

For Gun #1: P = -2*M2b*V1 + 2*M2b * V = 0
So, V1 = V

For Gun #2: P = -M2b*V2 + M2b * V = 0
So, V2 = V

For Gun #1: P = -2*M2b* V + 2*M2b * V
For Gun #2: P = -M2b* V + M2b * V

Now I'm officially lost...

13. Nov 25, 2008

### Kurdt

Staff Emeritus
Lets take a step back here. We can write the mass of the bullet and gun from #1 in terms of #2.

\begin{flalign*} 3m_{g2}v_1 + 2m_{b2}v &= 0 \\ m_{g2}v_2 + m_{b2}v &=0 \end{flalign*}

Can you take it from there? What is the ratio of the gun momentums.

14. Nov 25, 2008

### hansel13

3*Mg2*v1 + 2*Mb2*v = 0
M2g*v2 + M2b*v = 0

3*Mg2*v1 + 2 = 0
M2g*v2 + 1= 0

Mg2*v1 = 2/3
M2g*v2 = 1

P = 3*2/3 + 2 = 4
P = 1 + 1 = 2

So the ratio is 2:1.

15. Nov 25, 2008

### Kurdt

Staff Emeritus
Very good.

16. Nov 25, 2008

### hansel13

Thanks a lot. Definitely makes sense now.