# 3 problems involving linear momentum

1. Nov 2, 2005

### jrd007

1) Calculate the force exerted on a rocket, given that the propelling gases are expelled at a rate of 1500 kg/s with a speed of 4.0 x 10^4 m/s (@ the moment of take off) Answer: 6.0 x 10^7 N

Okay I can not figure how to get an equation to use... Though I did noticed that the rate x the speed in this problem does in fact give you 6.0 x 10^7. Does that play a role?
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2) A 12,600 kg railroad car travels alone on a level frictionless track with a constant speed of 18 m/s. A 5350 kg load, intially at rest, is dropped on to the car. What will be the car's new speed? Answer: 12.6 m/s

So... here was my approach...

(Before) KE1 + KE2 ( = 0) = (after) KE1 + KE2
(1/2)(12600 kg)((18 m/s)^2) + 0 = (12600 +5350)(v^2)
sq root of 114 = 10.6 m/s... I am off by 2 numbers... did I do something wrong?

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3) An atomic nucleus intially moving @ 420 m/s emits an alpha particle in the direction of it's velocity, and the remaining nucleus slows to 350 m/s. If the alpha particle has a mass of 4.0 u and the original nucleus has a mass of 222 u what speed does the alpha particle have when it is emitted. Answer: 4.2 x 10^3 m/s

I did the same approach as above with Kinetic energies... Cannot figure out what I am doing wrong. Please help me.

2. Nov 2, 2005

### verty

For the first one, it helps to remember that a newton is a kgm/s^2 (think m*a). In other words, you just multiply them together.

3. Nov 2, 2005

### verty

For part two, momentum before = momentum after

mv + mv = (m+m)v

You should be able to figure out number 3 now.

4. Nov 3, 2005

### jrd007

But vertigo, is 1500 kg/s considered a mass? I was confused because masses aren't usually per anything.

2) mv + mv = (m+m)v
(12600 kg)(18m/s) + (5350 kg)(0 m/s) = (17950 kg)(v)
v= 12.6

So for number 3 do I apply the same logic, with momentums?

5. Nov 3, 2005

### verty

1500 kg/s is a rate of change of mass, like power is W = J/s, a rate of change of energy. (kg/s)(m/s) = kgms^-2 = N

Do you understand that formula for number 2? Numbers 2 and 3 are very similar, read them again.

Last edited: Nov 3, 2005
6. Nov 3, 2005

### jrd007

Yes, I know the formula in # 2 with momentum.

So for number three I get this...

mv + mv = mv +mv
(222 u)(420 m/s) + (4.0 u)v = (222 u)(350 m/s)

But that doesn't work...

7. Nov 3, 2005

### jrd007

I also tried the problem like this:

(222 u)(420 m/s) + (4.0 u)v = (222 -4 u)(350 m/s)
And I get -4.2 x 10^3 m/s... it is the correct answer just negative... Must I consider the v to be negative, that way I get a positive answer?

8. Nov 3, 2005

### verty

What is the total momentum before? What is the total momentum after?

9. Nov 3, 2005

### jrd007

Isn't that what I just stated? I alpha particle has none after.

Before..................................After
(222 u)(420 m/s) + (4.0 u)v = (222 -4 u)(350 m/s)

10. Nov 3, 2005

### Fermat

The alpha particle does have a mometum afterwards.

You have a mass, 222, which splits into two separate masses, 218 and 4, with separate velocites.

The single mass is the before case. The two masses is the after case.

11. Nov 4, 2005

### jrd007

(222 u)(420 m/s) = (218 u)(350 m/s) + (4.0 u)(v)
I understand it now. Thank you!

v = 4.2 x 10^3 m/s!