# Homework Help: Finding Recoil Velocity of gun

1. May 11, 2012

### Molly1235

Hello, I am not sure if this thread is still active but I am doing the same kind of physics problem for homework, and I think I have come to a conclusion from all the helpful stuff on here, but just wondered if someone can check it over for me? This is not in my GCSE syllabus - my physics teacher just likes to set us O Level work, which is full of stuff no longer in our syllabus!!!

So the problem goes:
"A gun of mass 4.0kg fires a bullet of mass 10g at a speed of 60 m/s. What is the initital speed of recoil of the gun?"

First I thought I would have to convert both masses so they are in the same units, so I converted 4kg to 4000g.

"(Mass of gun)*(Speed of Gun) = (Mass of Bullet)*(Speed of Bullet)" Please correct me if I am wrong here.

So I substituted in the numbers.

4000*? = 10*60
4000*? = 600
600/4000 = 0.15

Speed of recoil = 0.15 m/s

I have no idea if this is correct, please please give me a pointer in the right direction?

Thanks,
Molly.

2. May 11, 2012

### Staff: Mentor

Your work looks perfect to me.

3. May 11, 2012

### I like Serena

Welcome to PF, Molly!

You're completely correct.

4. May 11, 2012

### Molly1235

Hiya. Yay, thank you both for your help - I'm really glad I managed to find some help! I have such a stupid Physics teacher who doesn't actually bother teaching us anything, so I have to learn everything via my textbook, and the internet if that fails! Also, can anyone tell me if there is a difference between 'initial speed of recoil' and 'initial speed'? I used the same equation for the questions that wanted initial speed rather than initial speed of recoil but not 100% sure this is correct. :)

5. May 11, 2012

### I like Serena

You're welcome.

The wording "initial speed" requires a context that specifies which speed is meant.
If you want a clear answer you need to give us more context for your question.

I can remark that typically you would have a process in which some object initially has a certain speed, the speed changes due to some influence, and you end up with a final speed.
The formulas you probably have, apply to such a process.

6. May 11, 2012

### Molly1235

Ok, well the question was "A man of mass 80kg jumps of a trolley of mass 320kg. If the initial speed of the man is 4m/s what is the initial speed of the trolley?" I used (Mass of trolley)*(Speed of trolley) = (Mass of man)*(Speed of man)

7. May 11, 2012

### I like Serena

Well, you're right!

It's called conservation of momentum.
Before the man jumps, your formula applies (both the man and the trolley have speed zero).
And after the man jumps, your formula still applies, from which you can deduce the speed of the trolley.

8. May 11, 2012

### Molly1235

Ok, thank you - yeah, the topic of the lesson was conservation of momentum, but my teacher was going on about how it doesn't actually exist, and Newton was wrong or something! Couldn't quite get my head round the whole theory, just trying to muddle my way through haha! All I could decipher from his ramblings was that everything at the start is speed 0.

My final question is "A railway truck, of mass 8000kg, is travelling at 18 m/s along a horizontal track when it bumps into, and joins onto two similar trucks which were at rest. What is the initial speed of the three trucks. How much energy is lost?"

I said the initial speed of the three trucks is 18 m/s as the other two trucks were at rest so their speed was 0, and that no energy is lost, as when they collide the momentum is surely conserved and just spread out through the 3 trucks?

9. May 11, 2012

### I like Serena

After the bump the speed of the 3 trucks combined will be less than the original speed of the first truck.
It can't be the same.

What do you think the total momentum is before the bump?
And what would the momentum be after the bump?

As for energy lost, you'll be needing another formula for that.
Do you have a formula for the kinetic energy of an object?
Something like $U_k = \frac 1 2 m v^2$?

10. May 11, 2012

### Molly1235

Erm, no haha! Never heard of that formula. Aww man, and I was doing so well. It says in my textbook that the total momentum before and after the collision will be the same...

11. May 11, 2012

### I like Serena

Yep, the total momentum before and after will be the same.
Momentum is mass times velocity.

The moving mass after is the mass of 3 trucks, instead of just 1 truck.
So what would happen to the speed to make sure the total momentum stays the same?

12. May 11, 2012

### Molly1235

Would it divide by 3? So the speed of each truck ends up 6m/s

13. May 11, 2012

### I like Serena

Yep. That's it!

As for energy being lost, perhaps your textbook says something about it?
Something like that when the objects bounce off each other, that (kinetic) energy is conserved?
But that when objects join, energy is lost as heat?

14. May 11, 2012

### Molly1235

Hmm, maybe that's more advanced level because my GCSE textbook says nothing at all about energy. Probably something off the O Level Spec that's not on the syllabus anymore!

Thanks so much for your help anyway! Much appreciated! :)

15. May 11, 2012

### I like Serena

Glad to be of help! :)